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$f\begin{bmatrix}x\\\\y\\\\z\end{bmatrix}=\begin{bmatrix}-5x-y+3z\\\\-18x-3y+9z\\\\-16x-3y+9z\end{bmatrix}$

Here is what I have so far

Let B={$\begin{bmatrix}1\\\\0\\\\0\end{bmatrix},\begin{bmatrix}0\\\\1\\\\0\end{bmatrix},\begin{bmatrix}0\\\\0\\\\1\end{bmatrix}$}

$f\begin{bmatrix}1\\\\0\\\\0\end{bmatrix}=\begin{bmatrix}-5\\\\-18\\\\-16\end{bmatrix}$ $f\begin{bmatrix}0\\\\1\\\\0\end{bmatrix}=\begin{bmatrix}-1\\\\-3\\\\-3\end{bmatrix}$ $f\begin{bmatrix}0\\\\0\\\\1\end{bmatrix}=\begin{bmatrix}3\\\\9\\\\9\end{bmatrix}$

Therefore $_BM_B(f)=\begin{bmatrix}-5&-1&3\\\\-18&-3&9\\\\-16&-3&9\end{bmatrix}$

so $charf(x)=det\begin{bmatrix}-5-x&-1&3\\\\-18&-3-x&9\\\\-16&-3&9-x\end{bmatrix}$

I'm not sure the next bit is right but when I calculated it I got

$x(3x-2)$ which would make $\lambda=0$ and $\lambda=\frac{2}{3}$

so when trying to find $E_0$ I got $\begin{bmatrix}0\\\\3z\\\\z\end{bmatrix}$ but I'm not sure what to do for $\lambda=\frac{2}{3}$ if it is correct

  • 3
    $x(3x-2)$ can't be right, the characteristic polynominal of a $3\times 3$ matrix is always a polynominal of degree 3.2012-12-04

1 Answers 1

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Your characteristic polynomial is $-x^2(x-1)$ so you obtain $0$ and $1$ as eigenvalues. The eigenspace of $0$ has merely dimension $1$ ($(0,3,1)^T$ is the only eigenvector with eigenvalue $0$), so your matrix is not diagonalizable.