How do I solve the following matrix equation?

Question

How do I solve

$$X^2=\begin{pmatrix} -1 &25 \\ -1& 21 \end{pmatrix}$$

where $X$ is a complex matrix? I met a lot of these exercises so please offer a detailed method. Thank you!

Answer 1

Without diagonalising, one way would be to simply let $a, b, c, d \in \Bbb C$ and $X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.

Then $X^2 = \begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2 + bc \end{pmatrix} = \begin{pmatrix} -1 & 25 \\ -1 & 21 \end{pmatrix}$.

Considering each equation in turn, we find:

$a^2+bc = -1 \implies b = -\frac{1+a^2}{c}$

$b(a+d) = 25 = -\frac{(a+d)(1+a^2)}{c} \implies c = -\frac{(a+d)(1+a^2)}{25}$

$c(a+d) = -1 = -\frac{(a+d)^2(1+a^2)}{25} \implies (a+d)^2 = \frac{25}{1+a^2} \implies d = -a \pm \frac{5}{\sqrt{1+a^2}}$

$d^2+bc = 21 = a^2 \mp \frac{10a}{\sqrt{1+a^2}} + \frac{25}{1+a^2} \pm 5\sqrt{1+a^2} \cdot -\frac{1}{5} \sqrt{1+a^2}$ $ = a^2 \mp \frac{10a}{\sqrt{1+a^2}}+\frac{25}{1+a^2} \mp (1+a^2)$

If we choose the plus sign, this becomes: $$22 = \frac{25}{1+a^2}-\frac{10a}{\sqrt{1+a^2}}$$ Multiply through by $1+a^2$ to give $$22 + 22a^2 = 25 - 10a \sqrt{1+a^2} \iff 22a^2+10a\sqrt{1+a^2}-3=0$$ Substitute $a = \sinh x$ to give: $$22\sinh^2x + 10 \sinh x \cosh x - 3 = 22\sinh^2x + 5 \sinh 2x - 3$$ $$=11\cosh 2x + 5 \sinh 2x - 14 = 6\cosh 2x + 5 e^{2x} - 14$$ $$= 8e^{2x} + 3e^{-2x} - 14 \iff 8e^{4x} - 14e^{2x}+3 = (2e^{2x}-3)(4e^{2x}-1) =0$$

Take $2x = \log{\frac{1}{4}} \iff x = -\log2 \iff a = -\frac{3}{4}$

Then substitute back to obtain $d = \frac{19}{4}$, $c = -\frac{1}{4}$, $b = \frac{25}{4}$

So one such solution is $$X = \frac{1}{4} \begin{pmatrix} -3 & 25 \\ -1 & 19 \end{pmatrix}$$ which is easily verified to satisfy $X^2 = \begin{pmatrix} -1 & 25 \\ -1 & 21 \end{pmatrix}$

---

Depending on the choices of signs you make (of which there are four combinations), the four solutions for $X$ are:

$$X = \pm \frac{1}{4} \begin{pmatrix} -3 & 25 \\ -1 & 19 \end{pmatrix}, \qquad X = \pm\frac{1}{2 \sqrt{6}} \begin{pmatrix} 1 & 25 \\ -1 & 23 \end{pmatrix}$$

Answer 2

If you diagonalize the matrix, you get a form $S\Lambda S^{-1}$, where $\Lambda$ is a diagonal matrix containing the eigenvalues of $X^2 =\begin{pmatrix} -1 & 25 \\-1 & 21\end{pmatrix}$; and $S$ the matrix which columns are formed by its eigenvectors. This decomposition will obey $X^2=S\Lambda S^{-1}.$

Now we get $\Lambda^{1/2}$, and see what happens if we raise $S\Lambda^{1/2} S^{-1}$ to the second power:

$$(S\Lambda^{1/2} S^{-1})\,\,(S\Lambda^{1/2} S^{-1})=S\Lambda^{1/2} (S^{-1}\,S)\Lambda^{1/2} S^{-1}= S\Lambda^{1/2} \Lambda^{1/2}S^{-1}=S\Lambda S^{-1}=X^2$$

Hence, $X = S\Lambda^{1/2} S^{-1}.$

---

To wrap it up,

\begin{align}\small X = S \Lambda^{1/2} S^{-1}&= \small{\begin{bmatrix}-0.7687560&-0.99884608\\-0.6395422&-0.04802616\end{bmatrix}\begin{bmatrix}4.44949&0\\0&0.4494897\end{bmatrix}\begin{bmatrix}0.07979307&-1.6595338\\-1.06256752&1.277250\end{bmatrix}}\\&= \begin{bmatrix}0.2041241&5.103104\\-0.2041241&4.694855\end{bmatrix}\end{align}