# Show that $\det(A) = 0$ or $\mathrm{tr}(A)=0$

Let $$A$$ and $$B$$ are two matrices in $$\mathcal{M}_{2,2}(\mathbb{R})$$ with $$B\neq0$$ and $$AB=-BA$$. Show that either $$\mathrm{tr}(A) = 0$$ or $$\det(A)=0$$.

I'm stuck since the order is 2. A few attempts are

1. From the hypothesis we imply that $$\mathrm{tr}(AB) = 0$$, thus $$(AB)^2=(\det AB)(AB)$$. Take $$\det$$ of both sides, we get $$\det AB = 0$$ or $$\det AB = 1$$. which leads to no thing.

2. Suppose $$\det A \neq 0$$, then multiply both side by $$A^{-1}$$ to the left we have $$B=A^{-1}(-B)A$$. This implies that $$\mathrm{tr}B= 0$$. But i'm still stuck afterward.

FIRST PROOF :

You have $$AB=-BA$$ so multiplying by $$A$$ on the left, you get $$A^2B = -ABA = BA^2$$.

Now, by Cayley-Hamilton theorem, you know that $$A^2- \mathrm{Tr}(A) A + \mathrm{det}(A) I = 0 \quad \quad (1)$$

Multiplying by $$B$$ on the right, you get $$A^2B- \mathrm{Tr}(A) AB + \mathrm{det}(A) B = 0, \quad \text{ i.e.} \quad BA^2+ \mathrm{Tr}(A) BA + \mathrm{det}(A) B = 0 \quad \quad (2)$$

And multiplying $$(1)$$ by $$B$$ on the left, you get $$BA^2- \mathrm{Tr}(A) BA + \mathrm{det}(A) B = 0 \quad \quad (3)$$

Now $$(2)-(3)$$ gives you $$\mathrm{Tr}(A)BA=0$$

If $$\mathrm{Tr}(A)=0$$, you are done. Otherwise, $$BA=0$$, so the equation $$(3)$$ gives you $$\mathrm{det}(A)B=0$$, and because $$B \neq 0$$, you deduce $$\mathrm{det}(A)=0$$. 

SECOND PROOF : Here is another way to solve the question. We suppose that $$\mathrm{det}(A) \neq 0$$, and we will prove that $$\mathrm{Tr}(A)=0$$.

Let's trigonalize $$A$$ in $$\mathbb{C}$$. There exists a basis $$\lbrace x,y \rbrace$$, and three complex numbers $$\lambda, \mu, \nu$$ such that $$Ax= \lambda x \quad \quad \text{ and } \quad Ay=\mu y + \nu x$$ (note that $$\lambda$$ and $$\mu$$ are the eigenvalues of $$A$$, and they are non-zero because $$\mathrm{det}(A) \neq 0$$).

Now apply $$B$$ to the first equation : you get $$BAx=\lambda Bx\quad \quad \text{ so} \quad A(Bx)=-\lambda(Bx)$$ If $$Bx \neq 0$$, then $$-\lambda$$ has to be an eigenvalue of $$A$$, and because $$\lambda \neq 0$$, that means that $$\lambda$$ and $$-\lambda$$ are the two disctincts eigenvalues of $$A$$. So $$\mathrm{Tr}(A)=0$$.

If on the contrary, $$Bx=0$$, then you must have $$By \neq 0$$ because $$B \neq 0$$. Applying $$B$$ to the second equation gives $$BAy=\mu By + 0\quad \quad \text{ so} \quad A(By)=-\mu(By)$$ so again, $$\mu$$ and $$-\mu$$ are the two only eigenvalues of $$A$$ and $$\mathrm{Tr}(A)=0$$.

• Thanks. I should start from the characteristics polynomial of A next times. But I still wonder that are there any other ways? Like use the properties of similar matrices? – aDmaL Aug 27 '20 at 12:54
• @aDmaL I edited my answer, to provide another way to think. There are probably many many ways to solve this kind of questions. – TheSilverDoe Aug 27 '20 at 13:41
• Really appreciate your effort to bring me another way to prove! – aDmaL Aug 27 '20 at 13:47
• It looks like a very simple problem. But it is tricky. Very ingenious way of solving the problem. Thank you for the nice method. – Lawrence Mano Aug 28 '20 at 7:37

Well, I manage to find out another answer for this question, it maybe helpful for future readers.

If we define $$A=\begin{pmatrix} a&b\\c&d \end{pmatrix}$$ and let $$M$$ be the matrix of $$T$$, where $$T$$ is a linear transformation defined by $$T:\mathcal{M}_{2,2}(\mathbb{R})\to\mathcal{M}_{2,2}(\mathbb{R}), T(X)=AX+XA.$$ Then, , we can find out that $$M= \begin{pmatrix} 2a&c&b&0\\b&a+d&0&b\\c&0&a+d&c\\0&c&b&2d \end{pmatrix}.$$ Moreover, the hypothesis is now equivalent to $$T(B)=0$$ , where $$B\neq0$$, which happens if and only if $$\det M = 0$$.

In the other hand we have $$\det M =4(a+d)^2(ad-bc)$$, thus, $$\det M = 0$$ iff either $$\mathrm{tr}(A)=a+d=0$$ or $$\det A = ad-bc=0$$.