Determining Eigenvectors from a 0 valued matrix

Question

So I need to find the eigenvectors for the following matrix: $$ \begin{pmatrix} 3 & 0\\ -1 & -2 \end{pmatrix} $$

And I've gotten as far as determining the eigenvalues (6 and 1)but seem to fall short when trying t obtain the corresponding eigenvectors (apparently they are (5,1) and (0,1)), getting as far as:

$$ \begin{pmatrix} 0 & 0\\ -1 & -5 \end{pmatrix} $$ Any help would be greatly appreciated.

Answer 1

Hint

Actually, the eigenvalues are $3$ and $-2$ since $$|\lambda I-A|=(\lambda-3)(\lambda+2)=0$$

Answer 2

Recall that the sum of the eigenvalues is the trace of the matrix, i.e. the sum of its diagonal entries: $$ \lambda_1+\lambda_2 = \operatorname{Tr}(A) = 3 - 2 = 1, $$ and that the product of the eigenvalues is the determinant: $$ \lambda_1\lambda_2 = \begin{vmatrix}3&0\\-1&-2\end{vmatrix} = -6. $$ By inspection we see that $\lambda_1 = 3$ and $\lambda_2=-2$. Solving the eigenvector equation for $\lambda_1$ we have $$ (A-3I)x = 0 \iff \left(\begin{array}{cc|c}0&0&0\\-1&-5&0 \end{array}\right)\iff -x_1-5x_2=0\iff x_2 = -\frac15 x_1. $$ So $v_1 = \frac1{\sqrt{26}}(5,-1)$ is an eigenvector corresponding to $\lambda_1$. Similarly for $\lambda_2$, $$ (A+2I)x = 0 \iff \left(\begin{array}{cc|c}5&0&0\\-1&0&0 \end{array}\right)\iff -x_1-5x_2=0\iff x_1=0. $$ So $v_2=(0,1)$ is an eigenvector corresponding to $\lambda_2$. We see immediately that $v_1$ and $v_2$ are orthogonal, i.e. $\langle v_1,v_2\rangle = \frac1{\sqrt{26}}(5\cdot0 + 1\cdot-1) = 0$, and are of unit norm, and hence form an orthonormal basis of $\mathbb R^2$. Moreover, letting $$P = \begin{pmatrix}\frac5{\sqrt{26}}&0\\\frac{-1}{\sqrt{26}}&1\end{pmatrix},\quad D = \begin{pmatrix}3&0\\0&-2\end{pmatrix}, $$ we see that $A=PDP^{-1}$ and hence $A^n = PD^nP^{-1}$ for all nonnegative integers $n$. So for each $t\in\mathbb R$ we may compute the matrix exponential \begin{align} e^{At} &= \sum_{n=0}^\infty \frac{A^n}{n!}\\ &= \sum_{n=0}^\infty \frac{PDt^nP^{-1}}{n!}\\ &= \sum_{n=0}^\infty \frac1{n!}\begin{pmatrix}\frac5{\sqrt{26}}&0\\\frac{-1}{\sqrt{26}}&1\end{pmatrix}\begin{pmatrix}(3t)^n&0\\0&(-2)^n\end{pmatrix}\left( \begin{array}{cc} \frac{\sqrt{26}}{5} & 0 \\ \frac{1}{5} & 1 \\ \end{array} \right)\\ &= \sum_{n=0}^\infty \frac1{n!}\left( \begin{array}{cc} 3^n t^n & 0 \\ \frac{1}{5} 2^n (-t)^n-\frac{3^n t^n}{5} & 2^n (-t)^n \\ \end{array} \right)\\ &= \left( \begin{array}{cc} e^{3 t} & 0 \\ -\frac{1}{5} e^{-2 t} \left(-1+e^{5 t}\right) & e^{-2 t} \\ \end{array} \right). \end{align}

Answer 3

To begin with, the eigenvalues aren’t $6$ and $1$ as you wrote. The determinant of your matrix is negative, so at the very least they must have opposite signs. As it happens, the matrix is triangular, so its eigenvalues are simply the values along its main diagonal: $3$ and $-2$.

You somehow ended up examining $A-3I$, though, which is the correct matrix for determining eigenvectors with eigenvalue $3$, but doesn’t follow from your eigenvalues of $6$ and $1$. Anyway, you could go through the usual process of row-reduction to compute the null space of this matrix, but you can skip all of that and find an eigenvector by inspection. The null space of a matrix is the orthogonal complement of its row space, which is obviously spanned by $(-1,-5)$, so $(5,-1)^T$ is an eigenvector of $A$ with eigenvalue $3$.

You can proceed similarly for the eigenvalue $-2$, or simply notice that the second column of $A$ is equal to $-2(0,1)^T$, so $(0,1)^T$ is an eigenvector of $A$ as well.