Solve square matrix equation.

Question

So I have to solve this equation.

$ XX^T-X^TX=2\begin{pmatrix}0 & 2\\\ 2 & 0\end{pmatrix}$

I wrote matrix $X\ $ as $X=\begin{pmatrix}a & b\\\ c & d\end{pmatrix}\ $ and $ XX^T-X^TX=\begin{pmatrix}b^2-c^2 & ac+bd-ab-cd\\\ ac+bd-ab-cd & c^2-b^2\end{pmatrix} $

so I had system of equations $ \begin{cases} (b+c)(b-c)=0 \\[2ex] (d-a)(b-c)=4 \\[2ex] \end{cases} $

One thing is certain $b=-c$ I have found 8 solution options. But I have diiferent ask. Does anyone have different option to solve it? My solution is to hard and overwhelming. I want to find just one solution.

Let's determine that $X∈M2(Q)$ – Kukoz Nov 19 '19 at 17:50 — Kukoz, Nov 19 '19 at 17:50

score 2 · Answer 1 · 2019-11-25T18:28:24.947

2

The OP's calculation is false (no longer since I corrected it).

@Siddhartha, you forgot a lot of solutions.

The general solution depends on $2$ parameters $a,c\not= 0$, that is,

$X=\begin{pmatrix}a&-c\\c&a-2/c\end{pmatrix}$.

edited Nov 25 '19 at 18:28

answered Nov 20 '19 at 12:13

Ah, sorry. I should have noticed. :) – Siddhartha Nov 20 '19 at 17:27

Siddhartha · Answer 2 · 2019-11-19T21:20:08.987

1

Here is the explanation:

As you noticed, $b = -c$. From 3rd equation we have $2b(d-a) = 4$. This means $b \neq 0 \neq c$. From this we have $d = a + {\frac {2}{b}}$. Similarly from 4th equation we have $d = - \left( a + {\frac {2}{b}} \right)$. Adding these two we get $2d = 0$. Thus $d = 0$. This proves $a = -{\frac {2}{b}} = {\frac {2}{c}}$. Now replace all entries by the relations of $c$ and we get your matrix.

\begin{pmatrix} 2/c & -c \\ c & 0 \end{pmatrix}

for any $c \in {\mathbb{R}}, c \neq 0$.

edited Nov 19 '19 at 21:20

answered Nov 19 '19 at 17:31

Siddhartha

798

But, $ d $ could be also $2$ or $1$. – Kukoz Nov 19 '19 at 17:51
Now check the solution. You cannot have $d = 2$ or $1$. – Siddhartha Nov 19 '19 at 21:20
So check $ a=3, b=-2, c=2, d=2$ – Kukoz Nov 20 '19 at 07:21

Solve square matrix equation.

2 Answers2