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I was trying to verify that $\Sigma_{1 \leq i,j \leq 2} \ x_{i,j} I_{i,j} \equiv Ax$, where $A = \begin{bmatrix} 1 & 1 & 1 &0\\ 1 & 1 & 0 &1\\ 1& 0& 1&1 \\ 0&1&1&1\end{bmatrix}$ and $I_{1,1} = \begin{bmatrix} 1&1 \\ 1&0\end{bmatrix}$, $I_{1,2}= \begin{bmatrix} 1&1 \\ 0&1\end{bmatrix} $, $I_{2,1}=\begin{bmatrix} 1&0 \\ 1&1\end{bmatrix}$, $I_{2,2}=\begin{bmatrix} 0&1\\ 1&1\end{bmatrix}$. But I get some dodgy result...

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  • $\begingroup$ you sum 2x2 matrices to get 4x4 matrix, are you sure? $\endgroup$ Nov 4, 2015 at 10:56
  • $\begingroup$ Not very... The column vectors of A are actually the $I_{i,j}$.. $\endgroup$
    – Naz
    Nov 4, 2015 at 11:04
  • $\begingroup$ oh.. I got it. Shall I post the solution? $\endgroup$
    – Naz
    Nov 4, 2015 at 11:06

1 Answer 1

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The columns of $A$ are actually $I_{i,j}$. Therefore $Ax = \begin{bmatrix} x_{11} & x_{12} & x_{21} & 0 \\ x_{11} & x_{12} & 0 & x_{22} \\ x_{11} & 0 & x_{21} & x_{22} \\ 0 & x_{12} & x_{21} & x_{22}\end{bmatrix}$. And the individual multiplications are of the form $\begin{bmatrix} x_{11}&x_{11} \\ x_{11}& 0\end{bmatrix}$ and if you now inspect the first column and the result just presented you will see that the two are identical. Therefore, the question is inaccurate, because it is not the sum that is equal to $Ax$.

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  • $\begingroup$ so you say that the sum of 2x2 matrices is concantination of nx1 matrices? $\endgroup$ Nov 4, 2015 at 11:20

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