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I'm having some problems answering a question set for my undergrad maths course. The question is:

Find the set $S=\left\{(x,y)\in\mathbb C\times\mathbb C:\begin{pmatrix}x&i\\iy&0\end{pmatrix}A^ T=\begin{pmatrix}0&0\\0&i+1\end{pmatrix}\right\}$ where $A=\begin{pmatrix}1+i&-1\\1&-i\end{pmatrix}$.

I evaluated $\begin{pmatrix}x&i\\iy&0\end{pmatrix}A^ T$ to equal $\begin{pmatrix}(1+i)x-i&x-i\\(-1+i)y&iy\end{pmatrix}$ but $(1+i)x-i$ and $x-i$ cannot both equal $0$?

Obviously i'm doing something wrong but my tutors aren't much help.

*Edit - the question was 'fixed' to

Find the set $S=\left\{(x,y)\in\mathbb C\times\mathbb C:\begin{pmatrix}x&i\\iy&0\end{pmatrix}^TA^ T=\begin{pmatrix}0&-5i\\i&2i\end{pmatrix}\right\}$

but I still can't solve it?

Hagen von Eitzen
  • 374,180

2 Answers2

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Actually, $$\begin{pmatrix}x & i\\iy & 0\end{pmatrix}A^T=\begin{pmatrix}x & i\\iy & 0\end{pmatrix}\begin{pmatrix}1+i & 1\\-1 & -i\end{pmatrix}=\begin{pmatrix}(1+i)x-i & x+1\\(-1+i)y & iy\end{pmatrix},$$

but in any case, there is no $x$ or $y$ that satisfies the given conditions. Hence, $S=\emptyset.$

Cameron Buie
  • 102,994
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As $\det(A)=(2-i)\ne0$, $A$ is invertible. Hence the defining equation of $S$ is equivalent to $$ \underbrace{\pmatrix{x&i\\ iy&0}}_M = \underbrace{\pmatrix{0&0\\ 0&i+1}}_B\ (A^\top)^{-1}.\tag{1} $$ Since the first row of $B$ is zero, the first row of $B(A^\top)^{-1}$ must be zero too. Yet the $(1,2)$-th entry of $M$ is nonzero, hence $(1)$ cannot be satisfied regardless of the value of $x$. Therefore $S$ is empty.

user1551
  • 139,064