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The eigenvectors of A (corresponding to the eigenvalue 2) = \begin{bmatrix} 1 & 2 \\ -1 & 4 \\ \end{bmatrix} can be solved by solving the matrix equation:

enter image description here

Here is my working: enter image description here

My answers were $4$ and $-1/2$ which were apparently wrong, though I am wondering if I made a typo when I put in $-1/2$ (I am unable to check further).

Xander Henderson
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coolkid3127
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  • Please include all the steps you took. Also what is the original matrix $A.$ – coffeemath May 19 '21 at 01:20
  • A is posted in the question after the =, so A is (1, 2, -1, 4). – coolkid3127 May 19 '21 at 01:20
  • Oh I didn't see that from the way you typed the question. But still it would be good if you include all your steps so someone can see where it went wrong, if it did. – coffeemath May 19 '21 at 01:24
  • Added working.. – coolkid3127 May 19 '21 at 01:32
  • I have closed your question for "lacking context". More specifically, you have included vital information about your question in images. Images are bandwidth intensive (a problem for some mobile users, particularly in parts of the world with poor internet infrastructure), less searchable, and inaccessible to people who access the site using screenreaders. Please read the MathJax tutorial and use the information there to format your question. – Xander Henderson May 21 '21 at 16:27

1 Answers1

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I don't know where the matrix $\begin{pmatrix} 4 & -2 \\ 1 & -1/2 \end{pmatrix}$ comes from.

The way I have seen it done is that $\begin{pmatrix} x \\ y \end{pmatrix}$ is a $\lambda$-eigenvector if

$\begin{pmatrix} 1-\lambda & 2 \\ -1 & 4-\lambda \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.

So when $\lambda = 2$ we want:

$\begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.

Of course $\begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -x+2y \\ -x+2y \end{pmatrix}$, so the vector works when $x=2y$.

So we want vectors $\begin{pmatrix} 2y \\ y \end{pmatrix}$ with $y\neq 0$.

Asinomás
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