I want to find the following matrix problem related to eigenvalue and eigenvectors

Question

$$ A=\begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\ \end{pmatrix} $$

$A$ has eigenvalue $ \lambda $ of multiplicative 2 and eigenvalue $ \mu $ of multiplicative $1$.

Let scalar product on $\Bbb {R}^3 $ be defined by $(x,y)=x^{t}y$. Let $w$ be the eigenvector of length $1$ for the eigenvalue $ \mu $ , and define a linear map $f : \Bbb {R}^3 \to \Bbb {R}^3 $ by

$$f(x)=x-(x, w)w $$

Then, how can we find the square matrix $B$ of size $3$ s.t. $f(x)=Bx $.

And also how to show that the image of $f$ coincides with the eigenspace of eigenvalue $ \lambda $?

Answer:

I got $ \lambda =1$ and $ \mu = 4$ .

Corresponding eigenvectors are $$v_1=\begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}, v_2=\begin{pmatrix} -1 \\ 0 \\ 1 \\ \end{pmatrix} , v_3=\begin{pmatrix} -1 \\ 1 \\ 0 \\ \end{pmatrix} $$

Therefore,

$$w=\frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}$$

$$f(x)=x-(x, w)w = x-x^t w^2$$

$$Bx=x-x^t w^2$$

How can we solve the above problem?

Answer 1

First of all, a warning: I wouldn't write $(x,w)w$ as $x^Tw^2$ because matrix multiplication is not associative. I would write it as $(x^Tw)w$ to make it clear.

Now, for your question. Notice that:

1. $v_2$ and $v_3$ are orthogonal to each $v_1$.
2. Any eigenvector $w$ which is an eigenvector of $\mu$ can be written as $\alpha v_1 + \beta v_2$.

This means that there are infinitely many possibilities for $w$, and you cannot find a general matrix $B$ as a single matrix, but rather as a function of $\alpha$ and $\beta$. Nevertheless, you can easily show that the image of $f$ is the eigenvalue of $\lambda$ by showing that:

1. The vectors $v_1,v_2,v_3$ form an orthogonal basis for $\mathbb R^3$
2. For any $x$, $f(x)$ is orthogonal to $v_2$ and $v_3$.

If you show $1$ and $2$, you can then use the fact that for an orthonormal basis (where vectors are all length $1$) $w_1,w_2,w_3$, you can write any $x$ as

$$x = (x, w_1)w_1 + (x, w_2)w_2 + (x,w_3)w_3$$