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Statement: Let $A$ be a $5 \times 4$-Matrix with row rank 3. Show that $Ax=0$ has nontrivial solutions.

This is a homework problem, as a hint I am given one formula (which we didn't discuss in class yet) but intuition wise I don't understand how it should help me at all to verify this statement. My tutor went about this as follows, consider: $$ F: \begin{cases} V & \longrightarrow W \\ v & \longmapsto A \circ v \end{cases} $$ know that: Row Rank of A = Column Rank of A = $\dim(\text{Im}(A))=\dim(\text{Im}(F))$ and make use of

$\dim(V)=\dim(\text{Kernel}(F))+\dim(\text{Im}(F)) \tag{given}$

I do understand the given Formula, but I don't understand how to apply it, since the Row rank of $A$ is equal to $3$, so is the Column Rank and therefore the dimension of the Image of $F$ is also equal to $3$. Can I make a statement about the dimension of $V$? I know that $v \in \mathbb{R}^4$ but that doesn't help me much with the dimension of $V$.

Some hints or insights, especial intuition wise insights, would be very helpful and appreciated.

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Shouldn't it be nontrivial solutions? – egreg Nov 26 '13 at 21:52
Oh you're right, excuse me please! My english is very troublesome today, I will update it immediately +1 – Spaced Nov 26 '13 at 21:53
up vote 1 down vote accepted

The nullity-rank theorem is stated for the general case and you can apply it to your situation:

Let $V=\mathbb{R}^4$, $W=\mathbb{R}^5$ and define $F(v)=Av$, for $v\in\mathbb{R}^4$. Then the image of $F$ is the column space, which has dimension $3$ because $A$ has rank $3$. Therefore $N(A)$, which is the solution set of $Ax=0$ and the kernel of $F$, has dimension $4-3=1$ and so contains nonzero vectors.

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Thank you very much, all it took was your phrasing for me to understand this problem with a special highlight on "and the kernel of $F$, has dimension $1$ and so contains nonzero vectors" thanks. – Spaced Nov 26 '13 at 22:04

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