If $A \in M_{2\times 2}(F)$ and $A^{2}=0$ then $tr(A)=0$

Question

I actually saw an statement here on stackxchange: "Let $A$ be a square matrix $n \times n$. $trace (A)=0$, if and only if, $A^{2}=0$."

So I could find a counterexample for →. And I think the statement should be correct only for $2\times 2$ square matrices.

Now I wanna know if we can claim: If $A\in M_{2\times 2}(F)$ and $A^{2}=0$ then $trace(A)=0$. Am I correct? If this makes sense, how do we prove it based on trace and multiplication definitions?

Answer 1

$A^2=\pmatrix{a&&b\\c&&d}^2=\pmatrix{a^2+bc&&ab+bd\\ac+dc&&bc+d^2}=\pmatrix{0&0\\0&0}$

$\implies b(a+d)=0$ and $ c(a+d)=0 \implies a+d=0 $ or $b=c=0$,

but if $b=c=0$ then $a^2+bc=0, bc+d^2=0\implies a=d=0\implies a+d=0$ anyways.

So indeed it can be shown based on the definitions of matrix trace and multiplication that

the trace of a $2\times2$ matrix whose square is the $0$ matrix is $0.$

Answer 2

In general, the statement "if $A^2 = 0$, then $\operatorname{tr}(A) = 0$" can be proven using the fact that the trace is the sum of the eigenvalues. One proof for the specific case of $2 \times 2$ matrices is as follows.

It's clear that if $A = 0$, then $A^2 = 0$ and $\operatorname{trace}(A) = 0$. So, suppose that $A^2 = 0$ but $A \neq 0$. Because $A^2 = 0$, $A$ cannot be an invertible matrix. It follows that the columns of $A$ are linearly independent. As a consequence, we can write $A$ in the form $A = xy^T$ for some vectors $x$ and $y$ (in particular, we can let $x$ be a non-zero column of $A$).

Verify that the trace of $A$ is also equal to $y^Tx$. On the other hand, note that $$ A^2 = (xy^T)(xy^T) = x(y^Tx)y^T = (y^Tx) \cdot xy^T = \operatorname{trace}(A) \cdot A. $$ So, $A^2 = 0$ implies that $\operatorname{trace}(A) = 0$.

Answer 3

Your statement is right but the following is not true for any $n≥2$ : "for a matrix A of size $n$, if $trace(A)=0$ then $A^2=0$.

If $A=0$ then it is easy to see that $trace(A)=0$

Suppose $A\ne 0$ but $A^2=0$ , then $\exists x\in F^2$ such that $Ax\ne 0$

You see $\{x,Ax\}$ forms a basis of $F^2$.(why?)

Let $c_1x+c_2Ax=0$ where $c_1,c_2\in F$ . Then $A(c_1x+c_2Ax)=0\implies c_1Ax+c_2A^2x=0$ . Then $c_1=0$ and $c_2=0$.

So ${x,Ax}$ linearly independent hence the basis of $F^2$.

Look at the matrix of $A$ with respect to this matrix it is $\begin{pmatrix} 0&0\\1&0 \end {pmatrix}$ as $A(x)=0\cdot x +1\cdot Ax$ and $A(Ax)=A^2x=0\cdot x +0\cdot Ax.$

So the $trace(A)=0$.