Question

Matrix Diagonalization (cont.)

Answer 1

I am honestly lost on this one @sillybilly123

Answer 2

V:

For \(\lambda = 1\), generate a few vectors on that \(\pi ~ : ~ 22x - 45 y + 17 z = 0 \) plane.

Eg:

- if x,y = 1,0 then \(\mathbf v_1 = <1, 0, - 22/17>\)

- if x,y = 0,1 then \(\mathbf v_2 = <0, 1, 45/17>\)

For \(\lambda = -2\) which lies along the line pointing \(\mathbf v_3 = <2,1,1>\) from O, that's its eigen-direction

So we can now [reverse(!)-]diagonalise:

\(A = S \Lambda S^{-1}\)

\(= \begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 1 \\
-22/17 & 45/17 & 1
\end{bmatrix} \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -2
\end{bmatrix} \begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 1 \\
-22/17 & 45/17 & 1
\end{bmatrix}^{-1}\)


\(= \begin{bmatrix}
-29/4 & 135/8 & -51/8 \\
-33/8 & 151/16 & -51/16 \\
-33/8 & 135/16 & -35/16
\end{bmatrix}\)

Next try out A on the eigenvectors to see that it works.

Lot easier starting here:

https://www.wolframalpha.com/input/?i=((1,0,2),(0,1,1),(-22%2F17,+45%2F17,1))+*+((1,0,0),(0,1,0),(0,0,-2))+*++inverse+((1,0,2),(0,1,1),(-22%2F17,+45%2F17,1))

Not yet convinced this is the most efficient way to go about this