I´m studying for a Linear Algebra exam and I´m having trouble with solving an exercice.
The questions are:
a) Being that A = 3I – 4Q. Determine the eigenvalues of the matrix A.
b) Again being A = 3I – 4Q. Determine the eigenvectors of the matrix A.
We are given the next information:
det(Q – λI) = (-3-λ) (-2-λ) (-2-λ) (3-λ) (4-λ)
$$ Q \begin{bmatrix} 1\\ 2\\ 0\\ 0\\ -1\\ \end{bmatrix} = \begin{bmatrix} -3\\ -6\\ 0\\ 0\\ 3\\ \end{bmatrix} $$
$$ Q \begin{bmatrix} 0\\ 0\\ -4\\ 3\\ 0\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ -12\\ 9\\ 0\\ \end{bmatrix} $$
$$ VQ(-2) = \begin{bmatrix} x\\ 2x\\ 0\\ 0\\ 5x\\ \end{bmatrix} + \begin{bmatrix} 2y\\ -y\\ 0\\ 0\\ 0\\ \end{bmatrix} $$
$$ VQ(4) = \begin{bmatrix} 0\\ 0\\ 3w\\ 4w\\ 0\\ \end{bmatrix} $$
I don't have problems with question a). But I don't have any idea on how to do b). After doing a), we know the eigenvalues but I still don't know the matrix A. How can I find the eigenvectors?
Thanks in advance for any help!
Best Answer
Your first condition states that $\begin{bmatrix}1&2&0&0&-1\end{bmatrix}^T$ is an eigenvector of $Q$ with eigenvalue $-3$. Your second condition states that $\begin{bmatrix}0&0&-4&3&0\end{bmatrix}^T$ is an eigenvector of $Q$ with eigenvalue $3$. And so on. Can you take it from here?