# Generalization of sum of outer product

Consider a matrix $$A \in \mathbb{R}^{d \times m}$$ such that $$m \geq d$$ and denote its columns i.e $$A_{:, i}$$ by $$a_i$$. Let $$AA^T$$ is invertible.

Now, consider the sum $$S(A) = \sum_{r=1}^m a_r a^T_r$$ which is an $$d \times d$$ matrix.

Note that each $$a_r a_r^T$$ is a rank 1 matrix. Note that when m = d, it follows from another question.

Is rank(S(A)) = d ?

Example: A = $$\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 1 \end{bmatrix}$$

S(A) = $$\begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix}$$

S in this example is full rank.

• It's not clear what you're asking. It seems to me that $S(A) = AA^T$. So, if $AA^T$ is invertible, of course $S(A)$ is invertible since it is the same matrix. – Omnomnomnom Apr 5 at 2:35
• Thanks. I did not see that. – listener Apr 6 at 1:31

## 1 Answer

as @Omnomnomnom pointed out: Looks like $$S(A) = AA^T$$.

• Since the answer is from OP himself/herself, I think it's a valid answer. – AspiringMathematician Apr 6 at 4:04