Find a homogeneous system that will give all the elements of B.

Question

If A = $\begin{bmatrix} 1 & -1 \\ 0 & -2 \end{bmatrix}$, set B = {X ∈ $M_2(\mathbb{R})$∣AX + $X^\intercal$A = 0}

Find a homogeneous system that will give all the elements of B.

My attempt:

AX + $X^\intercal$A = 0

AX = $-$ ($X^\intercal$A)

($\cfrac{1}{A}$) AX = $-$ ($X^\intercal$A)$(\cfrac{1}{A})$

X = $-$ $X^\intercal$

This would mean that all skew-symmetric matrices would satisfy the conditions.

However, I tried plugging in random skew-symmetric matrices but AX + $X^\intercal$A woouldn't equate to 0.

I may have erred in my solution: I just don't know where.

Answer 1

You cannot infer from the step $AX=-(X^TA)$ that $A^{-1}AX=-(X^TA)A^{-1}$.

When two square matrices are equal, say $X=Y$, you can left-multiply both sides by a matrix $M$ to get another equality $MX=MY$, or you can right-multiply both sides by $M$ to get $XM=YM$. However, you cannot infer that $MX=YM$, because matrix multiplication in general is not commutative.

Answer 2

Your step when apply inverse both side is not correct because you are multiplying from the left and from the right, use instead

$$\begin{bmatrix} 1 & -1 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} a& b \\ c& d \end{bmatrix}=- \begin{bmatrix} a& c \\ b& d \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & -2 \end{bmatrix} $$

that is

• $a-c=-a$
• $b-d=a+2c$
• $-2c=-b$
• $-2d=b+2d$