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
\[A=\begin{bmatrix}
2 & -1 & -1 \\
-1 &2 &-1 \\
-1 & -1 & 2
\end{bmatrix}.\] 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.

In this post, we explain how to diagonalize a matrix if it is diagonalizable.

As an example, we solve the following problem.

Diagonalize the matrix
\[A=\begin{bmatrix}
4 & -3 & -3 \\
3 &-2 &-3 \\
-1 & 1 & 2
\end{bmatrix}\] by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

(Update 10/15/2017. A new example problem was added.)

Here we explain how to diagonalize a matrix. We only describe the procedure of diagonalization, and no justification will be given.
The process can be summarized as follows. A concrete example is provided below, and several exercise problems are presented at the end of the post.

Diagonalization Procedure

Let $A$ be the $n\times n$ matrix that you want to diagonalize (if possible).

1. Find the characteristic polynomial $p(t)$ of $A$.
2. Find eigenvalues $\lambda$ of the matrix $A$ and their algebraic multiplicities from the characteristic polynomial $p(t)$.
3. For each eigenvalue $\lambda$ of $A$, find a basis of the eigenspace $E_{\lambda}$.
   If there is an eigenvalue $\lambda$ such that the geometric multiplicity of $\lambda$, $\dim(E_{\lambda})$, is less than the algebraic multiplicity of $\lambda$, then the matrix $A$ is not diagonalizable. If not, $A$ is diagonalizable, and proceed to the next step.
4. If we combine all basis vectors for all eigenspaces, we obtained $n$ linearly independent eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n$.
5. Define the nonsingular matrix
   \[S=[\mathbf{v}_1 \mathbf{v}_2 \dots \mathbf{v}_n] .\]
6. Define the diagonal matrix $D$, whose $(i,i)$-entry is the eigenvalue $\lambda$ such that the $i$-th column vector $\mathbf{v}_i$ is in the eigenspace $E_{\lambda}$.
7. Then the matrix $A$ is diagonalized as \[ S^{-1}AS=D.\]

Example of a matrix diagonalization

Now let us examine these steps with an example.
Let us consider the following $3\times 3$ matrix.
\[A=\begin{bmatrix}
4 & -3 & -3 \\
3 &-2 &-3 \\
-1 & 1 & 2
\end{bmatrix}.\] We want to diagonalize the matrix if possible.

Step 1: Find the characteristic polynomial

The characteristic polynomial $p(t)$ of $A$ is
\[p(t)=\det(A-tI)=\begin{vmatrix}
4-t & -3 & -3 \\
3 &-2-t &-3 \\
-1 & 1 & 2-t
\end{vmatrix}.\] Using the cofactor expansion, we get
\[p(t)=-(t-1)^2(t-2).\]

Step 2: Find the eigenvalues

From the characteristic polynomial obtained in Step 1, we see that eigenvalues are
\[\lambda=1 \text{ with algebraic multiplicity } 2\] and
\[\lambda=2 \text{ with algebraic multiplicity } 1.\]

Step 3: Find the eigenspaces

Let us first find the eigenspace $E_1$ corresponding to the eigenvalue $\lambda=1$.
By definition, $E_1$ is the null space of the matrix
\[A-I=\begin{bmatrix}
3 & -3 & -3 \\
3 &-3 &-3 \\
-1 & 1 & 1
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & -1 & -1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}\] by elementary row operations.
Hence if $(A-I)\mathbf{x}=\mathbf{0}$ for $\mathbf{x}\in \R^3$, we have
\[x_1=x_2+x_3.\] Therefore, we have
\begin{align*}
E_1=\calN(A-I)=\left \{\quad \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} \quad \right \}.
\end{align*}
From this, we see that the set
\[\left\{\quad\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix},\quad \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}\quad \right\}\] is a basis for the eigenspace $E_1$.
Thus, the dimension of $E_1$, which is the geometric multiplicity of $\lambda=1$, is $2$.

Similarly, we find a basis of the eigenspace $E_2=\calN(A-2I)$ for the eigenvalue $\lambda=2$.
We have
\begin{align*}
A-2I=\begin{bmatrix}
2 & -3 & -3 \\
3 &-4 &-3 \\
-1 & 1 & 0
\end{bmatrix}
\rightarrow \cdots \rightarrow \begin{bmatrix}
1 & 0 & 3 \\
0 &1 &3 \\
0 & 0 & 0
\end{bmatrix}
\end{align*}
by elementary row operations.
Then if $(A-2I)\mathbf{x}=\mathbf{0}$ for $\mathbf{x}\in \R^3$, then we have
\[x_1=-3x_3 \text{ and } x_2=-3x_3.\] Therefore we obtain
\begin{align*}
E_2=\calN(A-2I)=\left \{\quad \mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=x_3\begin{bmatrix}
-3 \\
-3 \\
1
\end{bmatrix} \quad \right \}.
\end{align*}
From this we see that the set
\[\left \{ \quad \begin{bmatrix}
-3 \\
-3 \\
1
\end{bmatrix} \quad \right \}\] is a basis for the eigenspace $E_2$ and the geometric multiplicity is $1$.

Since for both eigenvalues, the geometric multiplicity is equal to the algebraic multiplicity, the matrix $A$ is not defective, and hence diagonalizable.

Step 4: Determine linearly independent eigenvectors

From Step 3, the vectors
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}
-3 \\
-3 \\
1
\end{bmatrix} \] are linearly independent eigenvectors.

Step 5: Define the invertible matrix $S$

Define the matrix $S=[\mathbf{v}_1 \mathbf{v}_2 \mathbf{v}_3]$. Thus we have
\[S=\begin{bmatrix}
1 & 1 & -3 \\
1 &0 &-3 \\
0 & 1 & 1
\end{bmatrix}\] and the matrix $S$ is nonsingular (since the column vectors are linearly independent).

Step 6: Define the diagonal matrix $D$

Define the diagonal matrix
\[D=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 2
\end{bmatrix}.\] Note that $(1,1)$-entry of $D$ is $1$ because the first column vector $\mathbf{v}_1=\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}$ of $S$ is in the eigenspace $E_1$, that is, $\mathbf{v}_1$ is an eigenvector corresponding to eigenvalue $\lambda=1$.
Similarly, the $(2,2)$-entry of $D$ is $1$ because the second column $\mathbf{v}_2=\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}$ of $S$ is in $E_1$.
The $(3,3)$-entry of $D$ is $2$ because the third column vector $\mathbf{v}_3=\begin{bmatrix}
-3 \\
-3 \\
1
\end{bmatrix}$ of $S$ is in $E_2$.

(The order you arrange the vectors $\mathbf{v}_1, \mathbf{v_2}, \mathbf{v}_3$ to form $S$ does not matter but once you made $S$, then the order of the diagonal entries is determined by $S$, that is, the order of eigenvectors in $S$.)

Step 7: Finish the diagonalization

Finally, we can diagonalize the matrix $A$ as
\[S^{-1}AS=D,\] where
\[S=\begin{bmatrix}
1 & 1 & -3 \\
1 &0 &-3 \\
0 & 1 & 1
\end{bmatrix} \text{ and } D=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 2
\end{bmatrix}.\] (Here you don’t have to find the inverse matrix $S^{-1}$ unless you are asked to do so.)

Diagonalization Problems and Examples

Check out the following problems about the diagonalization of a matrix to see if you understand the procedure.

For a solution of this problem and related questions, see the post “Diagonalize a 2 by 2 Matrix $A$ and Calculate the Power $A^{100}$“.

For a solution, check out the post “Diagonalize the 3 by 3 Matrix if it is Diagonalizable“.

For a solution, see the post “Quiz 13 (Part 1) Diagonalize a matrix.“.

In the solution given in the post “Diagonalize the 3 by 3 Matrix Whose Entries are All One“, we use an indirect method to find eigenvalues and eigenvectors.

The next problem is a diagonalization problem of a matrix with variables.

The solution is given in the post↴
Diagonalize the Upper Triangular Matrix and Find the Power of the Matrix

A Hermitian Matrix can be diagonalized by a unitary matrix

This means that there exists a unitary matrix $U$ such that $U^{-1}AU$ is a diagonal matrix.

The solution is given in the post ↴
Diagonalize the $2\times 2$ Hermitian Matrix by a Unitary Matrix

More diagonalization problems

More Problems related to the diagonalization of a matrix are gathered in the following page:

Diagonalization of Matrices

The post How to Diagonalize a Matrix. Step by Step Explanation. first appeared on Problems in Mathematics.