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This is a relatively simple question and I've Google'd this topic but I can't seem to grasp the method. One site I visited helped be a bit, but it simply used substitution to solve the problem rather than elimination, so I feel as though it's very situational.

The question is the following:

The eigenvalues and eigenvectors of the matrix $\begin{bmatrix} 2 & -6 \\ 3 & -4 \end{bmatrix}$

They want me to diagonalize the matrix and find $S$ and $\Lambda$.

I have found the eigenvalues to be

$\Lambda = \begin{bmatrix} -1 + 3i & 0 \\ 0 & -1-3i \end{bmatrix} $

However, the method to find $S$, or the two eigenvectors, has me stuck. I've tried standard elimination. Using $-1 + 3i$, I got the following matrix after elimination

$\begin{bmatrix} 3+3i & -6 \\ 0 & 9-9i \end{bmatrix}$

I am not sure if I did the elimination incorrectly, but my process was to multiply the second row by $\frac{-3-3i}{3}$. Unfortunately, this isn't a singular matrix. Which step have I done wrong?

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    I'll be sure to check for that in the future. I've been studying for linear algebra today so I wasn't able to form a coherent question with adequate information. I apologize.2012-12-18

2 Answers 2

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First of all, what do you mean by your $\Lambda$? The eigenvalues are $-1\pm 3i$, as you seem to have discovered. To figure out the eigenvector corresponding to eigenvalue $\lambda$, letting $A$ be your original matrix, consider the matrix $A-\lambda I$. You will often (especially for $2\times 2$ matrices) be able to get the eigenvector by inspection.

With $\lambda=-1+3i$, for instance, $A-\lambda I=\begin{pmatrix}3-3i&-6\\3&-3-3i\end{pmatrix}.$ Looking at the second row, you see that it sends $(1+i,1)^t$ to zero. So does the first row (of course that's not an accident). So, the eigenvector for $\lambda=-1+3i$ is $(1+i,1)^t$.

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    Yup, I see my mistake. Thank you very much.2012-12-19
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It seems that you are trying to find the eigenvectors corresponding to the eigenvalue $-1-3i$ and not $-1+3i$. Your mistake is that after multiplying the second row by $\frac{-3-3i}{3}$ it becomes $(-3-3i, \ 6)$ and not $\dots$ Adding the first row it becomes $(0, \ 0)$ .

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    I am unable to see the content after "and not", but should I use the conjugate in this case?2012-12-19