Construct a $3 \times 3$ matrix $A$ with $C(A) \subset N(A)$ Question: Construct a $3 \times 3$ matrix $A$ with $C(A) \subset N(A)$. Where $C(A)$ is the column space of a matrix and $N(A)$ is its nullspace. Excluding the $0$ matrix.
I'm having trouble thinking this through and coming up with a structured argument to arrive at a solution.
What I've thought of so far:
Since I need $C(A) \subset N(A)$ then it would most likely help me if I can find a basis for $N(A)$. Because I need to be able to write any vector in $C(A)$ as linear combination of the basis of $N(A)$.
So how to do that? Well to find a basis for a null space of a matrix I would have to use the "free" variables not used as pivots in the matrix. With some messing around while writing this question out I got this matrix:
$$ 
\left[ \begin{array}\
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0 
\end{array} \right]
$$
So letting $x_3$ be the free variable I would get a basis vector of the form
$$
\left[ \begin{array}\
-1 \\
-1  \\
 1 
\end{array} \right]
$$
Which at least does reside in the $N(A)$, but I'm not able to write out the column space of my matrix from this vector.
What is the underlying idea I believe to accomplish this is to play with the matrix that satisfies $Ax = 0$. But I'm still having issues in getting the column space to come from linear combinations of these basis vectors.
How could I think through this properly? Right now it is just a bunch of throwing things at the wall with not too much order.
 A: The columnspace of a matrix $A$ is just the range of the linear map $x \mapsto Ax$, and the nullspace of $A$ is just the kernel. To say $C(A) \subseteq N(A)$ is to say, if we take any arbitrary vector in the range $C(A)$ (i.e. any vector of the form $Ax$), then multiplying by $A$ once more will produce the zero vector, i.e. $A^2x = 0$. This has to hold for all $x$, so
$$A^2 = 0.$$
Indeed, this condition is equivalent to $C(A) \subseteq N(A)$; if $A^2 = 0$ and we have $Ax \in C(A)$, then $A(Ax) = A^2x = 0$, so $Ax \in N(A)$.
This makes $A$ nilpotent, though not any non-zero nilpotent matrix will do (e.g. some satisfy $A^3 = 0$, but $A^2 \neq 0$). But, this gives you a great place to start. For example, the only eigenvalues of $A$ can be $0$ (if $Av = \lambda v$ for some $v \neq 0$, then $0 = A^2 v = A(\lambda v) = \lambda^2 v \implies \lambda^2 = 0 \implies \lambda = 0$), so if you're considering upper triangular matrices (such as your guess above), then every entry of the diagonal has to be $0$.
Indeed, it's a good idea to consider matrices of the form
$$\begin{bmatrix} 0 & * & * \\ 0 & 0 & * \\ 0 & 0 & 0 \end{bmatrix},$$
as every matrix is similar to an upper-triangular matrix. If there are any examples (which there definitely are), then there will be examples of this form!
A: HINT: Start by thinking about dimensions. What must $\dim C(A)$ be? Then start with a simple basis for a subspace of $\Bbb R^3$ of that dimension and go from there.
A: Let's note that $\dim C(A)+\dim N(A)=3$
We want $C(A)\subset N(A)$ so $\dim C(A)\lt \dim N(A)$ (note that we can't have $\dim C(A)=\dim N(A)$).
Now, $\dim N(A)=3$ is ruled out as $A$ must have a non-zero column.
$\dim N(A)=0$ would mean $A$ is  full rank so this case is also excluded.
$\dim N(A)=2$ is the only possibility. So consider:
$A=\begin{bmatrix}1&1&0\\-1 & -1 & 0\\0&0&0 \end{bmatrix}$
Note that $A_{.,1}$ (first column of $A$) forms a basis of $C(A)$ and since $AA_{.,1}=[0]_{3\times 1}$, it follows that $A_{.,1}\in C(A)$.
