Let $\mathrm{P}_2$ denote the vector space of polynomials of degree $2$ or less, and let $T : \mathrm{P}_2 \rightarrow \mathrm{P}_2$ be the derivative linear transformation, defined by
\[ T( ax^2 + bx + c ) = 2ax + b . \]

Is $T$ diagonalizable? If so, find a diagonal matrix which represents $T$. If not, explain why not.

Solution.

The standard basis of the vector space $\mathrm{P}_2$ is the set $B = \{ 1 , x , x^2 \}$. The matrix representing $T$ with respect to this basis is
\[ [T]_B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} . \]

The characteristic polynomial of this matrix is
\[ \det ( [T]_B – \lambda I ) = \begin{vmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 2 \\ 0 & 0 & -\lambda \end{vmatrix} = \, – \lambda^3 . \] We see that the only eigenvalue of $T$ is $0$ with algebraic multiplicity $3$.

On the other hand, a polynomial $f(x)$ satisfies $T(f)(x) = 0$ if and only if $f(x) = c$ is a constant. The null space of $T$ is spanned by the single constant polynomial $\mathbb{1}(x) = 1$, and thus is one-dimensional. This means that the geometric multiplicity of the eigenvalue $0$ is only $1$.

Because the geometric multiplicity of $0$ is less than the algebraic multiplicity, the map $T$ is defective, and thus not diagonalizable.

The post Is the Derivative Linear Transformation Diagonalizable? first appeared on Problems in Mathematics.

Consider the matrix $A=\begin{bmatrix}
a & -b\\
b& a
\end{bmatrix}$, where $a$ and $b$ are real numbers and $b\neq 0$.

(a) Find all eigenvalues of $A$.

(b) For each eigenvalue of $A$, determine the eigenspace $E_{\lambda}$.

(c) Diagonalize the matrix $A$ by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

Solution.

(a) Find all eigenvalues of $A$.

The characteristic polynomial $p(t)$ of the matrix $A$ is
\begin{align*}
p(t)&=\det(A-tI) = \begin{vmatrix}
a-t & -b\\
b& a-t
\end{vmatrix}\\[6pt] &=(a-t)^2+b^2.
\end{align*}

The eigenvalues of $A$ are roots of $p(t)$.
So we solve $p(t)=0$. We have
\begin{align*}
& \quad (a-t)^2+b^2=0\\
\Leftrightarrow & \quad (a-t)^2=-b^2\\
\Leftrightarrow &\quad a-t =\pm i b\\
\Leftrightarrow &\quad t= a \pm ib.
\end{align*}
Here $i=\sqrt{-1}$.

Thus, the eigenvalues of $A$ are $a\pm ib$.

(b) For each eigenvalue of $A$, determine the eigenspace $E_{\lambda}$.

We first determine the eigenspace $E_{\lambda}$ for $\lambda = a+ib$.
Recall that by definition $E_{\lambda}=\calN(A-\lambda I)$, the nullspace of $A-\lambda I$.

We compute
\begin{align*}
A-(a+ib)I=\begin{bmatrix}
-ib & -b\\
b& -ib
\end{bmatrix}
\xrightarrow{\frac{i}{b}R_1}
\begin{bmatrix}
1 & -i\\
b& -ib
\end{bmatrix}
\xrightarrow{R_2-bR_1}
\begin{bmatrix}
1 & -i\\
0& 0
\end{bmatrix}.
\end{align*}
Note that in the above row reduction, we needed the assumption $b\neq 0$.

It follows that the general solution of the system is $x_1=i x_2$.
Hence, we have
\[E_{a+ib} =\Span \left(\, \begin{bmatrix}
i \\
1
\end{bmatrix} \,\right).\]

Note that the other eigenvalue $a-ib$ is the complex conjugate of $a+ib$.
It follows that the eigenspace $E_{a-ib}$ is obtained by conjugating the eigenspace $E_{a+ib}$.
Thus,
\[E_{a-ib} =\Span \left(\, \begin{bmatrix}
-i \\
1
\end{bmatrix} \,\right).\]

(c) Diagonalize the matrix $A$

From part (b), we see that
\[\begin{bmatrix}
i \\
1
\end{bmatrix} \text{ and } \begin{bmatrix}
-i \\
1
\end{bmatrix}\] form an eigenbasis for $\C^2$.

So, we set
\[S=\begin{bmatrix}
i & -i\\
1& 1
\end{bmatrix} \text{ and } D=\begin{bmatrix}
a+ib & 0\\
0& a-ib
\end{bmatrix},\] and we obtain $S^{-1}AS=D$ by the diagonalization procedure.

The post Find Eigenvalues, Eigenvectors, and Diagonalize the 2 by 2 Matrix first appeared on Problems in Mathematics.

Diagonalize the $2\times 2$ matrix $A=\begin{bmatrix}
2 & -1\\
-1& 2
\end{bmatrix}$ by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

Solution.

The characteristic polynomial $p(t)$ of the matrix $A$ is
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
2-t & -1\\
-1& 2-1
\end{vmatrix}\\[6pt] &=(2-t)^2-1 =t^2-4t+3\\
&=(t-1)(t-3).
\end{align*}
It follows that the eigenvalues of $A$ are $\lambda=1, 3$ with algebraic multiplicities are both $1$.
Hence, the geometric multiplicities are $1$ and thus any nonzero vector in eahc eigenspace forms a eigenbasis.

Now let us find a eigenbasis for each eigenspace $E_{\lambda}=\calN(A-\lambda I)$.
For the eigenvalue $1$, we have
\[A-I=\begin{bmatrix}
1 & -1\\
-1& 1
\end{bmatrix}\xrightarrow{R_2+R_1} \begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}\] This yields that the eigenvectors corresponding to the eigenvalue $1$ are $x_2\begin{bmatrix}
1 \\
1
\end{bmatrix}$ with $x_2\neq 0$. Hence
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
1
\end{bmatrix} \in E_1\] is an eigenbasis for $E_1$.

Similarly, as we have
\[A-3I=\begin{bmatrix}
-1 & -1\\
-1& -1
\end{bmatrix} \xrightarrow{-R_1}\begin{bmatrix}
1 & 1\\
-1& -1
\end{bmatrix} \xrightarrow{R_2+R_1} \begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix},\] we see that
\[\mathbf{v}_2=\begin{bmatrix}
-1 \\
1
\end{bmatrix} \in E_3\] is an eigenbasis for $E_3$.

Let
\[S=\begin{bmatrix}
\mathbf{v}_1 & \mathbf{v}_2
\end{bmatrix}=
\begin{bmatrix}
1 & -1\\
1& 1
\end{bmatrix} \text{ and } D=\begin{bmatrix}
1 & 0\\
0& 3
\end{bmatrix}.\]

Then the diagonalization procedure yields that $S$ is nonsingular and $S^{-1}AS= D$.

The post Diagonalize a 2 by 2 Symmetric Matrix first appeared on Problems in Mathematics.

Consider the $2\times 2$ complex matrix
\[A=\begin{bmatrix}
a & b-a\\
0& b
\end{bmatrix}.\]

(a) Find the eigenvalues of $A$.

(b) For each eigenvalue of $A$, determine the eigenvectors.

(c) Diagonalize the matrix $A$.

(d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$.

Solution.

(a) Find the eigenvalues of $A$.

Since $A$ is an upper triangular matrix, eigenvalues are diagonal entries.
Hence $a, b$ are eigenvalues of $A$.

(b) For each eigenvalue of $A$, determine the eigenvectors.

Suppose now that $a\neq b$.
Let us find eigenvectors corresponding to the eigenvalue $a$.
We have
\begin{align*}
A-aI=\begin{bmatrix}
0 & b-a\\
0& b-a
\end{bmatrix}
\xrightarrow{R_2-R_1}
\begin{bmatrix}
0 & b-a\\
0& 0
\end{bmatrix}
\xrightarrow{\frac{1}{b-a}R_1}
\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}.
\end{align*}
It follows that the eigenvectors corresponding to $a$ are
\[x_1\begin{bmatrix}
1 \\
0
\end{bmatrix},\] where $x_1$ is any nonzero complex number.

Next, we find the eigenvectors corresponding to the eigenvalue $b$.
We have
\begin{align*}
A-bI=\begin{bmatrix}
a-b & b-a\\
0& 0
\end{bmatrix}
\xrightarrow{\frac{1}{a-b}R_1}
\begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}.
\end{align*}
Hence the eigenvectors corresponding to $b$ are
\[x_1\begin{bmatrix}
1 \\
1
\end{bmatrix},\] where $x_1$ is any nonzero complex number.

(c) Diagonalize the matrix $A$.

When $a=b$, then $A$ is already diagonal matrix. So let us consider the case $a\neq b$.
In the previous parts, we obtained the eigenvalues $a, b$, and corresponding eigenvectors
\[\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } \begin{bmatrix}
1 \\
1
\end{bmatrix}.\] Let $S=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}$ be a matrix whose column vectors are the eigenvectors.
Then $S$ is invertible and we have
\[S^{-1}AS=\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}\] by the diagonalization process.

Remark that this formula is also true even when $a=b$.

(d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$.

Using the result of the diagonalization in part (c), we have
\[A=S\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}S^{-1}.\] For each positive integer $k$, we have
\begin{align*}
A^k&=\left(\, S\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}S^{-1} \,\right)^k\\[6pt] &=S\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}^k S^{-1}=S\begin{bmatrix}
a^k & 0\\
0& b^k
\end{bmatrix}S^{-1}\\[6pt] &=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}
\begin{bmatrix}
a^k & 0\\
0& b^k
\end{bmatrix}
\begin{bmatrix}
1 & -1\\
0& 1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
a^k & b^k-a^k\\
0& b^k
\end{bmatrix}.
\end{align*}

In summary, we have the formula
\[A^k=\begin{bmatrix}
a^k & b^k-a^k\\
0& b^k
\end{bmatrix}.\] Click here if solved 62

The post Diagonalize the Upper Triangular Matrix and Find the Power of the Matrix first appeared on Problems in Mathematics.

Diagonalize the matrix
\[A=\begin{bmatrix}
1 & 1 & 1 \\
1 &1 &1 \\
1 & 1 & 1
\end{bmatrix}.\] Namely, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

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

Hint.

To diagonalize the matrix $A$, we need to find eigenvalues $A$ and bases of eigenspaces.

For a procedure of the diagonalization, see the post “How to Diagonalize a Matrix. Step by Step Explanation.“.

Below, we will find eigenvalues and eigenvectors without using the characteristic polynomial although you may use it.

Solution.

We use an indirect way to find eigenvalues and eigenvectors.
(We will not use the characteristic polynomial.)

Applying the elementary row operations, we have
\begin{align*}
A=\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}.
\end{align*}
Hence the solutions $\mathbf{x}$ of $A\mathbf{x}=\mathbf{0}$ satisfy
\[x_1=-x_2-x_3.\] Thus, every vector in the null space is of the form
\[\mathbf{x}=\begin{bmatrix}
-x_2-x_3 \\
x_2 \\
x_3
\end{bmatrix}=x_2\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\] for some scalars $x_2, x_3$.

It follows that
\[\left\{\, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \,\right\}\] is a basis of the null space $\calN(A)$.

Hence $0$ is an eigenvalue and the geometric multiplicity corresponding to $0$, which is the nullity of $A$, is $2$.
It follows that the algebraic multiplicity of the eigenvalue $0$ is either $2$ or $3$.
We see that it is $2$ shortly.

Note that by inspection we have
\[A\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}=3\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}.\] This yields that $3$ is an eigenvalue of $A$ and $\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}$ is a corresponding eigenvector.

The sum of algebraic multiplicities of all eigenvalues of $A$ is $3$.
Hence the algebraic multiplicity of $0$ must be $2$, and that of $3$ must be $1$.
In particular, the vector $\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}$ forms a basis of the eigenspace $E_3$.

In summary so far, we have eigenvalues $0, 3$ and basis vectors of eigenspaces 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.

Thus, we put
\[S=\begin{bmatrix}
-1 & -1 & 1 \\
1 &0 &1 \\
0 & 1 & 1
\end{bmatrix}\] and obtain
\[S^{-1}AS=D,\] where
\[D=\begin{bmatrix}
0 & 0 & 0 \\
0 &0 &0 \\
0 & 0 & 3
\end{bmatrix}.\]

Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

1. Find All the Eigenvalues of 4 by 4 Matrix
2. Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue
3. Diagonalize a 2 by 2 Matrix if Diagonalizable
4. Find an Orthonormal Basis of the Range of a Linear Transformation
5. The Product of Two Nonsingular Matrices is Nonsingular
6. Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not
7. Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials
8. Find Values of $a , b , c$ such that the Given Matrix is Diagonalizable
9. Idempotent Matrix and its Eigenvalues
10. Diagonalize the 3 by 3 Matrix Whose Entries are All One (This page)
11. Given the Characteristic Polynomial, Find the Rank of the Matrix
12. Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$
13. Determine Whether There Exists a Nonsingular Matrix Satisfying $A^4=ABA^2+2A^3$

The post Diagonalize the 3 by 3 Matrix Whose Entries are All One first appeared on Problems in Mathematics.

Determine whether the matrix
\[A=\begin{bmatrix}
1 & 4\\
2 & 3
\end{bmatrix}\] is diagonalizable.

If so, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

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

Solution.

To determine whether the matrix $A$ is diagonalizable, we first find eigenvalues of $A$.
To do so, we compute the characteristic polynomial $p(t)$ of $A$:
\begin{align*}
p(t)&=\begin{vmatrix}
1-t & 4\\
2& 3-t
\end{vmatrix}
=(1-t)(3-t)-8\\[6pt] &=t^2-4t-5=(t+1)(t-5).
\end{align*}

The roots of the characteristic polynomial $p(t)$ are eigenvalues of $A$.
Hence the eigenvalues of $A$ are $-1$ and $5$.

Since the $2\times 2$ matrix $A$ has two distinct eigenvalues, it is diagonalizable.

To find the invertible matrix $S$, we need eigenvectors.

Let us find the eigenvectors corresponding to the eigenvalue $-1$.
By elementary row operations, we have
\begin{align*}
&A-(-1)I=A+I=\begin{bmatrix}
2 & 4\\
2& 4
\end{bmatrix}\\[6pt] &\xrightarrow{R_2-R_1}
\begin{bmatrix}
2 & 4\\
0& 0
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_1}
\begin{bmatrix}
1 & 2\\
0& 0
\end{bmatrix}.
\end{align*}
It follows that the eigenvectors corresponding to $-1$ are of the form
\[a\begin{bmatrix}
-2 \\
1
\end{bmatrix}\] for any nonzero scalar $a$.

Similarly, we have
\begin{align*}
A-5I=\begin{bmatrix}
-4 & 4\\
2& -2
\end{bmatrix}
\xrightarrow{\frac{-1}{4}R_1}
\begin{bmatrix}
1 & -1\\
2& -2
\end{bmatrix}
\xrightarrow{R_2-2R_1}
\begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}.
\end{align*}
Hence the eigenvectors corresponding to $5$ are of the form
\[b\begin{bmatrix}
1 \\
1
\end{bmatrix}\] for any nonzero scalar $b$.

Thus, $\mathbf{u}=\begin{bmatrix}
-2 \\
1
\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}
1 \\
1
\end{bmatrix}$ are basis vectors of eigenspace $E_{-1}, E_{5}$, respectively.

Define
\[S:=\begin{bmatrix}
\mathbf{u} & \mathbf{v}
\end{bmatrix}
=\begin{bmatrix}
-2 & 1\\
1& 1
\end{bmatrix}.\] Then by the general procedure of the diagonalization, we have
\begin{align*}
S^{-1}AS=D,
\end{align*}
where
\[D:=\begin{bmatrix}
-1 & 0\\
0& 5
\end{bmatrix}.\]

Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

1. Find All the Eigenvalues of 4 by 4 Matrix
2. Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue
3. Diagonalize a 2 by 2 Matrix if Diagonalizable (This page)
4. Find an Orthonormal Basis of the Range of a Linear Transformation
5. The Product of Two Nonsingular Matrices is Nonsingular
6. Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not
7. Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials
8. Find Values of $a , b , c$ such that the Given Matrix is Diagonalizable
9. Idempotent Matrix and its Eigenvalues
10. Diagonalize the 3 by 3 Matrix Whose Entries are All One
11. Given the Characteristic Polynomial, Find the Rank of the Matrix
12. Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$
13. Determine Whether There Exists a Nonsingular Matrix Satisfying $A^4=ABA^2+2A^3$

The post Diagonalize a 2 by 2 Matrix if Diagonalizable first appeared on Problems in Mathematics.

Determine whether the matrix
\[A=\begin{bmatrix}
0 & 1 & 0 \\
-1 &0 &0 \\
0 & 0 & 2
\end{bmatrix}\] is diagonalizable.

If it is diagonalizable, then find the invertible matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

How to diagonalize matrices.

For a general procedure of the diagonalization of a matrix, please read the post “How to Diagonalize a Matrix. Step by Step Explanation“.

Solution.

We first determine the eigenvalues of the matrix $A$.
To do so, we compute the characteristic polynomial $p(t)$ of $A$.
We have
\begin{align*}
&p(t)=\det(A-tI)\\
&=\begin{vmatrix}
-t & 1 & 0 \\
-1 &-t &0 \\
0 & 0 & 2-t
\end{vmatrix}\\[6pt] &=(-1)^{3+3}(2-t)\begin{vmatrix}
-t & 1\\
-1& -t
\end{vmatrix} && \text{by the third row cofactor expansion}\\
&=(2-t)(t^2+1).
\end{align*}
Thus the eigenvalues of $A$ are $2, \pm i$.
Since the $3\times 3$ matrix $A$ has three distinct eigenvalues, it is diagonalizable.

To diagonalize $A$, we now find eigenvectors.
For the eigenvalue $2$, we compute
\begin{align*}
&A-2I=\begin{bmatrix}
-2 & 1 & 0 \\
-1 &-2 &0 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{-R_2}
\begin{bmatrix}
-2 & 1 & 0 \\
1 &2 &0 \\
0 & 0 & 0
\end{bmatrix}\\[6pt] &\xrightarrow{R_1 \leftrightarrow R_2}\begin{bmatrix}
1 & 2 & 0 \\
-2 &1 &0 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_2+2R_1}\begin{bmatrix}
1 & 2 & 0 \\
0 &5 &0 \\
0 & 0 & 0
\end{bmatrix}\\[6pt] &\xrightarrow{\frac{1}{5}R_2}\begin{bmatrix}
1 & 2 & 0 \\
0 &1 &0 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_1-2R_2}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}
Thus, the solutions $\mathbf{x}$ of $(A-2I)=\mathbf{0}$ satisfy $x=y=0$.
Hence the eigenspace is
\[E_2=\calN(A-2I)=\Span\left\{\, \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \,\right\}.\]

For the eigenvalue $i$, we compute
\begin{align*}
A-iI=\begin{bmatrix}
-i & 1 & 0 \\
-1 &-i &0 \\
0 & 0 & 2-i
\end{bmatrix}
\xrightarrow{iR_1}
\begin{bmatrix}
1 & i & 0 \\
-1 &-i &0 \\
0 & 0 & 2-i
\end{bmatrix}\\[6pt] \xrightarrow{\substack{R_2+R_1\\ \frac{1}{2-i}R_3}}
\begin{bmatrix}
1 & i & 0 \\
0 &0 &0 \\
0 & 0 & 1
\end{bmatrix}
\xrightarrow{R_2 \leftrightarrow R_3}
\begin{bmatrix}
1 & i & 0 \\
0 & 0 & 1\\
0 &0 &0
\end{bmatrix}.
\end{align*}
So the solutions $\mathbf{x}$ of $(A-iI)\mathbf{x}=\mathbf{0}$ satisfy
\[x=-iy \text{ and } z=0.\] Thus, the eigenspace is
\[E_i=\calN(A-iI)=\Span\left\{\, \begin{bmatrix}
1 \\
i \\
0
\end{bmatrix} \,\right\}.\]

Since $i$ and $-i$ are complex conjugate, their eigenspaces are also complex conjugate.
Hence the eigenspace for $-i$ is
\[E_{-i}=\Span\left\{\, \begin{bmatrix}
1 \\
-i \\
0
\end{bmatrix} \,\right\}.\]

From these computations, we have obtained eigenvalues $2, i, -i$ and eigenvector corresponding to these are
\[\mathbf{v}_{2}=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}, \mathbf{v}_i=\begin{bmatrix}
1 \\
i \\
0
\end{bmatrix}, \mathbf{v}_{-i}=\begin{bmatrix}
1 \\
-i \\
0
\end{bmatrix}.\]

Let
\[S=\begin{bmatrix}
\mathbf{v}_2 & \mathbf{v}_i & \mathbf{v}_{-i} \\
\end{bmatrix}=\begin{bmatrix}
0 & 1 & 1 \\
0 &i &-i \\
1 & 0 & 0
\end{bmatrix}\] and
\[D=\begin{bmatrix}
2 & 0 & 0 \\
0 &i &0 \\
0 & 0 & -i
\end{bmatrix}.\] Then $S$ is invertible and we have $S^{-1}AS=D$ by the diagonalization process.

The post Diagonalize the 3 by 3 Matrix if it 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.

Let $A$ and $B$ be $n\times n$ matrices.
Suppose that $A$ and $B$ have the same eigenvalues $\lambda_1, \dots, \lambda_n$ with the same corresponding eigenvectors $\mathbf{x}_1, \dots, \mathbf{x}_n$.
Prove that if the eigenvectors $\mathbf{x}_1, \dots, \mathbf{x}_n$ are linearly independent, then $A=B$.

Proof.

Since $A$ and $B$ have $n$ linearly independent eigenvectors $\mathbf{x}_1, \dots, \mathbf{x}_n$, they are diagonalizable.
Specifically, if we put $S=[\mathbf{x}_1, \dots, \mathbf{x}_n]$.

Then $S$ is invertible (as column vectors of $S$ are linearly independent) and we have
\[S^{-1}AS=D \text{ and } S^{-1}BS=D,\] where $D$ is the diagonal matrix whose diagonal entries are eigenvalues:
\[D=\begin{bmatrix}
\lambda_1 & 0 & \cdots & 0 \\
0 & \lambda_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n
\end{bmatrix}.\] It follows that we have
\[S^{-1}AS=D=S^{-1}BS,\] and hence $A=B$. This completes the proof.

The post If Two Matrices Have the Same Eigenvalues with Linearly Independent Eigenvectors, then They Are Equal first appeared on Problems in Mathematics.

Determine all $2\times 2$ matrices $A$ such that $A$ has eigenvalues $2$ and $-1$ with corresponding eigenvectors
\[\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } \begin{bmatrix}
2 \\
1
\end{bmatrix},\] respectively.

Solution.

Suppose that $A$ is a $2\times 2$ matrix having eigenvalues $2$ and $-1$ with corresponding eigenvectors
\[\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } \begin{bmatrix}
2 \\
1
\end{bmatrix},\] respectively.
Then since $A$ has two distinct eigenvalues, the matrix $A$ is diagonalizable.
As we know eigenvectors, we can diagonalize $A$ by the matrix
\[S:=\begin{bmatrix}
1 & 2\\
0& 1
\end{bmatrix}.\] That is, we have
\[S^{-1}AS=\begin{bmatrix}
2 & 0\\
0& -1
\end{bmatrix}.\] The inverse matrix of $S$ is given by
\[S^{-1}=\begin{bmatrix}
1 & -2\\
0& 1
\end{bmatrix}.\] It follows that we have
\begin{align*}
A&=S\begin{bmatrix}
2 & 0\\
0& -1
\end{bmatrix}S^{-1}\\[6pt] &=
\begin{bmatrix}
1 & 2\\
0& 1
\end{bmatrix}
\begin{bmatrix}
2 & 0\\
0& -1
\end{bmatrix}
\begin{bmatrix}
1 & -2\\
0& 1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
2 & -6\\
0& -1
\end{bmatrix}.
\end{align*}

Therefore, the only matrix satisfying the given conditions is
\[A=\begin{bmatrix}
2 & -6\\
0& -1
\end{bmatrix}.\] Click here if solved 26

The post Determine All Matrices Satisfying Some Conditions on Eigenvalues and Eigenvectors first appeared on Problems in Mathematics.