Is matrix $A^TA$ always symmetric?

Question

Through experience, I've seen that the following statement holds true: "$A^TA$ is always a symmetric matrix?", where $A$ is any matrix. However can this statement be proven/falsified?

Answer 1

Ideally we've already proved both $(A^T)^T=A$ and $(AB)^T=B^T A^T$. If not, prove these first. Then $(A^T A)^T=A^T (A^T)^T =A^TA$.

Answer 2

We know $(AB)^T=B^TA^T$, so $(A^TA)^T=A^T(A^T)^T=A^TA$ and hence $A^TA$ is always symmetric.

Answer 3

Another proof per element. Let $T$ be a transpose of $A$, meaning $A^T = T$. We want to proof that $R = AT$ is symmetric, i.e. $R_{i,j} = R_{j,i}$.

1. We know that $A$ has ($m\times n$) elements, and $T$ has ($n\times m$) elements.

2. Generally, a row of $A$ with an index ($k$) is the $k^{th}$ column of its transpose, $T$: $$A_k^{row} = T_k^{col}$$

3. Similarly, the $k^{th}$ column of $A$ is the $k^{th}$ row of its transpose: $$A_k^{col} = T_k^{row}$$

4. Per matrix multiplication, element $R_{i,j}$ is the dot product between the $i^{th}$ row of first matrix and the $j^{th}$ column of the second matrix: $$\;\; R_{i,j}= A_i^{row} \cdot T_j^{col}$$

5. And per equality in (2), the $T_j^{col}$ is the $A_j^{row}$: $$ R_{i,j}= A_i^{row} \cdot A_j^{row}$$

6. When switching indices $$ R_{j,i}= A_j^{row} \cdot T_i^{col}$$

7. According to (2) again, the $T_i^{col}$ is the $A_i^{row}$: $$ R_{j,i}= A_j^{row} \cdot A_i^{row}$$

8. From (5) and (7) we see that $R_{i,j} = R_{j,i}$ because dot product is commutative.

That also shows you a quick way to calculate a matrix multiplied by its transpose. Just calculate the dot product on the rows of the first matrix.

Example

$R = AA^T$

$$A = \begin{bmatrix} 1 & 3 & 5\\ 4 & 2 & 0 \end{bmatrix}$$ To calculate $AA^T$, just calculate the dot product of the rows of the first matrix, $A$ here.

• $R_{1,1} = A_1^{row} \cdot A_1^{row} = \begin{bmatrix}1 \\ 3 \\ 5\end{bmatrix} \cdot \begin{bmatrix}1 \\ 3 \\ 5\end{bmatrix} = [1 + 9 + 25] = [35] $
• $R_{1,2} = A_1^{row} \cdot A_2^{row} = \begin{bmatrix}1 \\ 3 \\ 5\end{bmatrix} \cdot \begin{bmatrix}4 \\ 2 \\ 0\end{bmatrix} = [4 + 6 + 0] = [10] $
• $R_{2,1} = A_2^{row} \cdot A_1^{row} = \begin{bmatrix}4 \\ 2 \\ 0\end{bmatrix} \cdot \begin{bmatrix}1 \\ 3 \\ 5\end{bmatrix} = [4 + 6 + 0] = [10] $
• $R_{2,2} = A_2^{row} \cdot A_2^{row} = \begin{bmatrix}4 \\ 2 \\ 0\end{bmatrix} \cdot \begin{bmatrix}4 \\ 2 \\ 0\end{bmatrix} = [16 + 4 + 0] = [20] $

$$R = \begin{bmatrix} 35 & 10\\ 10 & 20 \end{bmatrix}$$

$R = A^TA$

Do the same for rows of $A^T$, answer is below:

$$R = \begin{bmatrix} 17 & 11 & 5\\ 11 & 13 & 15\\ 5& 15 & 25\\ \end{bmatrix}$$

You can also apply the rule for the columns of the second matrix, doesn't matter.