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I want to let $\vec y$ become $\vec u_b \ \rho_b \ \vec y $ and $\vec u_c \ \rho_c \ \vec y $, after a series matrix multiplication,but i am not sure if my matrix multiplication is right or not.

$$ \mathbf u \mathbf \rho \mathbf y=\begin{bmatrix} \vec u_b & 0 \\ 0 & \vec u_c \\ \end{bmatrix} \begin{bmatrix} \rho_b & 0 \\ 0 & \rho_c \\ \end{bmatrix} \begin{bmatrix} \vec y & 0 \\ 0 & \vec y \\ \end{bmatrix}= \begin{bmatrix} \vec u_b \rho_b \vec y & 0 \\ 0 & \vec u_c \rho_c \vec y \\ \end{bmatrix} $$

Is this calculation right?If it is wrong,how do i improve it? I usually multiply the matrix whose elements are all scalar,but now their element are scalar and vector,so i am not sure this is right or wrong

By the way,$\vec u_b$ and $\vec u_c$ are $1$ by $N$ vector ,$\rho_b$ and $\rho_c$ are both scalar,$\vec y$ is a $N$ by $1$ matrix

$\mathbf u$ is $2$ by $N+1$ matrix,and $\mathbf y$ is $N+1$ by $2$ matrix.

It seems that there is something wrong,but i don't know how to improve this

XM551
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    Your $U$ matrix is actually $2\times N$ and $Y$ matrix is $N\times 2$. You can achieve the desired result by pre or post multiplying by your $\rho$ matrix. I.e. $\rho UY$ or $UY\rho$. If you desire for the product $U\rho Y$, you need $\rho=\begin{bmatrix}\rho_bI_N&0\0&\rho_cI_N\end{bmatrix}$, where $I_N$ is the $N\times N$ identity matrix. – Daryl Sep 26 '19 at 07:10
  • @Daryl So if i improve the $\rho$ like you said,my result can become what i want,i mean i can get $ \vec u_b \ \rho_b \vec y $ and $ \vec u_c \ \rho_c \vec y $ ? – XM551 Sep 26 '19 at 07:37
  • Yes. That’s right. – Daryl Sep 26 '19 at 20:53

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