# Let $A$ be a $3×4$ matrix. Estimate $\det(A'A)$ and $\det(AA')$

Let $A$ be a $3×4$ matrix. Estimate $\det(A'A)$ and $\det(AA')$.

I would first assume that $A$ has rank $3$. Then $A'A$ would be a $4\times 4$ matrix with rank $3$ and therefore it would have zero determinant. If the rank equals $4$ I can't say anything about the determinant only that it is not zero.

I can't estimate $\det(AA')$

Am I right?

• Is $A'$ the transpose of $A$? Note that $A$ is a 3x4 matrix, so its rank cannot be larger that 3. – Crostul Nov 13 '14 at 11:14
• I'm not sure what you mean with "if Rank=4 i cant say something about the det only that it is not Zero". The rank is not $4$ as you've correctly said. As for $\det\left(AA^T\right)$, it can take any non-negative value. – Git Gud Nov 13 '14 at 11:16

The rank of $A$ is at most $3$; if it is $3$, then $AA'$ has the same rank, so it's invertible and will have nonzero determinant. Otherwise the determinant is $0$.
The rank of $A'A$ is at most $3$, so this matrix always has zero determinant.
What the determinant of $AA'$ is, in case $\operatorname{rank}A=3$ cannot be “estimated”: just take $$A=\begin{bmatrix} a & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{bmatrix}$$ and $\det(AA')=a^2$. So any positive number can be the determinant.
Note however that all eigenvalues of $AA'$ are real (the matrix is symmetric) and non negative: indeed, if $AA'v=\lambda v$, then $$0\le(A'v)'(A'v)=v'AA'v=v'(\lambda v)=\lambda v'v$$ and therefore $\det(AA')\ge0$.
If $A$ is $3\times4$ with rank $3$, then (considering matrices as maps on column vectors) $A$ is surjective and $A'$ is injective, thus $\dim\ image(A'A) = 3 = rank(A'A)$. But since $A'A$ is $4\times4$, we must have $\det(A'A) = 0$, as you already noted. Now, from $rank\ A' = 3$ we get $ker\ A' = 0$, i.e. $x \neq 0 \Rightarrow A'x \neq 0$. Now assume there is a $y\neq0$ such that $AA'y = 0$. This implies $0 = y'AA'y = (A'y)'(A'y) = \|A'y\|^2 > 0$ since $A'y\neq0$ since $y\neq0$. This is a contradiction. This means that $ker\ AA' = 0$. This together with the fact that $AA'$ is square $3\times3$ implies that $AA'$ has full rank, namely $3$. So $AA'$ is invertible, and we must have $\det(AA') \neq 0$.