Find $P \in GL_3 (C) $ so that $^tP \cdot P* = A$.

Question

Find $P \in GL_3 (C) $ so that $^tP \cdot P* = A$.

$P*$ means the complex conjugated of $P$.

A is a hermitian matrix:

$ A = \begin{pmatrix} 1 & 0 &-i \\ 0&1&0 \\ i&0&2 \end{pmatrix}$

How can I get $P$?

Answer 1

Let's denote $^tP$ by $Q$, that is $Q = \ ^tP$. Then $P= \ ^tQ$ and $P^* = \ ^tQ^*$, also denoted by $Q^\dagger$. We need to find $Q$ so that $$Q \cdot Q^{\dagger}= A$$ As a general fact, this is possible for a given $A$ if and only if $A$ is hermitian and positive semi-definite. $A$ is clearly hermitian and its eigenvalues are $>0$ so yes, there exists such $Q$. Here is a possible approach: Consider the rows of $Q$ as three vectors in $\mathbb{C}^3$, $u_1$, $u_2$, $u_3$. The matrix $Q \cdot Q^{\dagger}$ has as element $(k,l)$ the dot product of the vectors $u_k$, $u_l$, that is $(u_k, u_l)$. So we need to find three vectors $u_1$, $u_2$, $u_3$ with prescribed dot products. We can now try to guess the vectors. $u_2$ is perpendicular to both $u_1$ and $u_3$. $u_1$ has norm squared $1$, and $u_3$ has norm squared $2$. Moreover, the dot product $(u_1, u_3) = i$. Let's take $u_1 = (1,0,0)$, $u_2 = (0,1,0)$ and $u_3 = (i, 0,1)$. All works OK now. We therefore take $$ Q = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1&0 \\ i & 0 & 1 \end{array} \right) $$ and we see that $$Q \cdot Q^{\dagger} = A$$ Therefore we can take $$ P =\ ^tQ =\left( \begin{array}{ccc} 1 & 0 & i \\ 0 & 1& 0 \\ 0 & 0 & 1 \end{array} \right) $$ Let's check: $$ \ ^t P \cdot P^* = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1&0 \\ i & 0 & 1 \end{array} \right) \cdot \left( \begin{array}{ccc} 1 & 0 & -i \\ 0 & 1& 0 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & -i \\ 0 & 1& 0 \\ i & 0 & 2 \end{array} \right)= A$$

Answer 2

We do a row reduction, followed by a column reduction as follows.

$$ \begin{pmatrix} 1 & 0 &-i \\ 0&1&0 \\ i&0&2 \end{pmatrix} \mapsto \begin{pmatrix} 1 & 0 &-i \\ 0&1&0 \\ 0&0&1 \end{pmatrix} \mapsto \begin{pmatrix} 1 & 0 &0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix} $$

Where the first operation is adding $(-i) (\text{Row 1})$ to Row 3. The second is adding $i(\text{column 1})$ to Column 3. The elementary matrix representing the first transformation is $$ Q= \begin{pmatrix} 1 & 0 &0 \\ 0&1&0 \\ -i &0&1 \end{pmatrix}. $$ While the second one is $Q^*$ where, by $Q^*$ I mean CONJUGATE TRANSPOSE. Thus we have $$QAQ^*=I \implies A=(Q^{-1})(Q^{-1})^*. $$ Note that $Q^{-1}$ is the elementary matrix corresponding to the reverse of operation corresponding to $Q$ (i.e adding $(i)\text{Row 1}$ to Row 3), and thus $$Q^{-1} = \begin{pmatrix} 1 & 0 &0 \\ 0&1&0 \\ i &0&1 \end{pmatrix}.$$

Answer 3

If, for you, $P^*$ denotes the complex conjugate, then clearly Math and Orange are right and I delete my post.