Eigenvalue problem

Question

If all the eigenvalues of $2 \times 2$ matrix $A$ are zero then prove that $A^2=0$.

$0$ is an eigenvalue of $A$. Then $AX=0$, again $A^2X=0$. Then $X$ is an eigenvector of both $A$ and $A^2$. Can we show that $A$ or $A^2$ must be zero?

Answer 1

It is not true in general. Consider the matrix

$$ A= \begin{pmatrix} 0&1&1\\ 0&0&1\\ 0&0&0 \end{pmatrix} $$

Added: but it is true for $2\times 2$ matrices by Cayley-Hamilton theorem because, in this case, the characteristic polynomial is $\lambda^2=0$.

Answer 2

Consider the characteristic polynomial of $A$

then if $A$=\begin{bmatrix} a& b\\ c&d\end{bmatrix}

then $A$ satisfies $A^2+\text{trace } A+\det A=0$

Since $0$ is the only eigen value so $\text{trace} A=\det A=0$

So $A^2=0$

Answer 3

Such a $2\times 2$-matrix is nilpotent, because all of its eigenvalues are zero. Every nilpotent matrix $A\in M_n(K)$ satisfies $A^n=0$, because its characteristic polynomial is $t^n$, and by Cayley-Hamilton. For $n=2$ we have $A^2=0$.

Answer 4

If $A=0$, then $A^2=0$.

Otherwise, let $v \in V$ with $v\ne0$ and $Av=0$ and $w \in V$ with $Aw\ne0$.

Then $v$ and $w$ are linearly independent and so form a basis for $V$.

Since $A^2v=0$, it is enough to prove that $A^2w=0$.

Write $Aw = \lambda v + \mu w$. Then $AAw = \mu Aw$.

If $Aw=0$, then $A^2w=0$.

If $Aw\ne0$, then $Aw$ is an eigenvector of $A$ with $\mu$ as eigenvalue and so $\mu=0$. Thus, $A^2w=AAw=0$.