# 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) Add to solve later

## 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. Add to solve later

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