Matrix multiplication after removing rows

I'm completely stuck on the following proof:

$A^{T}_{(j)}A_{(j)} = A^{T}A-a^{T}_{j}a_{j}$

Here $A_{(j)}$ represents the matrix $A$ when leaving out the $j$-th row. $a_{j}$ is the $j$-th row vector of $A$.

I think leaving out the j-th row corresponds to making that row be all 0's. So then each column in $A_{j}$ would receive no component from the j-th row of $A^{T}_{(j)}$ during the multiplication, but I'm having trouble decomposing it to $a_j$ in the first place. Can anyone help? I'd really appreciate it.

Using indices you may write: $$(A^T A)_{kl} = \sum_{m} A_{mk} A_{ml} = \sum_{m\neq j} A_{mk} A_{ml} + A_{jk} A_{jl} = \sum_{m\neq j} A_{mk} A_{ml} + (a_{j})_{k} (a_{j})_{l}$$
• k and l are indeed the column indices. The product of two matrices $B$ and $A$ may be written $(BA)_{kl} = \sum_m B_{km} A_{ml}$ which you use on $B=A^T$ (whence reverse indices). In the second and third sum I omit the j'th element (which corresponds to removing the j'th row in $A$. Commented Oct 12, 2016 at 21:15