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I need a little help with this question:

A is this self-adjoint matrix $$\pmatrix{ 1 & 0 & i \\ 0 & 2 & 0\\ −i & 0 & 1}$$

Compute the eigenvalues and a set of orthonormal eigenvectors of A.

I know how to compute the eigenvalues but my eigenvectors seem to be different. They normalise them for some reason

Any help would be appreciated, thanks

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    $\begingroup$ Wouldn't that be because they want ortho'normal' eigenvectors rather than just orthogonal eigenvectors? $\endgroup$ – jgon May 11 '15 at 19:59
  • $\begingroup$ oh i see. but for example (for the first eigenvalue of 0): any vector of the form (x,0,ix) is an eigenvector. But then they choose 1/root2 (1,0,i). how come? if you find the unit vector of (1,0,i): root of 1^2 + 0^2 + i^2 = 0 $\endgroup$ – jdhokia May 11 '15 at 20:04
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    $\begingroup$ $||(1,0,i)|| = \sqrt{(1,0,i)\cdot\overline{(1,0,i)}} = \sqrt{(1,0,i)\cdot(1,0,-i)} = \sqrt{2}$ $\endgroup$ – uranix May 11 '15 at 20:14
  • $\begingroup$ argh thanks :), i was using the dot product $\endgroup$ – jdhokia May 11 '15 at 21:00
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To find the eigenvalues you have to solve the equation: $$\det\pmatrix{ 1-\lambda & 0 & i \\ 0 & 2-\lambda & 0\\ −i & 0 & 1-\lambda}=0$$ and developing the determinant you find: $\lambda(2-\lambda)(\lambda-2)=0$, so the eigenvalues are: $\lambda_1=0$,$\lambda_2=2$ and $\lambda_2=2$.

The eigenspace of $\lambda_1$ is the subspace of vectors that verify the equation: $$ \pmatrix{ 1-\lambda_1 & 0 & i \\ 0 & 2-\lambda_1 & 0\\ −i & 0 & 1-\lambda_1}\pmatrix{x\\y\\z}=\pmatrix{0\\0\\0} $$

and thay have the form $$ \pmatrix{-iz\\0\\z} $$ so you can chose : $$v_1= \pmatrix{-i\\0\\1} $$ In the same way, the eigenspace of $\lambda_2=\lambda_3$ is: $$ \pmatrix{iz\\y\\z} $$ and you can chose the two orthogonal vectors:

$$v_2= \pmatrix{i\\0\\1} \qquad v_3= \pmatrix{0\\1\\0} $$

Now you can normalize these vectors with: $u_i=\dfrac{v_i}{||v_i||}$ (where the norm is defined as $||v||=\sqrt{\langle v,\bar v\rangle}$ in a complex vector space). So you find:

$$ ||v_1||=\sqrt{2}\qquad ||v_2||=\sqrt{2}\qquad ||v_3||=\sqrt{1} $$

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