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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!

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  • $\begingroup$ The eigenvectors of $A$ are the eigenvectors of $Q$, because $Qv=\lambda v\implies (3I-4Q)v=(3-4\lambda) v$. The information you are given clearly suffices to get the eigenvalues of $Q$. Does it suffice to find the eigenvectors of $Q$? $\endgroup$ – Wouter Jul 10 at 14:42
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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?

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  • $\begingroup$ To be honest with you, I don't really see it. I already knew what you said, but I don't know how to find the eigenvectors of A knowing its eigenvalues (which I already determined) and knowing the eigenvectors of Q and its relationship with A (A = 3I - 4Q, should I substitute something in here?). Thank you. $\endgroup$ – Gatsby Jul 10 at 15:27
  • $\begingroup$ Since$$Q.\begin{bmatrix}1\\2\\0\\0\\-1\end{bmatrix}=\begin{bmatrix}-3\\-6\\0\\0\\6\end{bmatrix}=(−3)×\begin{bmatrix}1\\2\\0\\0\\-1\end{bmatrix},$$the vector $\begin{bmatrix}1\\2\\0\\0\\-1\end{bmatrix}T$ is an eigenvector of $Q$ with eigenvalue $−3$, by the definition of eigenvector. And, since $A=3\operatorname{Id}-4Q$,$$A.\begin{bmatrix}1\\2\\0\\0\\-1\end{bmatrix}=3\times\begin{bmatrix}1\\2\\0\\0\\-1\end{bmatrix}-12\times\begin{bmatrix}1\\2\\0\\0\\-1\end{bmatrix}^T=-9\times\begin{bmatrix}1\\2\\0\\0\\-1\end{bmatrix}.$$ $\endgroup$ – José Carlos Santos Jul 10 at 18:10
  • $\begingroup$ I was clearly lacking some theoretical knowledge to be able to reach that conclusion. Obrigado :) $\endgroup$ – Gatsby Jul 11 at 13:20

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