What is the relation between $\det(A^TA)$ and $\det(AA^T)$ for an $m\times n$ matrix $A$? [closed]

Let $A$ be an $m\times n$ real matrix, and $\det(\cdot)$ the determinant. It is rather trivial that when $A$ is of size $1\times 1$, then $\det(A^TA)=\det(AA^T)$.

In general, what is the relation between $\det(A^TA)$ and $\det(AA^T)$?

Solution 1:

Summary: If $n \neq m$, one will be $0$, while the other will be non-zero iff the rows or the columns of $A$ are linearly independent. If $n = m$, then the determinants are equal and both will be non-zero iff $A$ is invertible.

For parts A, B and C below, assume $m \neq n$.

A) Proof of "one will be $0$":

$A B$'s columns are a linear combination of the columns of $A$. Thus if $A B$ has more columns than $A$, then the columns of $A B$ will not be linearly independent.

In $A A^T$ and $A^T A$, one of these will have more columns than $A$ or $A^T$ respectively. Thus for one of these the columns will not be linearly independent. Now both of these are square matrices, thus one of these will be singular, thus $\det(A A^T) = 0$ or $\det(A^T A) = 0$.

B) Proof of "the other will be non-zero only if the rows or the columns of $A$ are linearly independent", i.e. proof of "the other will be zero if the rows and the columns of $A$ are not linearly independent":

Now $A (c_1 {\bf x} + c_2 {\bf y}) = c_1 (A {\bf x}) + c_2 (A {\bf y})$, more generally a matrix times a linear combination of column vectors is a linear combination of the matrix times the column vectors.

The $i$-th column of $A B$'s columns will be $A$ times the $i$-th column of $B$. Thus if $i$-th column of $B$ is a linear combination of $B$'s other columns, then the $i$-th column of $A B$ will be a linear combination of $A B$'s other columns.

Thus if the columns of $B$ are not linearly independent, then the columns of $A B$ will not be linearly independent.

If the columns and rows of $A$ are not linearly independent, then the columns of $A$ and the columns of $A^T$ will not be linearly independent, thus the columns of $A A^T$ and $A^T A$ will not be linearly independent, thus both are singular, thus $\det(A A^T) = 0 = \det(A^T A)$.

C) Proof of "the other will be non-zero if the rows or the columns of $A$ are linearly independent", i.e. if the other is 0, then the rows and the columns of $A$ are not linearly independent.

The other is zero iff $\det(A^T A) = 0 = \det(A A^T) = 0$.

Assume that $\det(A^T A) = 0$, thus $A^T A$ is singular, thus there exists an non zero ${\bf x}$ such that $A^T A {\bf x} = {\bf 0}$.

Let ${\bf b} = A {\bf x}$, thus ${\bf b}^T = {\bf x}^T A^T$, thus ${\bf x}^T A^T A {\bf x} = {\bf b}^T {\bf b} = \|{\bf b}\|^2$. But ${\bf x}^T A^T A {\bf x} = {\bf x}^T {\bf 0} = 0$. Now $\|{\bf b}\|^2 = 0$ iff ${\bf b} = {\bf 0}$ for any vector ${\bf b}$. Thus $A {\bf x} = {\bf 0}$ where $x$ is non-zero. Thus the columns of $A$ are not linearly independent.

Similarly if $\det(A A^T) = 0$, then the columns of $A^T$ are not linearly independent, which implies the rows of $A$ are not linearly independent.

Thus if $\det(A^T A) = 0 = \det(A A^T) = 0$, then the rows and the columns of $A$ are not linearly independent.

D) Proof of "If $n = m$, then the determinants are equal and both will be non-zero iff $A$ is invertible."

Assume $m = n$, thus $A$ and $A^T$ are square matrices, thus $\det(A A^T) = \det(A) \det(A^T) = \det(A)\det(A) = \det(A)^2$.

Similarly $\det(A^T A) = \det(A)^2$.

Now $\det(A^T A) = 0$ iff $\det(A)^2 = 0$ iff $\det(A) = 0$ iff $A$ is singular.

Solution 2:

Case 1: $A$ is a square matrix

If $A$ is a square matrix, then $\det(A^T)=\det(A)$, so $$ \det(AA^T)=\det(A)\det(A^T)=(\det(A))^2 $$ and the same holds for $\det(A^TA)$.

Case 2: $A$ is not a square matrix

Suppose $A$ is $m\times n$, with $m\ne n$. If $m>n$, then the rank of $AA^T$ is at most $n$ and therefore $\det(AA^T)=0$. If $A$ has rank $n$, then $A^TA$ has rank $n$ (see later), so $\det(A^TA)\ne0$. Similarly if $m<n$.

Thus, the equality $\det(AA^T)=\det(A^TA)$ only holds if and only if the rank of $A$ is strictly less than $\min\{m,n\}$.

Exercise

The rank of $A$ equals the rank of $AA^T$ and of $A^TA$.

Solution 3:

The statement isn't true for generic rectangular matrices $m\neq n$ even though both expressions $\det(AA^T)$, $\det (A^T T)$ are well-defined.

For example, if $A$ is the row containing the entries $(a,b)$, then $AA^T$ is the $1\times 1$ matrix with the entry $a^2+b^2$ whose determinant is $a^2+b^2$.

On the other hand, $A^TA$ is the $2\times 2$ matrix with the rows $(a^2,ab)$, $(ab,b^2)$ whose determinant is zero. Not surprising because the rows aren't linearly independent. Note that the terms of this determinant, $a^2 b^2$, don't even have the same "units" as the terms in the first determinant. The vanishing of the "larger" among the two matrices holds in general.

For $m=n$, $\det A=\det A^T$ is well-defined and one may use $$\det(AA^T)=\det(A^T A) = \det(A)\det(A^T)=[\det(A)]^2$$ so the identity holds.