# Is matrix $A^TA$ always symmetric?

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?

• Do you know what $(A\cdot B)^T$ is? Aug 12, 2013 at 13:38
• Clearly your definition of symmetric is not (literally) to equal its own transpose. What is your definition? Aug 12, 2013 at 15:14

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

• These equivalencies look fine, but I don't understand how they prove symmetry. N. Osil's answer below seems to make more sense. Jun 2, 2022 at 15:19

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.

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

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

### $$R = A^TA$$

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.