Is $\mathbf{A}\mathbf{A}^T$ symmetric?

Question

Is matrix

$$\mathbf{B}=\mathbf{A}\mathbf{A}^T$$

necessarily symmetric (i.e., does $\mathbf{B}=\mathbf{B}^T$)?

Writing $$b_{ij}=a_{ij}a_{ji}$$ seems wrong because the $j$ on the LHS is different from the repeated/summed $j$ on the RHS. I'm not sure where to proceed in the proof or disproof from here.

I don't know how to proceed from here, either: $$\mathbf{B}^T=\left(\mathbf{A}\mathbf{A}^T\right)^T.$$

Once you are comfortable with that $(EF)^T = F^TE^T$ then use that but with $E=A$ and $F=A^T$. Couple that result with that $(A^T)^T=A$ and remark how it is equal to your original expression to complete the proof. — JMoravitz, Jan 19 '23 at 02:35
See also: https://math.stackexchange.com/questions/1279861/why-intuitively-is-the-order-reversed-when-taking-the-transpose-of-the-product?noredirect=1&lq=1 — JMoravitz, Jan 19 '23 at 02:37
@JMoravitz Not exactly my question, because that one doesn't deal with symmetric matrices, but related. — Geremia, Jan 19 '23 at 02:40
The other questions are more general than yours. They tell you exactly how you can proceed from $\mathbf{B}^T=\left(\mathbf{A}\mathbf{A}^T\right)^T.$ — David K, Jan 19 '23 at 03:00
@DavidK So more specific questions are considered duplicates of general questions? — Geremia, Jan 21 '23 at 18:56
My comment had more to do with an approach to mathematics than the closing of this question. The technique for transposing a product of matrices is one that is worth knowing; it's good for symmetric matrices and non-symmetric matrices. In this case, $(AA^T)^T=(A^T)^TA^T=AA^T,$ done. — David K, Jan 22 '23 at 00:33

score 3 · Answer 1 · answered Jan 19 '23 at 02:39

3

Yes.

$$\mathbf{B}^T=\left(\mathbf{A}\mathbf{A}^T\right)^T=\mathbf{A}\mathbf{A}^T=\mathbf{B},$$ from the fact $(\mathbf{A}\mathbf{B})^T=\mathbf{B}^T\mathbf{A}^T$ and $(\mathbf{A}^T)^T=\mathbf{A}$.

Q.E.D.

answered Jan 19 '23 at 02:39

Geremia

2,405

score 3 · Accepted Answer · answered Jan 19 '23 at 02:59

You can prove this with sums as well: Note that

$$b_{ij} = \sum_{k = 1}^n a_{ik}a_{jk} $$

and

$$ b_{ji} = \sum_{k = 1}^n a_{jk}a_{ik} $$

where the second factor in each summand uses that $(a_{ij})^T = a_{ji}.$ The sums are equal, and hence $B$ is symmetric.

Is $\mathbf{A}\mathbf{A}^T$ symmetric?

2 Answers2