# Diagonalize the 3 by 3 Matrix Whose Entries are All One

## Problem 483

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.

### More from my site

• Diagonalize a 2 by 2 Matrix if Diagonalizable 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 […]
• Quiz 13 (Part 1) Diagonalize a Matrix 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 $A$ and a diagonal matrix $D$ such that […]
• Two Matrices with the Same Characteristic Polynomial. Diagonalize if Possible. Let $A=\begin{bmatrix} 1 & 3 & 3 \\ -3 &-5 &-3 \\ 3 & 3 & 1 \end{bmatrix} \text{ and } B=\begin{bmatrix} 2 & 4 & 3 \\ -4 &-6 &-3 \\ 3 & 3 & 1 \end{bmatrix}.$ For this problem, you may use the fact that both matrices have the same characteristic […]
• Determine Dimensions of Eigenspaces From Characteristic Polynomial of Diagonalizable Matrix 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 […]
• Find Values of $a, b, c$ such that the Given Matrix is Diagonalizable For which values of constants $a, b$ and $c$ is the matrix $A=\begin{bmatrix} 7 & a & b \\ 0 &2 &c \\ 0 & 0 & 3 \end{bmatrix}$ diagonalizable? (The Ohio State University, Linear Algebra Final Exam Problem)   Solution. Note that the […]
• Maximize the Dimension of the Null Space of $A-aI$ Let $A=\begin{bmatrix} 5 & 2 & -1 \\ 2 &2 &2 \\ -1 & 2 & 5 \end{bmatrix}.$ Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix. Your score of this problem is equal to that […]
• Eigenvalues of a Hermitian Matrix are Real Numbers Show that eigenvalues of a Hermitian matrix $A$ are real numbers. (The Ohio State University Linear Algebra Exam Problem)   We give two proofs. These two proofs are essentially the same. The second proof is a bit simpler and concise compared to the first one. […]
• Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$ Let $A=\begin{bmatrix} 1 & -14 & 4 \\ -1 &6 &-2 \\ -2 & 24 & -7 \end{bmatrix} \quad \text{ and }\quad \mathbf{v}=\begin{bmatrix} 4 \\ -1 \\ -7 \end{bmatrix}.$ Find $A^{10}\mathbf{v}$. You may use the following information without proving […]

### 6 Responses

1. 06/27/2017

[…] 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 […]

2. 06/28/2017

[…] Diagonalize the 3 by 3 Matrix Whose Entries are All One […]

3. 06/28/2017

[…] Diagonalize the 3 by 3 Matrix Whose Entries are All One […]

4. 06/28/2017

[…] Diagonalize the 3 by 3 Matrix Whose Entries are All One […]

5. 08/02/2017

[…] Diagonalize the 3 by 3 Matrix Whose Entries are All One […]

6. 11/20/2017

[…] Diagonalize the 3 by 3 Matrix Whose Entries are All One […]

##### Find Values of $a, b, c$ such that the Given Matrix is Diagonalizable

For which values of constants $a, b$ and $c$ is the matrix \[A=\begin{bmatrix} 7 & a & b \\ 0...

Close