Let
\[A=\begin{bmatrix}
1-a & a\\
-a& 1+a
\end{bmatrix}\] be a $2\times 2$ matrix, where $a$ is a complex number.
Determine the values of $a$ such that the matrix $A$ is diagonalizable.

(Nagoya University, Linear Algebra Exam Problem)

Proof.

To find eigenvalues of the matrix $A$, we determine the characteristic polynomial $p(t)$ of $A$ as follows.
We have
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
1-a-t & a\\
-a& 1+a-t
\end{vmatrix}\\[6pt] &=(1-a-t)(1+a-t)+a^2\\
&=(1-t)^2-a^2+a^2=(1-t)^2.
\end{align*}

Thus, the eigenvalue of $A$ is $1$ with algebraic multiplicity $2$.

Let us determine the geometric multiplicity (the dimension of the eigenspace $E_1$).
We have
\begin{align*}
A-I=\begin{bmatrix}
-a & a\\
-a& a
\end{bmatrix}
\xrightarrow{R_2-R_1}
\begin{bmatrix}
-a & a\\
0& 0
\end{bmatrix} .
\end{align*}
If $a\neq 0$, then we further reduce it and get
\begin{align*}
\begin{bmatrix}
-a & a\\
0& 0
\end{bmatrix}\xrightarrow{\frac{-1}{a}R_1}
\begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}.
\end{align*}

Hence the eigenspace corresponding to the eigenvalue $1$ is
\begin{align*}
E_1=\calN(A-I)=\Span \left\{\, \begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right\},
\end{align*}
and its dimension is $1$, which is less than the algebraic multiplicity.
Thus, when $a\neq 0$, the matrix $A$ is not diagonalizable.

If $a=0$, then the matrix $A=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}$ is already diagonal, hence it is diagonalizable.

In conclusion, the matrix $A$ is diagonalizable if and only if $a=0$.

The post Determine the Values of $a$ such that the 2 by 2 Matrix is Diagonalizable first appeared on Problems in Mathematics.

Is every diagonalizable matrix invertible?

Solution.

The answer is No.

Counterexample

We give a counterexample. Consider the $2\times 2$ zero matrix.
The zero matrix is a diagonal matrix, and thus it is diagonalizable.
However, the zero matrix is not invertible as its determinant is zero.

More Theoretical Explanation

Let us give a more theoretical explanation.
If an $n\times n$ matrix $A$ is diagonalizable, then there exists an invertible matrix $P$ such that
\[P^{-1}AP=\begin{bmatrix}
\lambda_1 & 0 & \cdots & 0 \\
0 & \lambda_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n
\end{bmatrix},\] where $\lambda_1, \dots, \lambda_n$ are eigenvalues of $A$.
Then we consider the determinants of the matrices of both sides.
The determinant of the left hand side is
\begin{align*}
\det(P^{-1}AP)=\det(P)^{-1}\det(A)\det(P)=\det(A).
\end{align*}
On the other hand, the determinant of the right hand side is the product
\[\lambda_1\lambda_2\cdots \lambda_n\] since the right matrix is diagonal.
Hence we obtain
\[\det(A)=\lambda_1\lambda_2\cdots \lambda_n.\] (Note that it is always true that the determinant of a matrix is the product of its eigenvalues regardless diagonalizability.
See the post “Determinant/trace and eigenvalues of a matrix“.)

Hence if one of the eigenvalues of $A$ is zero, then the determinant of $A$ is zero, and hence $A$ is not invertible.

The true statement is:

Is Every Invertible Matrix Diagonalizable?

Note that it is not true that every invertible matrix is diagonalizable.

For example, consider the matrix
\[A=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}.\] The determinant of $A$ is $1$, hence $A$ is invertible.
The characteristic polynomial of $A$ is
\begin{align*}
p(t)=\det(A-tI)=\begin{vmatrix}
1-t & 1\\
0& 1-t
\end{vmatrix}=(1-t)^2.
\end{align*}
Thus, the eigenvalue of $A$ is $1$ with algebraic multiplicity $2$.
We have
\[A-I=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}\] and thus eigenvectors corresponding to the eigenvalue $1$ are
\[a\begin{bmatrix}
1 \\
0
\end{bmatrix}\] for any nonzero scalar $a$.
Thus, the geometric multiplicity of the eigenvalue $1$ is $1$.
Since the geometric multiplicity is strictly less than the algebraic multiplicity, the matrix $A$ is defective and not diagonalizable.

Is There a Matrix that is Not Diagonalizable and Not Invertible?

Finally, note that there is a matrix which is not diagonalizable and not invertible.
For example, the matrix $\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}$ is such a matrix.

Summary

There are all possibilities.

1. Diagonalizable, but not invertible.
   Example: \[\begin{bmatrix}
   0 & 0\\
   0& 0
   \end{bmatrix}.\]
2. Invertible, but not diagonalizable.
   Example: \[\begin{bmatrix}
   1 & 1\\
   0& 1
   \end{bmatrix}\]
3. Not diagonalizable and Not invertible.
   Example: \[\begin{bmatrix}
   0 & 1\\
   0& 0
   \end{bmatrix}.\]
4. Diagonalizable and invertible
   Example: \[\begin{bmatrix}
   1 & 0\\
   0& 1
   \end{bmatrix}.\]

The post True or False. Every Diagonalizable Matrix is Invertible first appeared on Problems in Mathematics.

Find all the eigenvalues and eigenvectors of the matrix
\[A=\begin{bmatrix}
10001 & 3 & 5 & 7 &9 & 11 \\
1 & 10003 & 5 & 7 & 9 & 11 \\
1 & 3 & 10005 & 7 & 9 & 11 \\
1 & 3 & 5 & 10007 & 9 & 11 \\
1 &3 & 5 & 7 & 10009 & 11 \\
1 &3 & 5 & 7 & 9 & 10011
\end{bmatrix}.\]

(MIT, Linear Algebra Homework Problem)
 

Solution.

Let
\[B=A-10000I,\] where $I$ is the $6 \times 6$ identity matrix. That is, we have
\[B=\begin{bmatrix}
1 & 3 & 5 & 7 &9 & 11 \\
1 & 3 & 5 & 7 &9 & 11 \\
1 & 3 & 5 & 7 &9 & 11 \\
1 & 3 & 5 & 7 &9 & 11 \\
1 & 3 & 5 & 7 &9 & 11 \\
1 & 3 & 5 & 7 &9 & 11 \\
\end{bmatrix}.\]

Since all the rows are the same, the matrix $B$ is singular and hence $\lambda=0$ is an eigenvalue of $B$.
Let us determine the geometric multiplicity of $\lambda=0$ (namely, the dimension of the null space of $B$).

We apply elementary row operations to $B$ and obtain
\begin{align*}
B\xrightarrow{\text{elementary row operations}}
\begin{bmatrix}
1 & 3 & 5 & 7 &9 & 11 \\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
\end{bmatrix}.
\end{align*}
Thus, if $B\mathbf{x}=\mathbf{0}$, then we have
\[x_1=-3x_2-5x_3-7x_4-9x_5-11x_6.\]

It follows from this that basis vectors of the eigenspace $E_0=\calN(B)$ are
\[\begin{bmatrix}
-3 \\
1 \\
0 \\
0 \\
0\\
0
\end{bmatrix}, \begin{bmatrix}
-5 \\
0 \\
1 \\
0 \\
0\\
0
\end{bmatrix}, \begin{bmatrix}
-7 \\
0 \\
0 \\
1 \\
0\\
0
\end{bmatrix}, \begin{bmatrix}
-9 \\
0 \\
0 \\
0 \\
1\\
0
\end{bmatrix}, \begin{bmatrix}
-11 \\
0 \\
0 \\
0 \\
0\\
1
\end{bmatrix},\] and hence the geometric multiplicity corresponding to $\lambda=0$ is $5$.

By inspection, we see that
\[B\mathbf{v}=36\mathbf{v},\] where
\[\mathbf{v}=\begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1\\
1
\end{bmatrix}.\] Thus, it yields that $\lambda=36$ is an eigenvalue of $B$ and $\mathbf{v}$ is a corresponding eigenvector.

Recall that the algebraic multiplicity of an eigenvalue is greater than or equal to the geometric multiplicity.

Also the sum of algebraic multiplicities of all eigenvalues of $B$ is equal to $6$ since $B$ is a $6\times 6$ matrix.

It follows from this observation that we determine that the algebraic multiplicity of $\lambda=0$ is $5$ and the algebraic and geometric multiplicities of $\lambda=36$ are both $1$.
Hence the vector $\mathbf{v}$ form a basis of the eigenspace $E_{36}$.

Now that we have determined eigenvalues and eigenvectors of $B$, we can deduce those of $A$ as follows.

In general, if $A=B+cI$, then the eigenvalues of $A$ are $\lambda+c$, where $\lambda$ are eigenvalues of $B$.
The eigenvectors for $A$ corresponding to $\lambda+c$ are exactly the eigenvectors for $B$ corresponding $\lambda$.
(See the post “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$” for a proof.)

In the current problem, we have $A=B+10000I$, and thus $c=10000$.
Therefore, the eigenvalues of $A$ are $10000, 10036$.
Eigenvectors corresponding to $10000$ are
\[x_2\begin{bmatrix}
-3 \\
1 \\
0 \\
0 \\
0\\
0
\end{bmatrix}+x_3\begin{bmatrix}
-5 \\
0 \\
1 \\
0 \\
0\\
0
\end{bmatrix}+x_4\begin{bmatrix}
-7 \\
0 \\
0 \\
1 \\
0\\
0
\end{bmatrix}+x_5\begin{bmatrix}
-9 \\
0 \\
0 \\
0 \\
1\\
0
\end{bmatrix}+x_6\begin{bmatrix}
-11 \\
0 \\
0 \\
0 \\
0\\
1
\end{bmatrix},\] where $(x_2, x_3, x_4, x_5, x_6)\neq (0, 0, 0, 0, 0, 0)$.

The eigenvector corresponding to $10036$ is
\[a\begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1\\
1
\end{bmatrix},\] where $a$ is any nonzero scalar.

The post Find All the Eigenvalues and Eigenvectors of the 6 by 6 Matrix first appeared on Problems in Mathematics.

Find all eigenvalues of the matrix
\[A=\begin{bmatrix}
0 & i & i & i \\
i &0 & i & i \\
i & i & 0 & i \\
i & i & i & 0
\end{bmatrix},\] where $i=\sqrt{-1}$. For each eigenvalue of $A$, determine its algebraic multiplicity and geometric multiplicity.

Proof.

We use an indirect method to find the eigenvalues of $A$ as follows.
Let $B=A+iI$. Then note that if $\lambda$ is an eigenvalue of $B$, then $\lambda-i$ is an eigenvalue of $A$.

Furthermore, the algebraic (respectively, geometric) multiplicity of $\lambda$ is the same as the algebraic (respectively, geometric) multiplicity of $\lambda-i$.

We reduced the matrix $B$ as follows:
\begin{align*}
B&=\begin{bmatrix}
i & i & i & i \\
i &i& i & i \\
i & i & i & i \\
i & i & i & i
\end{bmatrix}
\xrightarrow{\substack{R_2-R_1\\ R_3-R_1\\ R_4-R_1}}
\begin{bmatrix}
i & i & i & i \\
0&0&0&0\\
0&0&0&0\\
0&0&0&0
\end{bmatrix}
\xrightarrow{\frac{1}{i}R_1}
\begin{bmatrix}
1&1&1&1 \\
0&0&0&0\\
0&0&0&0\\
0&0&0&0
\end{bmatrix}.
\end{align*}

It follows that the rank of $B$ is $1$, and thus the nullity of $B$ is $3$ by the rank-nullity theorem.

Since the nullity is greater than $0$, it yields that $\lambda=0$ is an eigenvalue of $A$.

Recall that the eigenspace $E_0$ of $\lambda=0$ is the null space of $B$.
Hence the geometric multiplicity of $\lambda=0$ is the nullity of $B$, thus it is $3$.
It follows that the algebraic multiplicity of $\lambda=0$ is either $3$ or $4$.

Recall that the sum of all eigenvalues is equal to the trace of $B$, which is $\tr(B)=4i$.
Thus, we see that $\lambda=0$ is not the only eigenvalue.

There can be only one more eigenvalue of $B$ since the sum of algebraic multiplicities of all eigenvalues of $B$ is $4$.
It follows that $4i$ is an eigenvalue of $B$ with algebraic multiplicity $1$, and hence the geometric multiplicity is also $1$.
It yields tha the algebraic multiplicity of $\lambda=0$ must be $3$.

Another way to see that $4i$ is an eigenvalue of $B$ is to notice that the vector
\[\begin{bmatrix}
1\\1\\1\\1
\end{bmatrix}\] is an eigenvector. In fact, we have
\begin{align*}
B\begin{bmatrix}
1\\1\\1\\1
\end{bmatrix}=4i\begin{bmatrix}
1\\1\\1\\1
\end{bmatrix}.
\end{align*}
(Since all the entries are the same, it might not so difficult to notice this.)

In summary so far, we have obtained that eigenvalues of $B$ are $0$ and $4i$ with the algebraic/geometric multiplicities are $3$ and $1$, respectively.

Since $A=B-iI$, it follows that the eigenvalues of $A$ are $0-i=-i$ and $4i-i=3i$ with the algebraic/geometric multiplicities are $3$ and $1$, respectively.

Comment.

This is the second problem of Quiz 13 (Take Home Quiz) for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.
The first problem of Quiz 13 is “Quiz 13 (Part 1) Diagonalize a matrix.“.

List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

• Quiz 1. Gauss-Jordan elimination / homogeneous system.
• Quiz 2. The vector form for the general solution / Transpose matrices.
• Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent
• Quiz 4. Inverse matrix/ Nonsingular matrix satisfying a relation
• Quiz 5. Example and non-example of subspaces in 3-dimensional space
• Quiz 6. Determine vectors in null space, range / Find a basis of null space
• Quiz 7. Find a basis of the range, rank, and nullity of a matrix
• Quiz 8. Determine subsets are subspaces: functions taking integer values / set of skew-symmetric matrices
• Quiz 9. Find a basis of the subspace spanned by four matrices
• Quiz 10. Find orthogonal basis / Find value of linear transformation
• Quiz 11. Find eigenvalues and eigenvectors/ Properties of determinants
• Quiz 12. Find eigenvalues and their algebraic and geometric multiplicities
• Quiz 13 (Part 1). Diagonalize a matrix.
• Quiz 13 (Part 2). Find eigenvalues and eigenvectors of a special matrix

The post Quiz 13 (Part 2) Find Eigenvalues and Eigenvectors of a Special Matrix first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
2 & -1 & -1 \\
-1 &2 &-1 \\
-1 & -1 & 2
\end{bmatrix}.\] Determine whether the matrix $A$ is diagonalizable. If it is diagonalizable, then diagonalize $A$.
That is, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

 

We give two solutions.
The first solution is a standard method of diagonalization.
For a review of the process of diagonalization, see the post “How to diagonalize a matrix. Step by step explanation.”

The second solution is a more indirect method to find eigenvalues and eigenvectors.

Solution 1.

We claim that the matrix $A$ is diagonalizable.
One way to see this is to note that $A$ is a real symmetric matrix, and hence it is diagonalizable.

Alternatively, we can compute eigenspaces and check whether $A$ is not defective (namely, the algebraic multiplicity and the geometric multiplicity of each eigenvalue of $A$ are the same.

To diagonalize the matrix $A$, we need to find eigenvalues and three linearly independent eigenvectors.

We compute the characteristic polynomial of $A$ as follows:
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{bmatrix}
2-t & -1 & -1 \\
-1 &2-t &-1 \\
-1 & -1 & 2-t
\end{bmatrix}\\
&=(2-t)\begin{bmatrix}
2-t & -1\\
-1& 2-t
\end{bmatrix}
-(-1)\begin{bmatrix}
-1 & -1\\
-1& 2-t
\end{bmatrix}+(-1)\begin{bmatrix}
-1 & 2-t\\
-1& -1
\end{bmatrix} \\
&\text{(by the first row cofactor expansion)}\\
&=-t(t-3)^2.
\end{align*}
Since eigenvalues are the roots of the characteristic polynomial, eigenvalues of $A$ are $0$ and $3$ with algebraic multiplicity $1$ and $2$, respectively.

(If you did not confirm that $A$ is diagonalizable yet, then at this point we know that the geometric multiplicity of $\lambda=0$ is $1$ since the geometric multiplicity is alway greater than $0$ and less than or equal to the algebraic multiplicity. However the geometric multiplicity of $\lambda=3$ is either $1$ or $2$.)

Next, we determine the eigenspace $E_{\lambda}$ and its basis for each eigenvalue of $A$.

For $\lambda=3$, we find solutions of $(A-3I)\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A-3I&=\begin{bmatrix}
-1 & -1 & -1 \\
-1 &-1 &-1 \\
-1 & -1 & -1
\end{bmatrix}
\xrightarrow{\substack{R_2-R_1\\R_3-R_1}}
\begin{bmatrix}
-1 & -1 & -1 \\
0 &0 &0\\
0&0&0
\end{bmatrix}
\xrightarrow{-R_1}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &0\\
0&0&0
\end{bmatrix}.
\end{align*}
Hence a solution must satisfy $x_1=-x_2-x_3$, and thus the eigenspace is
\[ E_3=\left\{\, \mathbf{x} \in \R^3 \quad \middle| \quad \mathbf{x}=x_2\begin{bmatrix}
-1\\
1\\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1\\
0\\
1
\end{bmatrix} \text{ for any } x_2, x_3 \in \C \,\right\}.\] From this expression, it is straightforward to check that the set
\[\left\{\begin{bmatrix}
-1\\
1\\
0
\end{bmatrix}, \begin{bmatrix}
-1\\
0\\
1
\end{bmatrix} \,\right\}\] is a basis of $E_3$.
(Hence the geometric multiplicity of $\lambda=3$ is $2$, and $A$ is not defective and diagonalizable.)

For $\lambda=0$, we solve $(A-0I)\mathbf{x}=\mathbf{0}$, thus $A\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A&=\begin{bmatrix}
2 & -1 & -1 \\
-1 &2 &-1 \\
-1 & -1 & 2
\end{bmatrix}
\xrightarrow{R_1 \leftrightarrow R_2}
\begin{bmatrix}
-1 &2 &-1 \\
2 & -1 & -1 \\
-1 & -1 & 2
\end{bmatrix}
\xrightarrow{-R_1}
\begin{bmatrix}
1 & -2 & 1 \\
2 & -1 & -1 \\
-1 & -1 & 2
\end{bmatrix}\\
&\xrightarrow{\substack{R_2-2R_1\\ R_3+R_1}}
\begin{bmatrix}
1 & -2 & 1 \\
0 & 3 & -3 \\
0 & -3 & 3
\end{bmatrix}
\xrightarrow{R_3+R_2}
\begin{bmatrix}
1 & -2 & 1 \\
0 & 3 & -3 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{\frac{1}{3}R_2}
\begin{bmatrix}
1 & -2 & 1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix}\\
&\xrightarrow{R_1+2R_2}
\begin{bmatrix}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}
Hence any solution satisfies
\[x_1=x_3 \text{ and } x_2=x_3.\] Therefore, the eigenspace is
\[E_0=\left\{\, \mathbf{x} \in \R^3 \quad \middle| \quad x_3\begin{bmatrix}
1\\
1\\
1
\end{bmatrix} \text{ for any } x_3 \in \C \,\right\},\] and a basis of $E_0$ is
\[\left\{\, \begin{bmatrix}
1\\
1\\
1
\end{bmatrix} \,\right\}.\]

Let
\[\mathbf{u}_1=\begin{bmatrix}
-1\\
1\\
0
\end{bmatrix}, \mathbf{u}_2=\begin{bmatrix}
-1\\
0\\
1
\end{bmatrix}, \mathbf{u}_3=\begin{bmatrix}
1\\
1\\
1
\end{bmatrix}.\] Note that $\{\mathbf{u}_1, \mathbf{u}_2\}$ is a basis of $E_3$ and $\{\mathbf{u}_3\}$ is a basis of $E_0$.
Thus, it follows that the vectors $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ are linearly independent eigenvectors.
Put
\[S=[\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3]= \begin{bmatrix}
-1 & -1 & 1\\
1& 0& 1\\
0& 1 & 1
\end{bmatrix}.\] Since the columns of $S$ are linearly independent, the matrix $S$ is nonsingular. Then the procedure of the diagonalization yields that
\[S^{-1}AS=\begin{bmatrix}
\mathbf{3} & 0 & 0\\
0& \mathbf{3}& 0\\
0 & 0& \mathbf{0}
\end{bmatrix},\] where diagonal entries are eigenvalues corresponding to the eigenvector $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ with this order.

Solution 2.

The second solution uses a different method to find eigenvalues and eigenvectors.
Let $B=A-3I$. Then every entry of $B$ is $-1$.
We find eigenvalues $\lambda$ and eigenvectors of $B$. Then the eigenvalues of $A$ are $\lambda+3$ and eigenvectors are the same.

First we reduce the matrix $B$ as follows:
\begin{align*}
B&=\begin{bmatrix}
-1&-1&-1\\
-1&-1&-1\\
-1&-1&-1
\end{bmatrix}
\xrightarrow{\substack{R_2-R_1\\ R_3-R_1}}
\begin{bmatrix}
-1&-1&-1\\
0&0&0\\
0&0&0
\end{bmatrix}
\xrightarrow{-R_1}
\begin{bmatrix}
1&1&1\\
0&0&0\\
0&0&0
\end{bmatrix}.
\end{align*}
Hence the rank of $B$ is $1$, and the nullity of $B$ is $2$ by the rank-nullity theorem. It follows that $\lambda=0$ is an eigenvalue of $B$.
Note that the eigenspace $E_0$ corresponding to $\lambda=0$ is the null space of $A$. From the reduction above, we see that the null space consists of vectors
\[\mathbf{x}=x_2\begin{bmatrix}
-1\\1\\0
\end{bmatrix}
+x_3\begin{bmatrix}
-1\\0\\1
\end{bmatrix}\] for any complex numbers $x_2, x_3$.
It follows that
\[\begin{bmatrix}
-1\\1\\0
\end{bmatrix},
\begin{bmatrix}
-1\\0\\1
\end{bmatrix}\] are basis vectors of eigenspace $E_0$. Hence the geometric multiplicity of $\lambda=0$ is $2$.
(The algebraic multiplicity is either $2$ or $3$. We will see it must be $2$.)

Since all the entries of $B$ are $-1$, by inspection, we find that the vector
\[\begin{bmatrix}
1\\1\\1
\end{bmatrix}\] is an eigenvector corresponding to the eigenvalue $-3$.
In fact, we have
\begin{align*}
B\begin{bmatrix}
1\\1\\1
\end{bmatrix}
&=\begin{bmatrix}
-1&-1&-1\\
-1&-1&-1\\
-1&-1&-1
\end{bmatrix}
\begin{bmatrix}
1\\1\\1
\end{bmatrix}
=\begin{bmatrix}
-3\\-3\\-3
\end{bmatrix}
=-3\begin{bmatrix}
1\\1\\1
\end{bmatrix}.
\end{align*}

Since the algebraic multiplicity of $\lambda=0$ is either $2$ or $3$, and the sum of all the algebraic multiplicities is equal to $3$, the algebraic multiplicity of $\lambda=-3$ must be $1$ and that of $\lambda=0$ is $2$.
Hence the geometric multiplicity of $\lambda=-3$ is $1$. Thus
\[\begin{bmatrix}
1\\1\\1
\end{bmatrix}\] is a basis vector of $E_{-3}$.

In a nutshell, we have obtained that eigenvalues of $B$ are $0$ and $-3$ and basis vectors of $E_0$ and $E_{-3}$ are
\[\begin{bmatrix}
-1\\1\\0
\end{bmatrix},
\begin{bmatrix}
-1\\0\\1
\end{bmatrix} \text{ and } \begin{bmatrix}
1\\1\\1
\end{bmatrix},\] respectively.

Since $A=B+3I$, the eigenvalues of $A$ are $0+3=3$ and $-3+3=0$ and the corresponding eigenvectors are the same. Thus
\[\begin{bmatrix}
-1\\1\\0
\end{bmatrix},
\begin{bmatrix}
-1\\0\\1
\end{bmatrix} \text{ and } \begin{bmatrix}
1\\1\\1
\end{bmatrix}\] are the basis vectors of the eigenspace $E_3$ and $E_0$ of $A$, respectively.

As in solution 1, we put
\[S=\begin{bmatrix}
-1 & -1 & 1\\
1& 0& 1\\
0& 1 & 1
\end{bmatrix}.\] Then we have
\[S^{-1}AS=\begin{bmatrix}
\mathbf{3} & 0 & 0\\
0& \mathbf{3}& 0\\
0 & 0& \mathbf{0}
\end{bmatrix},\] where diagonal entries are eigenvalues corresponding to the eigenvector $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ with this order.

Comment.

This is the first problem of Quiz 13 (Take Home Quiz) for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

• Quiz 1. Gauss-Jordan elimination / homogeneous system.
• Quiz 2. The vector form for the general solution / Transpose matrices.
• Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent
• Quiz 4. Inverse matrix/ Nonsingular matrix satisfying a relation
• Quiz 5. Example and non-example of subspaces in 3-dimensional space
• Quiz 6. Determine vectors in null space, range / Find a basis of null space
• Quiz 7. Find a basis of the range, rank, and nullity of a matrix
• Quiz 8. Determine subsets are subspaces: functions taking integer values / set of skew-symmetric matrices
• Quiz 9. Find a basis of the subspace spanned by four matrices
• Quiz 10. Find orthogonal basis / Find value of linear transformation
• Quiz 11. Find eigenvalues and eigenvectors/ Properties of determinants
• Quiz 12. Find eigenvalues and their algebraic and geometric multiplicities
• Quiz 13 (Part 1). Diagonalize a matrix.
• Quiz 13 (Part 2). Find eigenvalues and eigenvectors of a special matrix

The post Quiz 13 (Part 1) Diagonalize a Matrix first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ matrix with the characteristic polynomial
\[p(t)=t^3(t-1)^2(t-2)^5(t+2)^4.\] Assume that the matrix $A$ is diagonalizable.

(a) Find the size of the matrix $A$.

(b) Find the dimension of the eigenspace $E_2$ corresponding to the eigenvalue $\lambda=2$.

(c) Find the nullity of $A$.

(The Ohio State University, Linear Algebra Final Exam Problem)
 

Hint/Definition.

• Recall that when a matrix is diagonalizable, the algebraic multiplicity of each eigenvalue is the same as the geometric multiplicity.
• The geometric multiplicity of an eigenvalue $\lambda$ is the dimension of the eigenspace $E_{\lambda}=\calN(A-\lambda I)$ corresponding to $\lambda$.
• The nullity of $A$ is the dimension of the null space $\calN(A)$ of $A$.

Solution.

(a) Find the size of the matrix $A$.

In general, if $A$ is an $n\times n$ matrix, then its characteristic polynomials has degree $n$.
Since the degree of $p(t)$ is $14$, the size of $A$ is $14 \times 14$.

(b) Find the dimension of the eigenspace $E_2$ corresponding to the eigenvalue $\lambda=2$.

Note that the dimension of the eigenspace $E_2$ is the geometric multiplicity of the eigenvalue $\lambda=2$ by definition.

From the characteristic polynomial $p(t)$, we see that $\lambda=2$ is an eigenvalue of $A$ with algebraic multiplicity $5$.
Since $A$ is diagonalizable, the algebraic multiplicity of each eigenvalue is the same as the geometric multiplicity.

It follows that the geometric multiplicity of $\lambda=2$ is $5$, hence the dimension of the eigenspace $E_2$ is $5$.

(c) Find the nullity of $A$.

We first observe that $\lambda=0$ is an eigenvalue of $A$ with algebraic multiplicity $3$ from the characteristic polynomial.

By definition, the nullity of $A$ is the dimension of the null space $\calN(A)$, and furthermore the null space $\calN(A)$ is the eigenspace $E_0$.
Thus, the nullity of $A$ is the same as the geometric multiplicity of the eigenvalue $\lambda=0$.

Since $A$ is diagonalizable, the algebraic and geometric multiplicities are the same. Hence the nullity of $A$ is $3$.

The post Determine Dimensions of Eigenspaces From Characteristic Polynomial of Diagonalizable Matrix first appeared on Problems in Mathematics.

Find the determinant of the following matrix
\[A=\begin{bmatrix}
6 & 2 & 2 & 2 &2 \\
2 & 6 & 2 & 2 & 2 \\
2 & 2 & 6 & 2 & 2 \\
2 & 2 & 2 & 6 & 2 \\
2 & 2 & 2 & 2 & 6
\end{bmatrix}.\]

(Harvard University, Linear Algebra Exam Problem)
 

Hint.

Computing the determinant directly by hand is tedious.
So use the fact that the determinant of a matrix $A$ is the product of all eigenvalues of $A$.

However, finding the eigenvalue of $A$ itself is as complicated as computing the determinant of $A$.
Instead, first determine the eigenvalues of $B=A-4I$.
Then use the fact that if $\lambda$ is an eigenvalue of $B$, then $\lambda+4$ is an eigenvalue of $A$.

For a proof of this fact, see the post “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$“.

Solution.

Let $B=A-4I$, where $I$ is the $5 \times 5$ identity matrix.
Then every entry of $B$ is $2$.

By elementary row operations, we can reduced the matrix $B$ into
\[B=\begin{bmatrix}
2 & 2 & 2 & 2 &2 \\
2 & 2 & 2 & 2 & 2 \\
2 & 2 & 2 & 2 & 2 \\
2 & 2 & 2 & 2 & 2 \\
2 & 2 & 2 & 2 & 2
\end{bmatrix}\to
\begin{bmatrix}
1 & 1 & 1 & 1 &1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}.\] Thus, the rank of $B$ is $1$, hence the nullity is $4$ by the rank-nullity theorem.
It follows that $0$ is an eigenvalue of $B$ and its geometric multiplicity is $4$.

Since all entries of $B$ are equal, we compute
\[B\begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1
\end{bmatrix}=10\begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1
\end{bmatrix}.\] This yields that $10$ is an eigenvalue of $B$ and the vector $[1\, 1\, 1\, 1\, 1]^{\trans}$ is an eigenvector corresponding to $10$.

Combining these observations, we see that the matrix $B$ has eigenvalues $0$ and $10$ with (algebraic) multiplicities $4$ and $0$, respectively.

Since $A=B+4I$, the eigenvalues of $A$ are $4$ and $14$ with algebraic multiplicities $4$ and $0$, respectively.
(See Problem Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$.)

The determinant of $A$ is the product of all the eigenvalues of $A$ (counting multiplicities). Thus we have
\[\det(A)=4^4\cdot 14=3584.\] Click here if solved 33

The post Determinant of Matrix whose Diagonal Entries are 6 and 2 Elsewhere first appeared on Problems in Mathematics.

Find all the eigenvalues and eigenvectors of the matrix
\[A=\begin{bmatrix}
3 & 9 & 9 & 9 \\
9 &3 & 9 & 9 \\
9 & 9 & 3 & 9 \\
9 & 9 & 9 & 3
\end{bmatrix}.\]

(Harvard University, Linear Algebra Final Exam Problem)

Hint.

Instead of computing the characteristic polynomial $p(t)=\det(A-tI)$ of $A$, consider the matrix $B=A+6I$.
Then use the relation between eigenvalues of $A$ and $B$.

See the problem “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$” for more about the relation.

Solution.

Let $B=A+6I$, where $I$ is the $4 \times 4$ identity matrix. Then every entry of $B$ is $9$. Consider the vector $\mathbf{v}=[1\, 1\, 1\, 1]^{\trans}$.
Then we have
\begin{align*}
B\mathbf{v}=\begin{bmatrix}
9 & 9 & 9 & 9 \\
9 &9 & 9 & 9 \\
9 & 9 & 9 & 9 \\
9 & 9 & 9 & 9
\end{bmatrix}\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}=36\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}=36\mathbf{v}.
\end{align*}

It follows that $36$ is an eigenvalue of $B$ and the vector $\mathbf{v}$ is an associated eigenvector.

We apply the elementary row operations and obtain
\[B\to \begin{bmatrix}
1 & 1 & 1 & 1 \\
0 &0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}.\] Hence the rank of $B$ is $1$, and it follows from the rank-nullity theorem that the nullity is $3$.
Thus, $0$ is an eigenvalue of $B$ and its geometric multiplicity is $3$.

The algebraic multiplicity is always greater than or equal to geometric multiplicity.
But the algebraic multiplicity of $0$ cannot be $4$, since $36$ is another eigenvalue.

As a result, the matrix $B$ has eigenvalues $36$ and $0$ with algebraic multiplicities $1$ and $3$.

The eigenvectors corresponding to $36$ are
\[a\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix},\] where $a$ is any nonzero complex number.
The eigenvector corresponding to $0$ are obtained by solving $B\mathbf{x}=\mathbf{0}$.
The solutions satisfy
\[x_1=-x_2-x_3-x_4.\] Hence the eigenvectors corresponding to $0$ are
\[\mathbf{x}=x_2\begin{bmatrix}
-1 \\
1 \\
0 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
0 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-1 \\
0 \\
0 \\
1
\end{bmatrix},\] where $(x_2, x_3, x_4)\neq (0,0,0)$.

Since $A=B-6I$, it follows that the matrix $A$ has eigenvalues $30$ and $-6$ with algebraic multiplicity $1$ and $3$. Their associated eigenvectors are exactly the eigenvectors for $B$.
(See Problem “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$” for a proof of this fact.)

Namely, the eigenvectors corresponding to $30$ are
\[a\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix},\] where $a$ is any nonzero complex number.
The eigenvectors corresponding to $-6$ are
\[x_2\begin{bmatrix}
-1 \\
1 \\
0 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
0 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-1 \\
0 \\
0 \\
1
\end{bmatrix},\] where $(x_2, x_3, x_4)\neq (0,0,0)$.

The post Eigenvalues and Eigenvectors of Matrix Whose Diagonal Entries are 3 and 9 Elsewhere first appeared on Problems in Mathematics.

Let $A$ be an $n \times n$ matrix and let $c$ be a complex number.

(a) For each eigenvalue $\lambda$ of $A$, prove that $\lambda+c$ is an eigenvalue of the matrix $A+cI$, where $I$ is the identity matrix. What can you say about the eigenvectors corresponding to $\lambda+c$?

(b) Prove that the algebraic multiplicity of the eigenvalue $\lambda$ of $A$ is the same as the algebraic multiplicity of the eigenvalue $\lambda+c$ of $A+cI$ are equal.

(c) How about geometric multiplicities?

Proof.

(a) $\lambda+c$ is an eigenvalue of $A+cI$.

Let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$. Then we have
\[A\mathbf{x}=\lambda \mathbf{x}.\] It follows that we have
\begin{align*}
(A+cI)\mathbf{x}&=A\mathbf{x}+c\mathbf{x}\\
&=\lambda \mathbf{x}+c\mathbf{x}\\
&=(\lambda+c)\mathbf{x}.
\end{align*}

Thus, we obtain
\[(A+cI)\mathbf{x}=(\lambda+c)\mathbf{x},\] where $\mathbf{x}$ is a nonzero vector.
Hence $\lambda+c$ is an eigenvalue of the matrix $A+cI$, and $\mathbf{x}$ is an eigenvector corresponding to $\lambda-c$.

In summary, if $\lambda$ is an eigenvalue of $A$ and $\mathbf{x}$ is an associated eigenvector, then $\lambda+c$ is an eigenvalue of $A+cI$ and $\mathbf{x}$ is an associated eigenvector corresponding to $\lambda+c$.

(b) Algebraic multiplicities are the same

Let $p(t)=\det(A-tI)$ be the characteristic polynomial of $A$.
Let $q(t)$ be the characteristic polynomial of the matrix $A+cI$.
Then we have
\begin{align*}
q(t)&=\det\left(\, (A+cI)-t \,\right)=\det\left(\, A-(t-c) \,\right)\\
&=p(t-c).
\end{align*}

Let $\lambda_1, \dots, \lambda_k$ be distinct eigenvalues of $A$ with algebraic multiplicities $n_1, \dots, n_k$, respectively.
Then we have
\[p(t)=\pm \prod_{i=1}^k (t-\lambda_i)^{n_i}.\] It follows that we have
\begin{align*}
q(t)&=p(t-c)\\
&=\pm \prod_{i=1}^k (t-c-\lambda_i)^{n_i}\\
&=\pm \prod_{i=1}^k \left(t-(\lambda_i+c)\right)^{n_i}.\\
\end{align*}
From the last equation, we read that the eigenvalues of the matrix $A+cI$ are $\lambda_i+c$ with algebraic multiplicity $n_i$ for $i=1,\dots, k$.
Thus, geometric multiplicities of $\lambda$ and $\lambda-c$ are the same.

(c) How about geometric multiplicities?

From part (a), we know that eigenvectors of $\lambda$ are eigenvectors of $\lambda-c$.
Reversing the argument, the eigenvectors of $\lambda+c$ are eigenvectors of $\lambda$.

Thus, the eigenvectors correspond one to one, and the eigenspace of $\lambda$ is the same as the eigenspace of $\lambda+c$.
Hence their geometric multiplicities are the same.

Applications

As applications of this problem, consider the following problems.

This is a problem of a Linear Algebra final exam at Harvard University.
For a solution of this problem, see the post “Eigenvalues and eigenvectors of matrix whose diagonal entries are 3 and 9 elsewhere“.

This is also a problem of a Linear Algebra final exam at Harvard University.
See the post “Determinant of matrix whose diagonal entries are 6 and 2 elsewhere” for a solution.

The post Eigenvalues and Algebraic/Geometric Multiplicities of Matrix $A+cI$ first appeared on Problems in Mathematics.

(a) Let
\[A=\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 &1 & 1 & 1 \\
0 & 0 & 0 & 0 \\
1 & 1 & 1 & 1
\end{bmatrix}.\] Find the eigenvalues of the matrix $A$. Also give the algebraic multiplicity of each eigenvalue.

(b) Let
\[A=\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 &1 & 1 & 1 \\
0 & 0 & 0 & 0 \\
1 & 1 & 1 & 1
\end{bmatrix}.\] One of the eigenvalues of the matrix $A$ is $\lambda=0$. Find the geometric multiplicity of the eigenvalue $\lambda=0$.

Solution (a). Eigenvalues of $A$ and algebraic multiplies

Eigenvalues and their algebraic multiplicities are determined by the characteristic polynomial $p(t)$ of $A$.
By definition, the characteristic polynomial of $A$ is $p(t)=\det(A-tI)$.
We have
\begin{align*}
&p(t)=\det(A-tI)\\
&=\begin{vmatrix}
-t & 0 & 0 & 0 \\
1 &1-t & 1 & 1 \\
0 & 0 & -t & 0 \\
1 & 1 & 1 & 1-t
\end{vmatrix}\\[6pt] &=-t\begin{vmatrix}
1-t & 1 & 1 \\
0 &-t &0 \\
1 & 1 & 1-t
\end{vmatrix} && \text{by the first row cofactor expansion}\\[6pt] &=-t\left(\, -t\begin{vmatrix}
1-t & 1\\
1& 1-t
\end{vmatrix} \,\right)&& \text{by the second row cofactor expansion}\\[6pt] &=t^2\left(\, (1-t)^2-1 \,\right)\\
&=t^2(t^2-2t)\\
&=t^3(t-2).
\end{align*}
Thus the characteristic polynomial is
\[p(t)=t^3(t-2).\] From this, the eigenvalues of $A$ are $0$ and $2$ with algebraic multiplicities $3$ and $1$, respectively.

Solution (b). Geometric multiplicity

We give two solutions for part (b).

First Solution (b). (Finding the rank first)

Recall that the geometric multiplicity of $\lambda$ is the dimension of the eigenspace $E_{\lambda}=\calN(A-\lambda I)$. That is, the geometric multiplicity of $\lambda$ is the nullity of the matrix $A-\lambda I$.

Let us now consider the case $\lambda=0$.
We first find the rank of $A-0 I=A$ as follows.
\begin{align*}
A-0 I= A=\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 &1 & 1 & 1 \\
0 & 0 & 0 & 0 \\
1 & 1 & 1 & 1
\end{bmatrix}
\xrightarrow{R_4-R_2}
\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 &1 & 1 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_1 \leftrightarrow R_2}
\begin{bmatrix}
1 &1 & 1 & 1\\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form.
Hence the rank of $A$ is $1$.
The rank-nullity theorem says that
\[\text{rank of $A$ } + \text{ nullity of $A$}=4.\] Thus, the nullity of $A=A-0I$ is $3$, and hence the geometric multiplicity of $\lambda=0$ is $3$.

Second Solution (b). (Finding a basis of the eigenspace)

In this solution, we find a basis of the eigenspace $E_0$.
By definition $E_0=\calN(A-0I)=\calN(A)$.
Thus, the eigenspace $E_0$ is the null space of the matrix $A$.
We solve the equation $A\mathbf{x}=\mathbf{0}$ as follows.
The augmented matrix of this equation is
\begin{align*}
[A\mid \mathbf{0}]= \left[\begin{array}{rrrr|r}
0 & 0 & 0 & 0 &0 \\
1 &1 & 1 & 1 &0 \\
0 & 0 & 0 & 0 &0\\
1 & 1 & 1 & 1 &0
\end{array} \right] \xrightarrow{R_4-R_2}
\left[\begin{array}{rrrr|r}
0 & 0 & 0 & 0 &0 \\
1 &1 & 1 & 1 &0 \\
0 & 0 & 0 & 0 &0\\
0 & 0 & 0 & 0 &0
\end{array} \right] \xrightarrow{R_1 \leftrightarrow R_2}
\left[\begin{array}{rrrr|r}
1 &1 & 1 & 1 &0 \\
0 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 0 &0\\
0 & 0 & 0 & 0 &0
\end{array} \right].
\end{align*}
Hence the solution satisfies
\[x_1=-x_2-x_3-x_4\] and the general solution is
\begin{align*}
\mathbf{x}=\begin{bmatrix}
-x_2-x_3-x_4 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=x_2\begin{bmatrix}
-1 \\
1 \\
0 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
0 \\
1 \\
0
\end{bmatrix}+x_4 \begin{bmatrix}
-1 \\
0 \\
0 \\
1
\end{bmatrix}.
\end{align*}
Therefore, the eigenspace is
\begin{align*}
&E_0=\calN(A)\\
&=\left\{\, \mathbf{x}\in \C^4 \quad \middle | \quad \mathbf{x}=x_2\begin{bmatrix}
-1 \\
1 \\
0 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
0 \\
1 \\
0
\end{bmatrix}+x_4 \begin{bmatrix}
-1 \\
0 \\
0 \\
1
\end{bmatrix}, \text{ for any } x_2, x_3, x_4\in \C \,\right\}\\[10pt] &=\Span\left\{\, \begin{bmatrix}
-1 \\
1 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
0 \\
1
\end{bmatrix} \,\right\}.
\end{align*}
Thus the set
\[\left\{\, \begin{bmatrix}
-1 \\
1 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
0 \\
1
\end{bmatrix} \,\right\}\] is a spanning set of $E_0$, and it is straightforward to check that the set is linearly independent.
Hence this set is a basis of $E_0$, and the dimension of $E_0$ is $3$.
The geometric multiplicity of $\lambda=0$ is the dimension of $E_0$ by definition.
Thus, the geometric multiplicity of $\lambda$ is $3$.

Comment.

These are Quiz 12 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

I could have combined these two problems into one problem, and just asked to find eigenvalues and algebraic/ geometric multiplicities for each eigenvalue.
The reason I didn’t do is that I wanted to rescue students who didn’t get a correct answer in part (a) for some reasons.
Also, since one of the eigenvalue is given in (b), students could use this information to double check their solutions in (a).
(At least if you didn’t get the eigenvalue $0$, you made a mistake somewhere.)

List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

• Quiz 1. Gauss-Jordan elimination / homogeneous system.
• Quiz 2. The vector form for the general solution / Transpose matrices.
• Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent
• Quiz 4. Inverse matrix/ Nonsingular matrix satisfying a relation
• Quiz 5. Example and non-example of subspaces in 3-dimensional space
• Quiz 6. Determine vectors in null space, range / Find a basis of null space
• Quiz 7. Find a basis of the range, rank, and nullity of a matrix
• Quiz 8. Determine subsets are subspaces: functions taking integer values / set of skew-symmetric matrices
• Quiz 9. Find a basis of the subspace spanned by four matrices
• Quiz 10. Find orthogonal basis / Find value of linear transformation
• Quiz 11. Find eigenvalues and eigenvectors/ Properties of determinants
• Quiz 12. Find eigenvalues and their algebraic and geometric multiplicities
• Quiz 13 (Part 1). Diagonalize a matrix.
• Quiz 13 (Part 2). Find eigenvalues and eigenvectors of a special matrix

The post Quiz 12. Find Eigenvalues and their Algebraic and Geometric Multiplicities first appeared on Problems in Mathematics.