Quiz 13 (Part 1) Diagonalize a Matrix

Introduction to Linear Algebra at the Ohio State University quiz problems and solutions

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 $A$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

 
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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.


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  1. 04/21/2017

    […] This is the second problem of Quiz 13 (Take Home Quiz) for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017. The first problem of Quiz 13 is “Quiz 13 (Part 1) Diagonalize a matrix.“. […]

  2. 04/21/2017

    […] Quiz 13 (Part 1) Diagonalize a matrix. […]

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    […] Quiz 13 (Part 1) Diagonalize a matrix. […]

  4. 04/26/2017

    […] Quiz 13 (Part 1). Diagonalize a matrix. […]

  5. 06/13/2017

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

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    […] Quiz 13 (Part 1). Diagonalize a matrix. […]

  8. 11/18/2017

    […] Quiz 13 (Part 1). Diagonalize a matrix. […]

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