Let $v \in \mathbb{R}^3$ be a nonzero vector .
Define a linear transformation
$$T: \mathbb{R}^3 \to \mathbb{R}^3 \ \ \ \text{by} \ \ \ T(x)= x – 2 \frac{x.v}{v.v} v$$
where $x.y $ denotes the standard inner product in $\mathbb{R}^3$
Find the eigenvalues of $T ?$
My attempt : I take ordered basis $e_1=(1,0,0) , e_2=(0,1,0)$ and $e_3=(0,0,1)$
since $v \in \mathbb{R}^3$ is a nonzero vector so i take $v=(1,0,0)$
Now $T(e_1)=e_1-2\frac{(e_1.v)}{||v||}v=(1,0,0)-2(1,0,0)=(-1,0,0)$
$T(e_2)=(0,1,0)-2=(0,1,0)$
$T(e_3)=(0,0,1)$
Now matrix of $T$ is $\begin{bmatrix} -1&0&0\\0&1&0 \\0&0&1 \end{bmatrix}$
Therefore the eigenvalue of $T$ are $-1$ and $1$
Is it true ?
Best Answer
Your eigenvalues are correct, but your reasoning is still wrong. You can't just define $v$, it is supposed to be arbitrary
ust find the eigenspaces: One is $\mathrm{span}({v})$: Just plug in $v$: $$ T(v) = v - 2v = -v. $$ So $v$ has eigenvalue $-1$.
The other one is $\mathrm{span}(v)^\perp$: Let $x \in \mathbb{R}^3$ such that $x.v = 0$. So $$ T(x) = x. $$ So all vectors in $\mathrm{span}(v)$ have eigenvalue 1. $\mathrm{span}(v)^\perp$ has dimension 2.
Operators like yours are called Householder transformations.