# Quiz 13 (Part 1) Diagonalize a Matrix ## Problem 385

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 $S^{-1}AS=D$. Add to solve later

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

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