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I have a $J \times J$ matrix $C$ that is upper triangular. Also, $C'C$ is positive definite. I also have a matrix $A$ formed by submatrices of size $J \times K$ as follows $$ A = \begin{bmatrix} A_1 \\ A_2 \\ A_3 \\ \vdots \\ A_N \end{bmatrix}, $$ where $A_i = [I_{J \times J}\ B_{i_{J \times (K - J)}}]$ for each $I$.
I create a new matrix $A_{new}$ such that $$ A_\textit{new} = \begin{bmatrix} CA_1 \\ CA_2 \\ CA_3 \\ \vdots \\ CA_N \end{bmatrix}. $$ Notice that $A_\textit{new}$ can be written as $$ A_\textit{new} = \begin{bmatrix} C & CB_1 \\ C & CB_2 \\ \vdots & \vdots \\ C & CB_N \end{bmatrix}. $$ Is there a way for me to write $A_\textit{new}'\cdot A_\textit{new}$ in terms of matrix operations or concatenations?

I have a loop where I need to obtain a matrix $A_\textit{new}$ for different values of $C$. This is extremely costly computationally as the matrices $B_i$ and $C$ are huge. Any help is appreciated!

Thank you!

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  • $\begingroup$ Write $Q=C'C$. Then $Q$ is a perfectly arbitrary positive definite martrix. Your target matrix breaks up into blocks of the form $\pmatrix{Q&QB_i\cr B_i'Q&B_i'QB_i\cr}$ where the $B_i$ are perfectly arbitrary. So it seems like your problem is equivalent to the problem of finding a fast way to compute $QB,B'Q$ and $B'QB$ where $Q$ is positive definite and otherwise $B$ and $Q$ are perfectly arbitrary. $\endgroup$
    – Steven Landsburg
    Commented Jun 13 at 23:56
  • $\begingroup$ Hi @StevenLandsburg. Thank you for checking my question. That is correct. $A_{new}'*A_{new} = \begin{bmatrix} N*Q& \sum_i QB_i \\ \sum_i B_i'Q & \sum_i B_i'QB_i \end{bmatrix}$. However, In this case the computation of the right lower block is computationally costly. Specially because in my case N is around 170k, K is 176 and J = 146. $\endgroup$
    – LauraH
    Commented Jun 14 at 13:44

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