# Is $AA^T$ a positive-definite symmetric matrix?

Let $A\in M_{n\times n}(\mathbb{R})$ be a matrix. Is it true that $AA^{T}$ is positive-definite?

Clearly $AA^{T}$ is symmetric. I have shown that a symmetric matrix $S\in M_{n\times n}(\mathbb{R})$ is positive-definite if and only if $S$ has only positive eigenvalues. Can this be helpful?

Hint: let $v$ be a non zero vector; then, setting $B=A^T$ for simplicity, $$v^TB^TBv=(Bv)^T(Bv)$$ is positive if and only if $Bv\ne 0$. How can you ensure that $Bv\ne0$ if and only if $v\ne0$?

Conversely, if $AA^T$ is positive definite, what can you say about the rank of $A$?

So, what's a necessary and sufficient condition so that $AA^T$ is positive definite?

• Is it also true for matrices over complex field? i.e Is $AA^*$ always positive definite? – Ziya May 7 '17 at 15:24
• @junk Positive semidefinite; $A$ needs to be invertible for $AA^*$ to be positive definite (if $A^*$ denotes the Hermitian transpose, of course). The proof is essentially the same. – egreg May 7 '17 at 15:31
• Does this also hold for non-square $A$? – daneel May 12 '18 at 17:58

$AA^T$ is not necessarily positive definite, but it is positive semi-definite, meaning that $\langle x, AA^Tx \rangle \ge 0$ for all vectors $x$. To see this, note that $\langle x, AA^Tx \rangle = \langle A^Tx, A^Tx \rangle = \Vert A^Tx \Vert^2 \ge 0$. A counter example to positive definiteness is provided, when $n = 2$, by taking

$A= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}; \tag{1}$

then

$AA^T = A, \tag{2}$

so if $x = (0, 1)^T$,

$\langle x, AA^Tx \rangle = 0. \tag{3}$

It is easy to generalize this example by taking $A$ to be a diagonal matrix in $M_{n \times n}(\Bbb R)$ with at least one zero on the diagonal; many other generalizations are also possible.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

• but if we take A to be non-singular, then it will positive definite, right? – blabla Dec 21 '16 at 0:39
• Yes, if $A$ is nonsingular, so is $A^T$, since $(A^T)^{-1} = (A^{-1})^T$. Thus for any $x$, $A^Tx \ne 0$, whence $\Vert A^Tx \Vert \ne 0$. But it is easy to see that $\langle A^T x, A^T x \rangle = \langle x, AA^T x \rangle$. – Robert Lewis Dec 21 '16 at 9:19
• Is it also true for matrices over complex field? i.e Is $AA^*$ always positive definite? – Ziya May 7 '17 at 15:28
• @Ziva: if by $A^\ast$ you mean what is customarily denoted by $A^\dagger$, the Hermitian adjoint of $A$, the answer is affirmative. If $A^\ast$ is merely the complex conjugate of $A$, I doubt it, but have no counter examples to offer at present. – Robert Lewis May 7 '17 at 19:32

For a matrix to be positive definite we need $x^TMx \geq0$. Consider the case where the matrix A is not full rank so therefore has more rows than columns. It should be clear there exist a vector $x$ where $x \neq 0$, such that $x^TA = 0$. Thus we have $x^TAA^Tx = 0$ Therefore $AA^T$ cannot be strictly positive definite. (But you can show it is always semi-positive definite.)

• This is imprecise, and somewhat wrong, even though the idea is OK. 1. "A is not full rank" should be "A is not full row rank". 2. "therefore has more rows than columns" is wrong, because "not full (row) rank" does not necessarily imply that it "has more rows than columns". Zero matrix is an obvious counter-example, but you can take any matrix with more than one row, and more columns than rows (so called wide matrix) and all of the rows mutually equal. It will not have a full (row) rank, but it will also have more columns than rows. – Vedran Šego Mar 28 '14 at 16:59