# For every *non-square* matrix prove that $AA^t$ or/and $A^tA$ is singular

For every non-square matrix prove that $AA^t$ or/and $A^tA$ is singular.

Like the title, I want to prove this and I tried to think of ways to prove it but I couldn't think of some..

I know by this answer that $AA^t$ is symmetric but I cant make the connection.

If someone has at least a hint to that It'll be great if you could write it in the comment section so I could give it a shot!

• Use the fact that $rank(AB) \leq rank(A)$ for any matrices $A,B$ for which the product is defined. Additionally, rank($A$) = rank($A^t$). Finally, look at the dimensions of $AA^t$ and $A^tA$ and you will see that at least one must be rank deficient, i.e. singular. Oct 24 '16 at 21:54
• Hint: If $A$ is an $n\times m$ matrix with $n < m$, then $A$ must not be one-to-one. I.e., there is a $v \neq 0$ such that $Av = 0$. Then the same is true of $A^t A$ (just try applying it to $v$ as before). Oct 24 '16 at 21:59
• @KenDuna So because $rank(A^tA)\leq rank(A)$ (and $AA^t$) I need to prove that at least one of them is less than $rank(A)$, because Invertible matrix has full rank, correct? Oct 24 '16 at 22:13
• @Zik332 A square matrix is singular iff it is not one-to-one iff it is not onto. Oct 24 '16 at 22:23
• @Zik332 As I pointed out, $v$ is in its kernel. So it is not one-to-one. Oct 24 '16 at 22:40

To elaborate on my hint, suppose $A$ is an $n \times m$ matrix and $n \neq m$. It must be that $rank(A^t) = rank(A) \leq \min(n,m) < \max(n,m)$.
Using the fact that $rank(AB) \leq rank(A)$ for any $A,B$ for which the product is defined, we have that:
$$rank(AA^t) \leq rank(A) < \max(n,m)$$ $$rank(A^tA) \leq rank(A^t) < \max(n,m).$$
But it must be the case that the dimensions of $AA^t$ or $A^tA$ is $\max(n,m)$. Therefore at least one of them does not have full rank. For square matrices, not having full rank is equivalent to being singular.