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Let $V=M_2(\mathbb{R})$ with $\left\langle A,B\right\rangle:=\text{tr}\left(AB^*\right)$ and $A=\left(\begin{array}{cc} 4 & 1 \\ -3 & 3 \end{array}\right)$ find $B\in M_2(\mathbb{R})$ such that $B$ will be orthonormal with respect to the inner product $\left\langle A,B\right\rangle:=\text{tr}\left(AB^*\right)$ What I have thought about is that the matrix $B$ have to fulfill:

a. $$\left(\begin{array}{cc} 4 & 1 \\ -3 & 3 \end{array}\right)\cdot \left(\begin{array}{cc} x_1 & x_2 \\ x_3 & x_4 \end{array}\right)=0$$ b.$$x_1\cdot \bar{x_1}+x_4\cdot \bar{x_4}=1$$

But I get to more equations than veriables

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  • $\begingroup$ There is something wrong. If you solve the first matrix equation for $x_1, x_2, x_3, x_4$ you'll easily notice that $x_1 = x_2 = x_3 = x_4 = 0$. $\endgroup$ – Von Neumann Aug 31 '16 at 13:12
  • $\begingroup$ If I understand the question correctly? you should find $B$ so that ${\rm tr} A B^*=0$ and ${\rm tr} B B^*=1$. This leads to two different equations than yours. (and there are many solutions) $\endgroup$ – H. H. Rugh Aug 31 '16 at 13:15
  • $\begingroup$ The field is $\mathbf{R}$, so your adjoint matrix is the transpose. That eliminates the conjugate part. If you go through the normal property, you will find that your matrix $B$ is symmetric. That eliminates 2 more variables. $\endgroup$ – IAmNoOne Aug 31 '16 at 13:17
  • $\begingroup$ @H.H.Rugh you are right I was mistaken about part 1 $\endgroup$ – newhere Aug 31 '16 at 13:18
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    $\begingroup$ I changed the content of the body to match your title. Your question is completely different from the content. $\endgroup$ – IAmNoOne Aug 31 '16 at 13:38
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$B=\left(\begin{matrix} x_1 & x_2 \\ x_3 & x_4 \end{matrix}\right)$ orthonormal mean that :

1) $\langle A, B\rangle =tr(AB^*)=0$

2) $\langle B, B\rangle =tr(BB^*)=1$

so $2)$ implies that \begin{eqnarray} 1&=& tr(BB^*)\\ &=& tr\left[\left(\begin{matrix} x_1 & x_2 \\ x_3 & x_4 \end{matrix}\right)\left(\begin{matrix} \bar x_1 & \bar x_3 \\ \bar x_2 & \bar x_4 \end{matrix}\right) \right]\\ &=& tr\left[\left(\begin{matrix} |x_1|^2+|x_2|^2 & \alpha \\ \beta & |x_3|^2+|x_4|^2 \end{matrix}\right) \right]\\ &=& |x_1|^2+|x_2|^2+|x_3|^2+|x_4|^2 \end{eqnarray} and $1)$ mean that : \begin{eqnarray} 0&=& tr(AB^*)\\ &=& tr\left[\left(\begin{matrix} 4 & 1 \\ -3 & 3 \end{matrix}\right)\left(\begin{matrix} \bar x_1 & \bar x_3 \\ \bar x_2 & \bar x_4 \end{matrix}\right) \right]\\ &=& tr\left[\left(\begin{matrix} 4\bar x_1+\bar x_2 & \alpha' \\ \beta' & 3(\bar x_4-\bar x_3) \end{matrix}\right) \right]\\ &=& 4\bar x_1+\bar x_2 +3(\bar x_4-\bar x_3) \end{eqnarray} so your final system is : $$ \left\{ \begin{array}{l} |x_1|^2+|x_2|^2+|x_3|^2+|x_4|^2=1\\ 4 x_1+ x_2 +3( x_4- x_3)=0 \end{array} \right. $$ so the set of solution of this equation is a manifold of dimension 3.

we can write it as : $$ x_2=3(x_3-x_4)-4x_1 $$ so the set of solution is : $$ S=\left\{ \frac{1}{\sqrt{|a|^2+| 3(c-d)-4a|^2+|c|^2+|d|^2}}\left(\begin{matrix} a & 3(c-d)-4a\\ c&d \end{matrix}\right) \textrm{ such that } a,b,c \in \mathbb{C}-\{0\}\right\} $$

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  • $\begingroup$ after the comments above this is the set of equations. for the second equation I set $x_2=x_3=0$ and got $4x_1=-3x_4$ but it can not satsfity the first equation too $\endgroup$ – newhere Aug 31 '16 at 13:31
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    $\begingroup$ can you see now the explicit set of solution ? $\endgroup$ – Hamza Aug 31 '16 at 13:45

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