# Symmetric Matrix and Its Eigenvalues, Eigenspaces, and Eigenspaces

## Problem 42

Let $A$ be a $4\times 4$ real symmetric matrix. Suppose that $\mathbf{v}_1=\begin{bmatrix} -1 \\ 2 \\ 0 \\ -1 \end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $1$ of $A$.
Suppose that the eigenspace for the eigenvalue $2$ is $3$-dimensional.

(a) Find an orthonormal basis for the eigenspace of the eigenvalue $2$ of $A$.

(b) Find $A\mathbf{v}$, where
$\mathbf{v}=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}.$

(The University of Tokyo Linear Algebra Exam)

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## Hint.

1. A symmetric matrix is diagonalizable. Hence the sum of dimensions of eigenspaces is $4$.
2. Show that the eigenspaces for eigenvalues $1$ and $2$ are orthogonal.
3. To obtain an orthonormal basis from any basis, use Gram-Schmidt process and then normalize the lengths.
4. For (b), express the vector $\mathbf{v}$ as a linear combination of a basis consisting of eigenvectors.

## Proof.

### (a) Find an orthonormal basis for the eigenspace of the eigenvalue $2$ of $A$.

Note that since $A$ is symmetric, it is diagonalizable. Thus $\C^4$ is a direct sum of eigenspaces of $A$.
Since the eigenspace $E_2$ for the eigenvalue $2$ has dimension $3$, the eigenspace $E_1$ for the eigenvalue $1$ must have dimension $1$.

Thus $\mathbf{v}_1$ is a basis for $E_1$. We also see that there is no other eigenvalues. Hence $C^4=E_1\oplus E_2$.

We fist claim that $E_1$ and $E_2$ are orthogonal.

Let $x\in E_1$ and $y \in E_2$.
Then we calculate the inner product
\begin{align*}
(x, y)&=(Ax, y)=\bar{x}^{\trans}\bar{A}^{\trans}y = \bar{x}^{\trans}Ay = (x, Ay)=(x,2y)=2(x,y)
\end{align*}
Thus we have $(x,y)=0$, hence $x$ and $y$ are orthogonal.

Since $\mathbf{v}_1$ is a basis for $E_1$, if $\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in E_2$, we have
$-x_1+2x_2-x_4=0.$ Solving this we see that
$x=s\begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix}+t \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}+u\begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}.$ Thus a basis of $E_3$ is
$\left\{\, \begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\, \right\}.$ Let us call these vectors $b_1, b_2, b_3$.
The note that $b_1\cdot b_2=0$, $b_2\cdot b_3=0$, but $b_1\cdot b_3=-2$.
Hence they are not orthogonal.

We apply Gram-Schmidt orthogonalization process.
Let $b_3’=b_3+ab_1$ be a vector that is orthogonal to $b_1$ for some $a$.
Taking the inner product with $b_1$, we get
$0=b_1\cdot b_3+ab_1\cdot b_1=-2+a5$.
Thus $a=2/5$ and
$b_3’=b_3+(2/5)b_1=\frac{1}{5}\begin{bmatrix} -1 \\ 2 \\ 0 \\ 5 \end{bmatrix}.$ Then $b_1, b_2, b_3’$ are orthogonal vectors.

Now we normalize them so that the lengths of them become $1$.
We divide each vector with its length and obtain a orthonormal basis for $E_3$
$\left \{\,\frac{1}{5}\begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \frac{1}{30}\begin{bmatrix} -1 \\ 2 \\ 0 \\ 5 \end{bmatrix} \,\right\}.$

### (b) Find $A\mathbf{v}$

We express $\mathbf{v}$ as a linear combination of orthogonal basis $\{\mathbf{v}_1, b_1 b_2, b_3′ \}$ of $C^4=E_1\oplus E_2$.
Let
$\mathbf{v}=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}=c_1\begin{bmatrix} -1 \\ 2 \\ 0 \\ -1 \end{bmatrix}+c_2\begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix}+c_3\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}+c_4\begin{bmatrix} -1 \\ 2 \\ 0 \\ 5 \end{bmatrix}.$ (I scaled $b_3’$ to $5b_3’$. This does not change the orthogonality.)
We solve this for $c_i$. (There are several ways to do this.)
We take the inner product of this linear combination with $\mathbf{v}_1$ and obtain
$-1=6c_1$, thus $c_1=-1/6$.

Similarly taking inner product with other basis vectors, we obtain $c_2=2/5$, $c_3=0$, and $c_4=-1/30$.
Then we compute
\begin{align*}
A\mathbf{v} &= A\left( \frac{-1}{6}\begin{bmatrix}
-1 \\
2 \\
0 \\
-1
\end{bmatrix}+\frac{2}{5}\begin{bmatrix}
2 \\
1 \\
0 \\
0
\end{bmatrix}+\frac{-1}{30}\begin{bmatrix}
-1 \\
2 \\
0 \\
5
\end{bmatrix} \right)
\6pt] &= \frac{-1}{6}A\begin{bmatrix} -1 \\ 2 \\ 0 \\ -1 \end{bmatrix}+\frac{2}{5}A\begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix}+\frac{-1}{30}A\begin{bmatrix} -1 \\ 2 \\ 0 \\ 5 \end{bmatrix} \\[6pt] &= \frac{-1}{6}\begin{bmatrix} -1 \\ 2 \\ 0 \\ -1 \end{bmatrix}+\frac{4}{5}\begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix}+\frac{-2}{30}\begin{bmatrix} -1 \\ 2 \\ 0 \\ 5 \end{bmatrix} \\[6pt] &=\frac{1}{6}\begin{bmatrix} 11 \\ 2 \\ 0 \\ -1 \end{bmatrix}. \end{align*} Therefore the answer is \[A\mathbf{v}=\frac{1}{6}\begin{bmatrix} 11 \\ 2 \\ 0 \\ -1 \end{bmatrix}.

## Comment.

At first glance, there seems insufficient amount of information to answer the question, but everything we need to solve this problem was given.
This is one of the final exam problems of linear algebra course taught at the University of Tokyo.
I like this problem because the statement is concise yet it requires some essential knowledge/skills of linear algebra.

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