How do you diagonalize this matrix and find P and D such that A = PDP^-1?
1 1 4 0 -4 0 -5 -1 -8
I3 = 3x3 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?
$\endgroup$ 31 Answer
$\begingroup$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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