How to Diagonalize a Matrix. Step by Step Explanation.

Linear algebra problems and solutions

Problem 211

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.)
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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.

Problem. Diagonalize \[A=\begin{bmatrix}
1 & 2\\
4& 3
\end{bmatrix}\] and compute $A^{100}$.

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

Problem. 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 $S^{-1}AS=D$.

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

Problem. 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 $A$.

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

Problem. Diagonalize the matrix
\[A=\begin{bmatrix}
1 & 1 & 1 \\
1 &1 &1 \\
1 & 1 & 1
\end{bmatrix}.\]

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.

Problem.
Diagonalize the complex matrix
\[A=\begin{bmatrix}
a & b-a\\
0& b
\end{bmatrix}.\] Using the result of the diagonalization, compute $A^k$ for each $k\in \N$.

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

Theorem. If $A$ is a Hermitian matrix, then $A$ can be diagonalized by a unitary matrix $U$.

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

Problem.
Diagonalize the Hermitian matrix
\[A=\begin{bmatrix}
1 & i\\
-i& 1
\end{bmatrix}\] by a unitary matrix.

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

Related Question.

More Problems related to the diagonalization of a matrix:


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9 Responses

  1. 12/11/2016

    […] the post how to diagonalize a matrix for a review of the diagonalization […]

  2. 04/21/2017

    […] 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.” […]

  3. 06/14/2017

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

  4. 06/25/2017

    […] 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 […]

  5. 06/27/2017

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

  6. 08/07/2017

    […] follows from the general procedure of the diagonalization that $P$ is a nonsingular matrix and [P^{-1}AP=D,] where $D$ is a diagonal matrix […]

  7. 08/07/2017

    […] The solution is given in the post How to Diagonalize a Matrix. Step by Step Explanation […]

  8. 10/11/2017

    […] When $a=b$, then $A$ is already diagonal matrix. So let us consider the case $aneq b$. 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. […]

  9. 10/15/2017

    […] It follows that the matrix [U=begin{bmatrix} mathbf{u}_1 & mathbf{u}_2 end{bmatrix}=frac{1}{sqrt{2}}begin{bmatrix} 1 & 1\ i& -i end{bmatrix}] is unitary and [U^{-1}AU=begin{bmatrix} 0 & 0\ 0& 2 end{bmatrix}] by diagonalization process. […]

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