Q:

Dimension of the image of a linear transformation

$$A=\begin{bmatrix}
0 &0& *& *\\
0 &0& *& *\\
0 &0& 0& *\\
0 &0& 0& 0
\end{bmatrix}$$
(* denotes a non-zero element)
If $A$ is considered as a linear transformation on $\mathbb{R}^4$, what is the dimension of $Image(A)$ and $Image(A^2)$
By observation I think that dimension of $Image(A)$ and $Image(A^2)$ are 2 and 1 respectively. I observe the columns and how the basis elements are mapped. Is this answer correct?

A:

The dimension of the image of a matrix when regarded as a linear map is its rank.
So if $*$ denotes a non zero element then the last column of $A$ is not a multplite of the third since it has a $*$ in the third place and the third column has a zero. Hence they are linearly independent hence rank of $A$ is 2. Here is $A^2$:
$$\begin{bmatrix} 0 & 0 & 0 & * \\ 0 & 0 & 0 & * \\
0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
so rank $A^2$ is 1 hence dimension of image of $A^2$ is indeed 1.

