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Question: Calculate the eigenvectors and eigenvalues of the following matrix $$\begin{pmatrix}3&-3\\0&-2\\ \end{pmatrix}$$

My attempt:

I have calculated the eigenvalues to be $\lambda = 3$ and $\lambda =-2$ and I have managed to get the eigen vector for $\lambda = 3$ to be \begin{pmatrix}1\\0\\ \end{pmatrix} However I get \begin{pmatrix}1\\\frac{5}{3}\\ \end{pmatrix}

but the correct answer seems to be \begin{pmatrix}0\\1\\ \end{pmatrix} Where am I going wrong?

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  • $\begingroup$ Are you sure that the original matrix is the correct one? $\endgroup$ – StackTD Apr 20 '16 at 15:43
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    $\begingroup$ I'm not so sure your calculation is correct. The characteristic polynomial is $\lambda^{2}-5\lambda-2$, and the corresponding eigenvalues are roots of this polynomial, $(5\pm\sqrt{33})/2$. $\endgroup$ – parsiad Apr 20 '16 at 15:47
  • $\begingroup$ I couldn't get your eigenvalues. $\endgroup$ – velut luna Apr 20 '16 at 15:48
  • $\begingroup$ I too am getting different eigenvalues. They should be -0.37228 and 5.37228 (calculated with Octave). I found the characteristic polynomial to be $\lambda^{2}-5\lambda-2$ $\endgroup$ – Fausto Arinos Barbuto Apr 20 '16 at 15:55
  • $\begingroup$ trace is the sum of e-vals, det of A is the product of e-vals. and $1+4 \neq -2+3$ $\endgroup$ – Tiger Blood Apr 20 '16 at 15:56
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The eigenvalues of a triangular matrix are simply the diagonal entries, so your eigenvalues $3$ and $-2$ are correct. As $$\begin{pmatrix} 3 & -3 \\ 0 & -2 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix} = 3 \begin{pmatrix} 1 \\ 0 \end{pmatrix},$$ the vector $\begin{pmatrix}1 \\ 0 \end{pmatrix}$ is an eigenvector corresponding to $\lambda = 3$.

As $$\begin{pmatrix} 3 & -3 \\ 0 & -2 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 5/3 \end{pmatrix} = \begin{pmatrix} -2 \\ -10/3 \end{pmatrix} = -2 \begin{pmatrix} 1 \\ 5/3 \end{pmatrix},$$ the vector $\begin{pmatrix}1 \\ 5/3\end{pmatrix}$ is an eigenvector corresponding to $\lambda = -2$.

So, your answers are correct. Note that $\begin{pmatrix}0 \\ 1\end{pmatrix}$ is not an eigenvector of this matrix, because $$\begin{pmatrix} 3 & -3 \\ 0 & -2 \\ \end{pmatrix} \begin{pmatrix}0 \\ 1\end{pmatrix} = \begin{pmatrix}-3 \\ -2\end{pmatrix} $$ which is not a scalar multiple of $\begin{pmatrix}0 \\ 1\end{pmatrix}$.

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