What does $\ker(A) \cap \operatorname{im}(A)$ say about the rank and nullity of $A$

linear algebramatricesmatrix-rank

I'm trying to investigate what $\ker(A) \cap \operatorname{im}(A)$ implies for $\operatorname{rank}(A)$ and $\operatorname{null}(A)$. Specifically, if we have some vector, say
$$
\newcommand{\im}{\operatorname{im}}
\newcommand{\rank}{\operatorname{rank}}
\newcommand{\null}{\operatorname{null}}
v = \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}
$$
in $\ker(A) \cap \im(A)$, what does that imply about $\rank(A)$ and $\null(A)$?

Here is some work I've done so far:

  • $A$ must be a $n \times 4$ matrix. That means $A$ has four columns, and so $\dim(A) = 4$, and so $\rank(A) + \null(A) = 4$ by the Rank-Nullity Theorem.
  • Clearly, $\rank(A)$ cannot be $0$, because $\rank(A) = \dim(\im(A)) > 0$, since we know the given vector $v$ is in $\im(A)$ (which means $\im(A)$ cannot have dimension $0$). This implies $\null(A)$ cannot be $4$. Likewise, $\null(A)$ cannot be $0$ because $v \in \ker(A)$.
  • In fact, $A$ must be a $4 \times 4$ matrix. This is because $v \in \ker(A)$ implies $Av$ is a valid matrix product (so $v$ has the same number of rows as $A$'s columns; that is, $4$), and because $v \in \im(A)$ implies that the product $Aw$ for some vector $w$ has length $4$, which implies $A$ has $4$ rows as well.

However, I haven't been able to find examples of (or prove the impossibility of) when $\rank(A) = 1, 2, 3$. Could anyone finish this problem?

Best Answer

Let $B$ be the following matrux: $$B=\left(\begin{array}{cccc} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0 \end{array}\right).$$ Note that the nullspace is spanned by $(1,0,0,0)$ (transpose where appropriate) and the range by $(1,0,0,0)$, $(0,1,0,0)$, and $(0,0,1,0)$. So the nullspace is contained in the range, and $B$ has rank $3$.

Now, "my" intersection is spanned by $(1,0,0,0)$, and you want it to have $(1,2,3,4)$ instead. So let $C$ be given by $$C=\left(\begin{array}{cccc} 1&0&0&0\\ 2&1&0&0\\ 3&0&1&0\\ 4&0&0&1 \end{array}\right), \quad\text{so }C^{-1}= \left(\begin{array}{rccc} 1&0&0&0\\ -2&1&0&0\\ -3&0&1&0\\ -4&0&0&1 \end{array}\right).$$

Now let $A=CBC^{-1}$. Since $C^{-1}$ sends $(1,2,3,4)$ to $(1,0,0,0)$, and $B$ maps it to $(0,0,0,0)$, we have that $(1,2,3,4)$ lies in the nullspace of $A$. On the other hand, an easy calculation shows that $A$ sends $(0,1,0,0)$ to $(1,2,3,4)$, so $(1,2,3,4)$ lies in the intersection of the nullspace and ranges of $A$.

Taking $A^2$ you get a matrix with rank 2 whose range equals the nullspace. Here $(1,2,3,4)$ is the image of $(0,0,1,0)$. Finally taking $A^3$ you get. matrix of rank 1 whose range is contained in the nullspace, and here the image of $(0,0,0,1)$ is $(1,2,3,4)$.

You can get any particular vector on the intersection doing something similar. So the only thing you can deduce from the intersection being nontrivial is that the matrix has rank less than $4$ and more than $0$. Similar statements hold for any dimension: you just get that the matrix is nonzero and not of full rank. All other ranks are possible.

Even more generally, for any $n$ and any $k$, $1\leq k\leq \frac{n}{2}$, given any $k$ linearly independent vectors $v_1,\ldots v_k$, you can find an $n\times n$ matrix $A$ with $v_1,\ldots,v_k\in N(A)\cap R(A)$, and $k\leq\mathrm{rank}(A)\leq n-k$.

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