Finding a matrix that satisfy a givven inner product 
Let $V=M_2(\mathbb{R})$
   with $\left\langle A,B\right\rangle:=\text{tr}\left(AB^*\right)$   and let $A=\left(\begin{array}{cc} 1 & -3 \\ -2 & 2 \end{array}\right)$ find $B\in M_2(\mathbb{R})$ such that $B$ is orthonormal and orthogonal to $A$

So we need the both $||B||=1$ and $\text{tr}\left(\left(\begin{array}{cc} 1 & -3 \\ -2 & 2 \end{array}\right)\left(\begin{array}{cc} a & c \\ b & d \end{array}\right)\right)=0$
So we have $\sqrt{\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\left(\begin{array}{cc} a & c \\ b & d \end{array}\right)}=1$
$a^2+b^2+c^2+d^2=1$ and $a-3b-2c+2d=0$ how should I continue?
 A: It is well known that the inner product $\langle A,B\rangle=\operatorname{tr}(AB^T)$ on $M_n(\mathbb R)$ is equal to the usual inner product $\langle\operatorname{vec}(A),\operatorname{vec}(B)\rangle$ on $\mathbb R^{n^2}$. So, essentially, you just need to find a unit vector in $\mathbb R^4$ that is orthogonal to $\mathbf a=\operatorname{vec}(A)=(1,-2,-3,2)^T$. (Note that the usual convention to vectorise a matrix is column-major; therefore the second element of $\mathbf a$ is the second element on the first column of $A$, not the first row.) As every $\mathbf b\perp \mathbf a$ must be of the form $\mathbf b=(2x+3y-2z,\ x,\ y,\ z)^T$, the general solution is given by
$$
B=\frac1{\sqrt{(2x+3y-2z)^2+x^2+y^2+z^2}}\pmatrix{2x+3y-2z&y\\ x&z}
$$
for any real numbers $x,y,z$ with $x^2+y^2+z^2\ne0$.
A: $||B||=1\implies a^2+b^2+c^2+d^2=1$. $A\perp B=0\implies a-3b-2c+2d=0$. Pick $a,b,c,d$ in such a way that $a=\sqrt{0.5}\cos\phi,b=\sqrt{0.5}\sin\phi,c=\sqrt{0.5}\cos\theta,d=\sqrt{0.5}\sin\theta$ so that the $1$-st equality is taken care of. If we pick $\theta,\phi$ in such a way that $a=3b$ and $c=d$ we're done. $\phi=\arctan\left({3^{-1}}\right)$ and $\theta={\pi\over4}$ do the job.
