ionic haven
I am asked to verify the following statement:
If $A \in \mathcal{M}{m}(\mathbb{K})$ satisfies $\textrm{im}(A) \subseteq \textrm{ker}(A)$, then for all $x \in \mathbb{K}^m$ it verifies that $A^2x = \theta{\mathbb{R}^m}$
I managed to prove that $\forall x \in \textrm{im}(A)$, $A^2x = \theta$ because all vectors in the kernel verify that by multiplying the matrix by them we get $\theta$, but don't know how to prove or disprove that this would work for every vector in $\mathbb{K}^m$
inland tangleBOT
forest latchBOT
Spiwocoal
halcyon solar
This statement is not true. If $\textrm{im}(A) \subseteq \textrm{ker}(A)$, then $A$ is called a "nilpotent matrix," and it means that $A^k = \mathbf{0}$ for some positive integer $k$. However, this does not necessarily mean that $A^2x = \theta_{\mathbb{R}^m}$ for all $x \in \mathbb{K}^m$.
To see why this is the case, consider the following example. Let $A \in \mathcal{M}_2(\mathbb{R})$ be the matrix
$$A = \begin{bmatrix} 0 & 1 \ 0 & 0 \
\end{bmatrix}.$$
It is easy to see that $\textrm{im}(A) = \textrm{span}(\begin{bmatrix}
1 \ 0
\end{bmatrix}) \subseteq \textrm{ker}(A) = \textrm{span}(\begin{bmatrix} 1 \ 0 \end{bmatrix})$, so $A$ is a nilpotent matrix. However, we have $A^2 = \begin{bmatrix}
0 & 0 \
0 & 0 \
\end{bmatrix}$, so $A^2 x \neq \theta_{\mathbb{R}^m}$ for all $x \in \mathbb{K}^m$.
In general, for a nilpotent matrix $A \in \mathcal{M}m(\mathbb{K})$, there exists a positive integer $k$ such that $A^k = \mathbf{0}$, but this does not imply that $A^2x = \theta{\mathbb{R}^m}$ for all $x \in \mathbb{K}^m$.