[Math] How to diagonalize this matrix and find P and D such that A = PDP^-1

diagonalizationeigenvalues-eigenvectorslinear algebramatrices

question

 1   1   4
 0  -4   0
-5  -1  -8

I3 = 3×3 identity matrix

             λ   0   0
λI3  =       0   λ   0
             0   0   λ


    λ-1     -1      -4
=   0        λ+4     0
    5        1       λ+8

Rule of Sarrus to find determinant:

λ–1     -1      -4       λ–1    -1      
0       λ+4      0       0       λ+4
5       1        λ+8     5       1

Eigenvalues:
λ = -3, λ = -4

Eigenvectors:

for λ = -3:

-1
 0
 1

for λ = -4:

 4
 0
-5

The thing is, when I put this in any online matrix calculator, it comes up with different eigenvectors.

       -4   0   0
D   =   0  -4   0
        0   0  -3

I am not coming up with any of the multiple choice answers, but I know it has to be one of them. Any ideas?

Best Answer

If you solve $(A+4I)x=0$ by reducing the augmented matrix, you obtain $\begin{bmatrix}1&1/5&4/5&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$.

Letting $y=5s$ and $z=5t$ gives $x=-s-4t$, so

the vectors $\begin{bmatrix}-1\\5\\0\end{bmatrix}$ and $\begin{bmatrix}-4\\0\\5\end{bmatrix}$ are eigenvectors corresponding to $\lambda=-4$.

Therefore we could take these vectors as two columns of $P$; but since

the vectors $\begin{bmatrix} 1\\-9\\1\end{bmatrix}$ and $\begin{bmatrix}0\\-4\\1\end{bmatrix}$ are linear combinations of these vectors, we can use them as the columns corresponding to $\lambda=-4$ instead.

Therefore C) is correct.

A quicker way to work this problem is to find the eigenvalues $\lambda=-3$ and $\lambda=-4$,

conclude that C) is the only possible correct answer,

and then check to see if $Av_1=-4v_1$, $Av_2=-4v_2$, and $Av_3=-3v_3$ where

$P=\begin{bmatrix}v_1 &|&v_2&|&v_3\end{bmatrix}$.

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