Using wolfram and symbolab, diagonalization of $\begin{pmatrix}8&12&-18\\ 4&18&-20\\ 4&13&-15\end{pmatrix} \rightarrow A=PDP^{-1}\rightarrow \begin{pmatrix}8&12&-18\\ 4&18&-20\\ 4&13&-15\end{pmatrix}=\begin{pmatrix}1&3&2\\ 1&2&4\\ 1&2&3\end{pmatrix}\begin{pmatrix}2&0&0\\ 0&4&0\\ 0&0&5\end{pmatrix}\begin{pmatrix}-2&-5&8\\ 1&1&-2\\ 0&1&-1\end{pmatrix}$
I'm struggling to derive the correct $P_2, P_3$, of $P= ({P_1, P_2 , P_3})$. I know the answer, but how do we get there?
Working for when $\lambda=4$ is:
$RREF \begin{pmatrix}8-4&12&-18\\ \:4&18-4&-20\\ \:\:4&13&-15-4\end{pmatrix}\rightarrow \begin{pmatrix}1&0&-\frac{3}{2}\\ 0&1&-1\\ 0&0&0\end{pmatrix}$
$\begin{pmatrix}1&0&-\frac{3}{2}\\ 0&1&-1\\ 0&0&0\end{pmatrix} \vec{v}=\vec{0}$ :
$v_1-\frac{3}{2}v_3=0 \rightarrow v_3=\frac{2}{3}v_1$
$v_2 - v_3=0\rightarrow v_3=v_2$
Thus, clearly: $\vec{v}= v_3\begin{pmatrix}\frac{2}{3}\\ 1\\ 1\end{pmatrix}: v_3$ is any constant $\in \mathbb{R}$
Qn 1: Why is the answer $\begin{pmatrix}3\\\:2\\ \:2\end{pmatrix}$ instead?
Working for when $\lambda=5$ is:
$RREF \begin{pmatrix}8-5&12&-18\\ \:4&18-5&-20\\ \:\:4&13&-15-5\end{pmatrix}\rightarrow \begin{pmatrix}1&0&-\frac{2}{3}\\ 0&1&-\frac{4}{3}\\ 0&0&0\end{pmatrix}$
$\begin{pmatrix}1&0&-\frac{2}{3}\\ 0&1&-\frac{4}{3}\\ 0&0&0\end{pmatrix} \vec{v}=\vec{0}$ :
$v_1-\frac{2}{3}v_3=0 \rightarrow v_3=\frac{3}{2}v_1$
$v_2 -\frac{4}{3}v_3=0\rightarrow v_3=\frac{3}{4}v_2$
Thus, clearly: $\vec{v}= v_3\begin{pmatrix}\frac{3}{2}\\ \frac{4}{3}\\ 1\end{pmatrix}: v_3$ is any constant $\in \mathbb{R}$
Even if I scaled it up by 3, $\vec{v}= v_3\cdot 3\begin{pmatrix}\frac{3}{2}\\ \frac{4}{3}\\ 1\end{pmatrix}=\begin{pmatrix}\frac{9}{2}\\ 4\\ 3\end{pmatrix}$
Qn 2: Why is the answer $\begin{pmatrix}2\\\:4\\ \:3\end{pmatrix}$ instead?
Working for when $\lambda=2$ is:
RREF:
$\begin{pmatrix} 1&0&-1\\ 0&1&-1\\ 0&0&0 \end{pmatrix} \vec{v}=\vec{0}$ :
$v_1- v_3=0 \rightarrow v_1= v_3$
$v_2 - v_3=0\rightarrow v_2=v_3$
Thus, clearly: $\vec{v}= v_3 \begin{pmatrix} 1 \\ 1\\ 1\end{pmatrix}: v_3$ is any constant $\in \mathbb{R}$
Taking $v_3 = 1$, we get $$P_1 = \begin{pmatrix}1\\ 1\\ 1\end{pmatrix}$$
Working for when $\lambda=4$ is:
RREF:
$\begin{pmatrix}1&0&-\frac{3}{2}\\ 0&1&-1\\ 0&0&0\end{pmatrix} \vec{v}=\vec{0}$ :
$v_1-\frac{3}{2}v_3=0 \rightarrow v_1=\frac{3}{2}v_3$
$v_2 - v_3=0\rightarrow v_2=v_3$
Thus, clearly: $\vec{v}= v_3\begin{pmatrix}\frac{3}{2}\\ 1\\ 1\end{pmatrix}: v_3$ is any constant $\in \mathbb{R}$
Taking $v_3 = 2$, we get $$P_2 = \begin{pmatrix}3\\ 2\\ 2\end{pmatrix}$$
orking for when $\lambda=5$ is:
RREF:
$\begin{pmatrix}1&0&-\frac{2}{3}\\ 0&1&-\frac{4}{3}\\ 0&0&0\end{pmatrix} \vec{v}=\vec{0}$ :
$v_1-\frac{2}{3}v_3=0 \rightarrow v_1=\frac{2}{3} v_3$
$v_2 - \frac{4}{3} v_3=0\rightarrow v_2= \frac{4}{3} v_3$
Thus, clearly: $\vec{v}= v_3 \begin{pmatrix} \frac{2}{3}\\ \frac{4}{3} \\ 1\end{pmatrix}: v_3$ is any constant $\in \mathbb{R}$
Taking $v_3 = 3$, we get $$P_3 = \begin{pmatrix}2 \\ 4\\ 3\end{pmatrix}$$
Define $$ P = \left[ \matrix{P_1 & P_2 & P_3 \cr} \right] = \left[ \matrix{ 1 & 3 & 2 \cr 1 & 2 & 4 \cr 1 & 2 & 3 \cr} \right] $$
It is easy to verify that $$ P^{-1} A P = D = \left[ \matrix{ 2 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 5 \cr} \right] $$