Can any 2x2 matrix be written like sum of two squared matrices?

Question

How to prove that for any $A$ is a $2\times2$ matrix with real elements exist $B$ and $C$ so that $A=B^2+C^2$? So far, I used Cayley-Hamilton theorem and I have: $A =$ $\frac{1}{Tr(A)}A^2 + \frac{det(A)}{Tr(A)}I_n$. I know that I need a positive trace, so I choose $A_1 = A + tI_n$ and $\lim_{t\to∞}(A + tI_n) = \infty$

Answer 1

First, we note that if $A$ has real eigenvalues, it is similar to a matrix in Jordan canonical form.

If $A = \pmatrix{\lambda_1\\&\lambda_2}$ with $\lambda_1 \leq \lambda_2$, then: if $\lambda_1 \geq 0$, we have $$ A = \pmatrix{\sqrt{\lambda_1}\\&\sqrt{\lambda_2}}^2 + 0^2 $$ If $\lambda_1 < 0$, we have $$ A = \pmatrix{0 & -\sqrt{|\lambda_1|}\\ \sqrt{|\lambda_1|} & 0}^2 + \pmatrix{0&0\\0&\sqrt{\lambda_2 - \lambda_1} }^2 $$ If $A = \pmatrix{\lambda&1\\0&\lambda}$ with $\lambda > 0$, take $$ A = \pmatrix{\sqrt{\lambda} & 1/(2\sqrt{\lambda})\\0&\sqrt{\lambda}}^2 + 0^2 $$ If $A = \pmatrix{\lambda&1\\0&\lambda}$ with $\lambda < 0$, take $$ A = \pmatrix{1&1/2\\0&1}^2 + \pmatrix{0&-\sqrt{1-\lambda}\\ \sqrt{1-\lambda}&0}^2 $$

If $A$ has complex eigenvalues, it is similar to a matrix of the form $$ \pmatrix{a&-b\\b&a} $$ For this matrix, we can take $$ A = \pmatrix{c&-d\\d&c}^2 + 0^2 $$ where $c,d$ are such that $(c + di)^2 = a+bi$.