# Finding eigenvalues

Find all eigenvalues and eigenvectors:

a.) $\pmatrix{i&1\\0&-1+i}$

b.) $\pmatrix{\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta}$

For a I got: $$\operatorname{det} \pmatrix{i-\lambda&1\\0&-1+i-\lambda}= \lambda^{2} - 2\lambda i + \lambda - i - 1$$

For b I got: $$\operatorname{det} \pmatrix{\cos\theta - \lambda & -\sin\theta \\ \sin\theta & \cos\theta - \lambda}= \cos^2\theta + \sin^2\theta + \lambda^2 -2\lambda \cos\theta = \lambda^2 -2\lambda \cos\theta +1$$

But how can I find the corresponding eigenvalues for a and b?

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You must solve $\det(A - \lambda I) = 0$. –  Pragabhava Nov 20 '12 at 21:51
@Pragabhava yes i know, I did that on my attempt above. But how can I solve the characteristic equations I got so that i can find the eigenvalues. –  Q.matin Nov 20 '12 at 21:54
You forgot to write \operatorname{det} in front of the matrices. Also, the tag (differential-equations) is misleading, maybe (linear-algebra) is better. –  AD. Nov 20 '12 at 21:54
@AD. what do you mean by that? –  Q.matin Nov 20 '12 at 21:55
I can edit it... –  AD. Nov 20 '12 at 21:56

For $a$ you can note that the matrix in case is upper triangular, or use the fact the the quadratic formula is also valid over $\mathbb{C}$.

For $b$ the last equality you have is not true, how did the $\cos(\theta)$ coefficient of $\lambda$ disappeared ? you should apply the quadratic formula in this case too and use a simple trigonometric identity.

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Thanks, I know for a that it is upper triangular so then the eigenvalues are i and -1 + i, but I want to know how to solve the characteristic polynomial I have above, just in case I might need to know how to do it in the future. And for b I edited it, i forgot to put the cos. And can you show me what trig identites I should use in order to find the eigenvalues I want –  Q.matin Nov 20 '12 at 21:59
I guess now that $a$ is OK ? just the quadratic formula, try it. For $b$ - what to you get under the square root in the quadratic formula ? can you say now what trig identity to use ? (I promise it is one of the more known ones) –  Belgi Nov 20 '12 at 22:01
Actually you dont have to show me part a . I feel real dumb now for asking it. But can you still show me how to do part b? –  Q.matin Nov 20 '12 at 22:01
This a second degree polynomial over $\mathbb{R}$, you need to find it roots...where are you stuck ? –  Belgi Nov 20 '12 at 22:02
I got for b $\sqrt{(\lambda^2 cos^2\theta - 1)}$ so the trig I should use to replace it is $sin^2\theta$ ? –  Q.matin Nov 20 '12 at 22:05

You can find eigen values by putting $\det(A-\lambda E)=0$.

If you want to find corresponding eigenvectors, too, try solving this equation:

$Av=\lambda v$, where v is an eigenvector for $\lambda$ in this equation. In other termss: $Av_i=\lambda_i v_i$

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