$$A=\begin{bmatrix} 1&-2&-8\\ 0&-1&0\\ 0&0&-1 \end{bmatrix}$$ $$P=\begin{bmatrix} 1&-4&1\\ 1&0&0\\ 0&1&0 \end{bmatrix}$$
Confirm that P diagonalizes A. I want to do this by first finding the eigen values and vectors of A then show how these match up with P's columns.
because its an upper triangular $det(A- \lambda I = (1-\lambda)(-1-\lambda)(-1-\lambda)=0$ easily so eigenvalues are 1 and -1 whose multiplicity is 2. Which means -1 must have two free valuables meaning a geometric multiplicity of 2 to be diagonalizable.
So i row reduce $A-I$ and get $$\begin{bmatrix} 0&-2&-8\\ 0&-2&0\\ 0&0&-2 \end{bmatrix}$$ which if if multiplied $R_2$ and $R_3$ by $-\frac{1}{2}$ to get two leading 1s $$\begin{bmatrix} 0&-2&-8\\ 0&1&0\\ 0&0&1 \end{bmatrix}$$ then reducing $R_1$ $$\begin{bmatrix} 0&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}$$ I get that $X_1 = t$ so for row 2 $$t+X_2 = 0$$ $$x_2 = -t $$ and like wise $$X_3 = t$$
Which is a vector $t\begin{bmatrix} 1\\ -1\\ -1 \end{bmatrix}$ which doesn't correspond with any column of P even if i change t around for a different value.
Also if i don't multiply $R_{1,2}$ by $-\frac{1}{2}$ and reduce by only multiples of row 2 and row 3 i ge that $$\begin{bmatrix} 0&-0&0\\ 0&-2&0\\ 0&0&-2 \end{bmatrix}$$ in which $X_1 = t \quad X_{2,3}=-\frac{1}{2}t$
$t\begin{bmatrix} 1\\ -\frac{1}{2}\\ -\frac{1}{2} \end{bmatrix}$ which almost make sense weren't for row 1 being a half.
The matrix
$$\begin{bmatrix} 0&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}$$
Gives $x_1$ as a free variable, and the other two lines give $x_2 = 0$ and $x_3 = 0$. The eigenspace is therefore:
$$\begin{bmatrix} x_1 \\ 0 \\ 0 \end{bmatrix} = x_1\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$$