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?

  • $\begingroup$ What you say are the eigenvalues of the matrix is, in fact, the Jordan Normal Form of it, which in this case is the diagonal matrix with the eigenvalues in the main diagonal. Also, you talk of $\,S\,$: what is that? Then you talk of a matrix obtained by elimination...elimination of what (hopefully not of whom!)? Try to be more accurate and explicative in your question. $\endgroup$ – DonAntonio Dec 18 '12 at 23:29
  • $\begingroup$ 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. $\endgroup$ – Goyatuzo Dec 18 '12 at 23:53

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 $(1+i,1)^t$.

  • $\begingroup$ I think this is the part that is confusing me. I am not good at finding patterns in the complex space as I am with real numbers. Where did you get $(1+i, 1)^t$ from, and how does the second row send it to 0? $\endgroup$ – Goyatuzo Dec 18 '12 at 23:56
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    $\begingroup$ @Goyatuzo I asked myself "what do I need to multiply $3$ by in order to cancel $-3-3i$?" The answer was $1+i$. $\endgroup$ – Jonathan Dec 19 '12 at 0:00
  • $\begingroup$ Was that a typo for $3-3i$ and not $-3-3i$? If so, then I think I see the pattern. $\endgroup$ – Goyatuzo Dec 19 '12 at 0:02
  • $\begingroup$ Nope, not a typo. I mean that $3(1+i)+(-3-3i)(1)=0$. $\endgroup$ – Jonathan Dec 19 '12 at 0:14
  • $\begingroup$ Yup, I see my mistake. Thank you very much. $\endgroup$ – Goyatuzo Dec 19 '12 at 0:17

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 $(0, \ 0)$ .

  • $\begingroup$ I am unable to see the content after "and not", but should I use the conjugate in this case? $\endgroup$ – Goyatuzo Dec 19 '12 at 0:00

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