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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 $\det(A - \lambda I) = 0$.2012-11-20
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    @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.2012-11-20
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    You forgot to write \operatorname{det} in front of the matrices. Also, the tag (differential-equations) is misleading, maybe (linear-algebra) is better.2012-11-20
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    @AD. what do you mean by that?2012-11-20
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    I can edit it...2012-11-20
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    @Q.matin you wrote something like matrix=scalar, you meant |matrix|=scalar2012-11-20
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    @Belgi That was what I meant, don't bring new notions now.. :))2012-11-20
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    @Q.matin **Hint** The first one: $$\det(\cdot) = (i-\lambda)(-1+i-\lambda) = 0 \, \Rightarrow \, \lambda = \begin{cases} \quad \, i\\-1+i\end{cases}$$ For the second, $\lambda$ will depend on the value of $\theta$, i.e. $\lambda = \lambda(\theta)$.2012-11-20
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    @Pragabhava wow i feel real dumb! For part a. Idk what I was thinking. Thanks !2012-11-20
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    @AD. you brought new notation :P2012-11-20
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    @Belgi No I took it from Pragabhava's comment ;)2012-11-20

3 Answers 3

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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 want2012-11-20
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    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)2012-11-20
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    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?2012-11-20
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    This a second degree polynomial over $\mathbb{R}$, you need to find it roots...where are you stuck ?2012-11-20
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    I got for b $\sqrt{(\lambda^2 cos^2\theta - 1)}$ so the trig I should use to replace it is $sin^2\theta$ ?2012-11-20
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    @Q.matin Note that $\lambda$ is the variable, you are solving for it (its like "x"). denote the polynomial as $ax^2+bx+c$ then $a=1$, $b=-2\cos(\theta)$ ,$c=1$2012-11-20
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    ah, that is where i got it wrong. Thanks!!2012-11-20
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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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Basic tools

For $2 \times 2$ matrices the characteristic polynomial is: $$ p(\lambda) = \lambda^{2} - \lambda\, \text{tr }\mathbf{A} + \det \mathbf{A} $$

The roots of this function are the eigenvalues, $\lambda_{k}$, k=1,2$.

The eigenvectors solve the eigenvalue equation $$ \mathbf{A} u_{k} = \lambda_{k} u_{k} $$


Case 1

$$ \mathbf{A} = % \left( \begin{array}{cc} i & 1 \\ 0 & -1+i \\ \end{array} \right) $$ The trace and determinant are $$ \text{tr } \mathbf{A} = -1 + 2 i, \qquad \det \mathbf{A} = -1 - i $$ The characteristic polynomial is $$ p(\lambda) = \lambda ^2+(1-2 i) \lambda +(-1-i) $$ The roots of this polynomial are the eigenvalues: $$ \lambda \left( \mathbf{A} \right) = \left\{ i, -1 + i \right\} $$

First eigenvector: $$ \begin{align} \left( \mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) u_{1} &= \mathbf{0} \\[3pt] % \left( \begin{array}{cc} i & 1 \\ 0 & -1+i \\ \end{array} \right) - \left( -1 + i \right) % \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) \left( \begin{array}{cc} u_{x} \\ u_{y} \\ \end{array} \right) &= \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) \\ % \left( \begin{array}{cc} u_{x} + u_{y}\\ 0 \\ \end{array} \right) &= \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) % \end{align} $$ The result is $$ u_{1} = \left( \begin{array}{cc} u_{x} \\ u_{y} \\ \end{array} \right) = \alpha \left( \begin{array}{r} 1 \\ -1 \\ \end{array} \right), \quad \alpha \in \mathbb{C} $$

Second eigenvector: $$ \begin{align} \left( \mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) u_{2} &= \mathbf{0} \\[3pt] % \left( \begin{array}{cr} 0 & 1 \\ 0 & -1 \\ \end{array} \right) % \left( \begin{array}{cc} u_{x} \\ u_{y} \\ \end{array} \right) &= \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) \\ % \end{align} $$ The result is $$ u_{2} = \alpha \left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right), \quad \alpha \in \mathbb{C} $$


Case 2

$$ \mathbf{A} = % \left( \begin{array}{cr} \cos (\theta ) & -\sin (\theta ) \\ \sin (\theta ) & \cos (\theta ) \\ \end{array} \right) $$ The trace and determinant are $$ \text{tr } \mathbf{A} = 2 \cos \theta, \qquad \det \mathbf{A} = \cos^{2} \theta + \sin^{\theta} = 1 $$ The characteristic polynomial is $$ p(\lambda) = \lambda ^2+(1-2 i) \lambda +(-1-i) $$ The roots of this polynomial are the eigenvalues: $$ \lambda \left( \mathbf{A} \right) = \left\{ \cos \theta -i \sin \theta ,\cos \theta +i \sin \theta \right\} $$

First eigenvector: $$ \begin{align} \left( \mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) u_{1} &= \mathbf{0} \\[3pt] % \left( \begin{array}{cr} i \sin \theta & -\sin \theta \\ \sin \theta & i \sin \theta \\ \end{array} \right) % \left( \begin{array}{cc} u_{x} \\ u_{y} \\ \end{array} \right) &= \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) \\ % \left( \begin{array}{cc} -u_{y} \sin \theta +i u_{x} \sin \theta \\ u_{x} \sin \theta +i u_{y} \sin \theta \end{array} \right) &= \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) % \end{align} $$ The result is $$ u_{1} = \left( \begin{array}{cc} u_{x} \\ u_{y} \\ \end{array} \right) = \alpha \left( \begin{array}{r} i \\ -1 \\ \end{array} \right), \quad \alpha \in \mathbb{C} $$

Second eigenvector: $$ \begin{align} \left( \mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) u_{2} &= \mathbf{0} \\[3pt] % \left( \begin{array}{rr} -i \sin \theta & -\sin \theta \\ \sin \theta & -i \sin \theta \\ \end{array} \right) % \left( \begin{array}{c} u_{x} \\ u_{y} \\ \end{array} \right) &= \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) \\ % \end{align} $$ The result is $$ u_{2} = \alpha \left( \begin{array}{c} i \\ 1 \\ \end{array} \right), \quad \alpha \in \mathbb{C} $$