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I am given the matrix A and asked to orthogonally diagonalize it.

$$A=\begin{pmatrix} 1 & -i \\ i & 1 \\ \end{pmatrix} $$

While doing this I got $\lambda = 0,2.$ Then I found the eigenvectors corresponding to the eigenvalues to be $$\begin{pmatrix} i \\ 1 \\ \end{pmatrix} $$ and $$\begin{pmatrix} -i \\ 1 \\ \end{pmatrix}, $$ respectively. While trying to divide each eigenvector by its norm I run into a problem, take $V_1$ for example $\|V_1\| = 0$ so I obviously can't divide by $ V_1$ by its norm to make it orthogonal. My question is, is $A$ even orthogonally diagonizable at all? Or if the $\|V_1\| = 0$ does this mean that it is already orthogonal and I can just use my eigenvectors as is to generate the matrix $U$ such that $A=UDU^*?$

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    $\begingroup$ $||V_1|| \neq 0$. Same goes for norm of $V_2$. You have to use absolute values around the squared entries. $\endgroup$ – layabout Jun 14 '18 at 18:47
  • $\begingroup$ You mean to unitarily diagonalize it, no? $\endgroup$ – Mathematician 42 Jun 14 '18 at 18:48
  • $\begingroup$ You are not working with a real inner product, but a Hermitian inner product, i.e. $\left\langle x,y \right\rangle=\sum_{i=1}^n x_i\overline{y_i}$. $\endgroup$ – Mathematician 42 Jun 14 '18 at 18:50
  • $\begingroup$ The inner product is always positive definate. That is $\langle x,x\rangle \ge 0$ and only equals $0$ if $x = 0.$ (and the norm is $\sqrt {\langle x,x\rangle}$) To meet this criterion, we need a slightly different inner product definition when working with vectors over a complex field. $\langle x,y\rangle = x^\dagger y$ ($\dagger$ is the conjugate transpose) will do. So the norm $(i,1) = \sqrt {(-i\cdot i + 1\cdot 1)}$ $\endgroup$ – Doug M Jun 14 '18 at 18:52
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The norm of a vector $v$ is

$$ \|v\|^2 = (v^*)^Tv = \sum_k v_k^* v_k $$

where $^*$ denotes the complex conjugate. So in your case

$$ \|v_1\|^2 = \pmatrix{i & 1}^* \pmatrix{i \\ 1} = \pmatrix{-i & 1} \pmatrix{i \\ 1} = 1 + 1 = 2 $$

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  • $\begingroup$ Thank you very much. I had forgotten that there were special properties for cases like this. $\endgroup$ – Moseph Jun 14 '18 at 19:09
  • $\begingroup$ @Moseph Happy to help $\endgroup$ – caverac Jun 14 '18 at 19:20
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It happens that $\|(a,b)\|=\sqrt{|a|^2+|b|^2}$. Therefore, the norm of both vectors that you mentioned is $\sqrt2$.

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