All nilpotent $2 \times 2$ matrices satisfy $A^{2}=0$

Question

I have problems to show that if $A$ is a $2 \times 2$ matrix and if there exists some positive integer such that $A^{n}=0$ then $A^{2}=0$. I only showed that $A$ is a singular matrix but nothing else. Thanks any help will be appreciated.

Answer 1

If $A$ is a nilpotent $2 \times 2$- matrix, then $0$ is the only eigenvalue of $A$.

($Ax = \mu x$ implies $0=A^2x= \mu^2x$).

Hence the char. Polynomial ist $p(\mu)=\mu^2$. Now invoke Cayley-Hamilton.

A direct proof is also possible: let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, compute $A^2$ an determine $a,b,c$ and $d$ via $A^2=0$.

Answer 2

If $A^n =O$, then $f(x) = x^n$ is an annihilating polynomial of $A$. Hence the minimal polynomial of $A$ is $m(x)= x^k$ for some $k\leqslant n$. By Hamilton-Cayley theorem, the characteristic polynomial $p(x) $ of $A$ annihilates $A$. Since $p(x) =\det(A-xI)$, $\deg p = 2$. Since $m \mid p$, $\deg m \leqslant 2$, i.e. $k \leqslant 2$. Then either $m(x) =x$ or $m(x)=x^2$. Therefore $m(A) = A = O$ or $m(A) = A^2 = O$. Either case, $A^2 =O$.

Answer 3

Let A be $2 \times 2$ matrix such that $A^n$=0, for some n. Then clearly the matrix A satisfies the polynomial p(x)=$x^n$. So characteristic polynomial of A will annihilate p(x ) so using division algorithm it will be $x^2$ (since a degree characteristic polynomial is equal to the order of matrix ). So by Cayley Hamilton theorem, matrix A should satisfy the characteristic polynomial so $A^2=0$.

Answer 4

The only eigenvalue of $A$ is $0$: if $Av=\lambda v$, then $A^nv=\lambda^nv$.

Therefore $$ a_{11}+a_{22}=0 $$ Also the determinant is $0$, so $a_{11}a_{22}-a_{12}a_{21}=0$.

Hence $a_{22}=-a_{11}$ and $a_{12}a_{21}=-a_{11}^2$.

Now compute $A^2$.

Answer 5

Let $A$ be any matrix of any dimension $n$. Then $$ \mathbb{R}^n \supseteq \text{Im}(A) \supseteq \text{Im}(A^2) \supseteq \text{Im}(A^3) \supseteq \ldots $$

If any containment in this chain is ever an equality, then the rest are also equalities.

If $A$ is nilpotent, then this chain reaches zero when we get to $\text{Im}(A^n)$. By the above paragraph, it must strictly decrease at each step. Consequently, if $A$ is $m \times m$ for $m < n$ it must be zero by step $m$.

Answer 6

Let $(\lambda, v)$ be an eigenvalue-eigenvector pair for $A$. Then

$$ 0 = 0v = A^nv = \lambda^nv.$$

This implies that $\lambda = 0$. Since $\lambda$ is an arbitrary eigenvalue, the characteristic polynomial of $A$ is $p_A(x) = x^2$. Thus, the possibilities for the minimal polynomial are

$$m_A(x) = x\qquad m_A(x) = x^2$$

and so $A = 0$ or $A^2 = 0$.

Answer 7

$A^2-Tr(A)A+det(A)I=0 \Rightarrow$

$ A^n-Tr(A)A^{n-1}+det(A)A^{n-2}=0\cdot A^{n-2=}=0\Rightarrow $

$-Tr(A)A^{n-2}=0\Rightarrow A^{n-2}=0.... \Rightarrow A^2=0$

(if $Tr(A)=0$ then from the initial relation $A^2=0$)

Answer 8

If the eigenvalues of $A$ are $\lambda$s, the eigenvalues of $A^n$ are $\lambda^n$s therefore according to Cayley-Hamilton's theorem we obtain $$\lambda^n=0\Rightarrow \lambda=0$$therefore all the eigenvalues of $A$ are zero and the characteristic would become$$\lambda^2=0$$again by applying Cayley-Hamilton's theorem we finally have$$A^2=0$$

Answer 9

There must be some vector $v_0$ such that $Av_0=0$. If that's not intuitively obvious, here's a proof:

We have that $A^n=0$ for all $v$. We can rewrite that as $A(A^{n-1}v)=0$ for all $v$. Thus, we can pick any vector to be $v$, and defining $v_0=A^{n-1}v$ will give us a vector such that $Av_0=0$.

Since we're dealing with 2-dimensional space, there must be some $v_1$ such that $\{v_0,v_1\}$ is a basis for the space. So we can write $Av_1$ in terms of those vectors: $Av_1=c_0v_0+c_1v_1$ for some $c_0,c_1$. If we define $v_2=Av_1$, by the linearity of matrix multiplication, we have we have that $Av_2=A(c_0v_0+c_1v_0)=c_0Av_0+c_1Av_1=0+c_1(c_0v_0+c_1v_1)=c_1Av_1$. So there is some real number such that $Av_2=c_1v_2$ (that is, $v_2$ is an eigenvector with eigenvalue $c_1$). So now $A^nv_1=A^{n-1}(Av_1)=A^{n-1}v_2=c_1^{n-1}v_2$. Since we have to have $A^b=0$, we have to have $c_1^{n-1}v_2=0$. And if $c_1^{n-1}v_2=0$, then it follows that $c_1v_2=0$. So $A^2$ sends both basis vectors to the zero vector, hence it is the zero matrix.

This proof can be adapted to show that for any matrix of size $k \times k$, if $A^n=0$, then $A^k=0$.