0
$\begingroup$

Let $ \sigma_1 $ be the 2-norm of $\mathbf A$; there exist unit length vectors $\mathbf x_1 \in \mathbb{C}^m,\space \mathbf x_1^*\mathbf x_1 = 1 $ and $ \mathbf y_1 \in \mathbb{C}^n,\space \mathbf y_1^*\mathbf y_1 = 1$ , such that $ \mathbf A \mathbf x_1 = \sigma_1 \mathbf y_1. $ Define the unitary matrices $ \mathbf V_1, \mathbf U_1 $ so that their first column is $ \mathbf x_1, \mathbf y_1 $, respectively:$ \mathbf V_1 = [\mathbf x_1\space \hat{\mathbf V}_1],\space \mathbf U_1 = [\mathbf y_1\space \hat{\mathbf U}_1] $

How can I conformally partition $\mathbf A$ so that I can perform the following matrix muliplication? $$ \mathbf U_1^* \mathbf A \mathbf V_1 $$ Thanks in advance for your time and effort.

Best,

Tri

1 Answers 1

0

Recall that for conformal matrix multiplication, the number of columns of the first matches the number of rows of the second. Similarly, in block-matrix multiplication, the partition of the columns of the first must match the partition of the rows of the second.

So, we must patition $A$ into $$ A = \pmatrix{\mathbf a_1^* \\ \hat{\mathbf{ A}}_1} $$

  • 0
    Thank you for your reply. I had been trying something similar, but without the transpose. Given that $\mathbf A = \sigma_1 \mathbf y_1 \mathbf x_1^* $, please may you clarify what $\mathbf a_1^* $ and $ \hat{\mathbf A}_1$ are. – 2017-01-15
  • 0
    By the way shouldn't it be the number of COLUMNS of the first matches the number of ROWS of the second?! – 2017-01-15
  • 0
    I said it wrong for some reason, you're right. – 2017-01-15
  • 0
    $\mathbf a_1^*$ is just the first row of $A$. So, in the case that $A$ has that form, we have $$ \mathbf a_1^* = \sigma_1 \mathbf y_{1}(1) \mathbf x_1^* $$ where $ \mathbf y_{1}(1)$ is the first entry of $ \mathbf y_{1}$ – 2017-01-15
  • 0
    Thanks a bunch. I thought as much but wasn't sure. The thing is $$\mathbf U_1^* \mathbf A \mathbf V_1$$ should equal $$ \pmatrix{\sigma_1 & \mathbf w^* \\ 0 & \mathbf B}. $$ I can't seem to reproduce the zero entry in the bottom left block of the matrix. Any ideas? – 2017-01-15
  • 0
    Thank you for your help. I've just realised that the source of my woes lies elseware. The zero is implied by the fact that $\mathbf U_1 $ is unitary, is it not? $$ \hat{\mathbf U}_1 \sigma_1 \mathbf y_1 = 0 $$ – 2017-01-15