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Problem Statement: Let $A=\begin{bmatrix} 2 && 1 \\ 1 && 2 \end{bmatrix}$. Find an orthonormal basis for $\mathbb{C}^2$ with respect to the Hermitian form $\bar{x}^TAy$.

I am trying to figure out this proof, but not sure if what I am currently trying to do will help me. I was goin to start out by taking an arbitrary basis $B=\left\{v_1,v_2\right\}$ and write two arbitrary vectors in $\mathbb{C}^2$ as linear combinations of basis elements: $$v=x_1v_1+x_2v_2$$ $$w=y_1v_1+y_2v_2$$

Then break down the Hermitian form $$\langle v,w \rangle=\langle x_1v_1+x_2v_2, y_1v_1+y_2v_2 \rangle=\bar{x_1}\langle v_1, y_1v_1+y_2v_2 \rangle+\bar{x_2}\langle v_2, y_1v_1+y_2v_2 \rangle=\bar{x_1}y_1\langle v_1, v_1 \rangle+\bar{x_1}y_2\langle v_1, v_2 \rangle+\bar{x_2}y_1\langle v_2, v_1 \rangle+\bar{x_2}y_2\langle v_2, v_2 \rangle=\sum_{i=1,j=1}^2 \bar{x}_iy_j\langle v_i, v_j\rangle=\bar{x}^TAy$$

Then I want to use the fact that we're given $A$ in order to define $v_1,v_2$, but I am confused with one thing: In my notes, I have that $$a_{ij}=\langle v_i,v_j\rangle$$ but my notes also say for any orthonormal basis, that $$\langle v_i,v_j\rangle:=\begin{cases} 1\ \mathrm{if}\ i=j \\ 0\ \mathrm{if}\ i\neq j \end{cases}$$ which conflicts with the fact that $A=\begin{bmatrix} 2 && 1 \\ 1 && 2 \end{bmatrix}$ because that would mean that $$\langle v_i,v_j\rangle:=\begin{cases} 2\ \mathrm{if}\ i=j \\ 1\ \mathrm{if}\ i\neq j \end{cases}$$

Is there a more straightforward approach to this problem?

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The straight-forward approach is Gram-Schmidt.

Start with the standard (canonical) basis $\{e_1,e_2\}$. Our goal is to get an orthonormal basis $\{v_1,v_2\}$.

First, we take $$ v_1 = \frac {1}{\sqrt{\langle e_1,e_1 \rangle}} e_1 = \frac 1 {\sqrt 2}e_1 $$ Now, take $$ w_2 = e_2 - \langle e_2,v_1 \rangle v_1 = e_2 - \frac 1 2 \langle e_2,e_1 \rangle e_1 = e_2 - \frac 12 e_1 $$ Noting that $$ \langle w_2,w_2\rangle = \frac 12 - \frac 12 - \frac 12 + 2 = \frac 32 $$ and set $$ v_2 = \frac{1}{\sqrt{\langle w_2,w_2 \rangle}}w_2 = \sqrt{\frac{2}{3}}(e_2 - \frac 12 e_1) $$ so, now we have an orthonormal basis.

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  • $\begingroup$ I am confused how this is with repsect to the Hermitian form $\bar{x}^TAy$. Why did we select the basis that we did? It seems that the matrix $A$ was pointless in this case? or did we use the fact that $a_{11}=2$ to define $\langle e_1,e_1 \rangle=2$ and $a_{21}=1$ to define $\langle e_2,e_1 \rangle=1$? $\endgroup$ – yung_Pabs Feb 7 '16 at 0:38
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    $\begingroup$ I didn't define $\langle e_1,e_1 \rangle$, I computed it using the definition from your question (without showing my work). Check that I did so correctly. By the "standard basis", I mean specifically that $$ e_1 = \pmatrix{1\\0}, \quad e_2 = \pmatrix{0\\1} $$ $\endgroup$ – Omnomnomnom Feb 7 '16 at 0:43
  • $\begingroup$ So are you computing the inner product with $a_{ij}=⟨v_i,v_j⟩$ using our given matrix $A$? $\endgroup$ – yung_Pabs Feb 7 '16 at 0:47
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    $\begingroup$ Yes!${}{}{}{}{}{}{}{}{}$ $\endgroup$ – Omnomnomnom Feb 7 '16 at 0:50

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