Linear Algebra Proof: Invertible matrices, construction of the field on complex numbers

Let $x$, $y$ $\in \mathbb{R}$ and let $Z = \begin{bmatrix} x & -y \\ y & x \end{bmatrix}$.

Prove that $Z$ is invertible if and only if $x \neq 0$ and $y \neq 0$.

I believe that this is constructing a field over the field of complex numbers, but I am struggling to do so. Does anyone have any recommendations or way forward?

• There is an ismorphism from the matrices in the from of $Z$ to the the complex numbers $(x+iy)$ And, you could prove that, and ultimately will want to prove that. But, that is really not necessary to prove the proposition above. – Doug M May 30 '17 at 21:34

3 Answers

A $2\times2$ matrix is invertible if and only if it's determinant is not 0. This matrix has the determinant $x^2 + y^2$, which is $0$ if and only if both $x$ and $y$ are zero. Therefore it is invertible if and only if $x \neq 0$ or $y \neq 0$.

If $x=y=0$, then $Z$ is the null matrix, which is not invertible.

Otherwise, the matrix$\begin{bmatrix}\frac x{x^2+y^2}&\frac y{x^2+y^2}\\\frac {-y}{x^2+y^2}&\frac x{x^2+y^2}\end{bmatrix}$ is the inverse of $Z$.

Another perspective: $$\mathbf{A}^{-1} = \left[ \begin{array}{cr} x & -y \\ y & x \end{array} \right]^{-1} = \frac{\text{adj}\mathbf{A}}{\det \mathbf{A}}= \frac{1}{x^{2}+y^{2}} \left[ \begin{array}{rr} -x & y \\ -y & -x \end{array} \right]$$ When does the inverse not exist? When $x^{2}+y^{2} = 0$.