I have been managing to do problems similar to this one that involves stretches and reflections okay, but I'm struggling with this one that involves projections.
I have a problem as follows:
$T$ projects a vector onto the line $2x+3y = 0$
My attempt which I'm fairly certain is incorrect:
Algebra brings us to $y = -\frac{2}{3}x$
Start off with $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $u$ is $\begin{bmatrix} 3 \\ -2 \end{bmatrix}$
projection gives $\frac{3}{13}$*$\begin{bmatrix} 3 \\ -2 \end{bmatrix}$
so I just assumed $\lambda = \frac{3}{13}$. My answer is way off so I just stopped here.
The answer is supposed to be $\lambda = 1$, $\begin{bmatrix} 3 \\ -2 \end{bmatrix}$ and $\lambda = 0$, $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$
Could I get pointers toward the right direction?
$T$ transform a $2$ dimensional space onto a line, that is spanned by the vector $v=\frac1{\sqrt{3^2+2^2}}\begin{bmatrix} 3 \\ -2 \end{bmatrix}$.
Given a vector $x$, its projection is $(x^Tv)v=(vv^T)x$
Let $w$ be a unit vector that is perpendicular to $v$.
We can write $vv^T= \begin{bmatrix} v & w\end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix} v & w\end{bmatrix}^T= 1vv^T+0ww^T$ which is the eigenvalue decomposition.
Hence the eigector is $v$ with eigenvalue $1$ and $w$ with eigenvalue $0$.
More geometrical interpretation: Given a vector, decompose it into a direction that is along $v$ and a direction that is along $w$. For the direction along $v$, we do not change its length and we discard the other direction.