# Show that if A^2 is the zero matrix, then the only eigenvalue of A is 0.

The aim of this question is to prove the statement for only the eigenvalue of $A$ to be zero.

The concept behind this question is the knowledge of eigenspace and eigenvalue.

Suppose that a non-zero value $\lambda$ is an eigenvalue of the vector $A$ and the corresponding eigenvector = $\vec{ x }$.

As given in the question statement, we have:

$A^2=0$

We can write that:

$\vec{ 0} =\ \left[ \begin{matrix} 0 & 0\\0 & 0\\ \end{matrix} \right]\ \vec{x}$

$\vec{ 0} = A^2 \vec{x}$

$\vec{ 0} = A \lambda \vec{x}$

$\vec{ 0} = \lambda^2 \vec{x}$

This is proved as:

Let us suppose a vector $v$ such that it is a non-zero vector and fulfills the following condition:

$A \times v = \lambda v$

Thus we can write that:

$= A^2 \times v$

$= A \times \left( A \times v \right)$

$= A \left( \lambda v \right)$

$= \lambda \left( A \times v \right)$

$=\lambda^{2 } v ≠0$

And therefore we can say that $A^2 ≠ 0$

As $\vec{x} ≠ \vec{0}$ , this concludes that $\lambda^2$ = 0  and therefore the only possible eigenvalue is $\lambda = 0$.

Otherwise then $A$ would be invertible, and so would $A^2$ since it is the product of invertible matrices.

## Numerical Results

$A \times v = \lambda v$

Thus, we can write:

$= A^2 \times v$

$= A \times \left( A \times v \right)$

$= A \left( \lambda v \right)$

$= \lambda \left( A \times v \right)$

$=\lambda^{2 } v ≠0$

And therefore, we can say that $A^2 ≠ 0$

## Example

Find the basis for the given eigenspace, corresponding to the given eigenvalue:

$A =\ \left[ \begin{matrix} 4 & 1\\3 & 6\\ \end{matrix} \right]\ ,\lambda=3 , \lambda = 7$

For given $\lambda = 3$ will be equal to $A -\ 3I$

This will be:

$\left[ \begin{matrix} 1 & 1\\3 & 3\\ \end{matrix} \right]\ \sim \left[ \begin{matrix} 1 & 1\\0 & 0\\ \end{matrix} \right]\$

So the basis for the given eigenspace, corresponding to the given eigenvalue $\lambda = 3$ is:

$= \left[\begin{matrix} 1 \\ -1 \\ \end{matrix} \right]$

For given $\lambda = 7$ will be equal to $A -\ 7 I$

This will be:

$\left[ \begin{matrix} -3 & 1\\3 & -1\\ \end{matrix} \right]\ \sim \left[ \begin{matrix} -3 & 1\\0 & 0\\ \end{matrix} \right]\$

So the basis for the given eigenspace, corresponding to the given eigenvalue $\lambda = 7$ is:

$= \left[\begin{matrix} 1 \\ 3 \\ \end{matrix} \right]$

So the basis for the given eigenspace, corresponding to the given eigenvalue $\lambda = 3$ and $\lambda = 7$  are:

$Span = \left[\begin{matrix} 1 \\ -1 \\ \end{matrix} \right]$

$Span = \left[\begin{matrix} 1 \\ 3 \\ \end{matrix} \right]$