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Is any legal multiplication by identity matrix equal to the original matrix,

$$ A_{n\times m} I_m = A_{n\times m} $$$$ I_n A_{n\times m} = A_{n\times m} $$

I mean is it always right?

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1 Answer

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Yes, this is why it is called the identity matrix. Any matrix multiplied by the identity matrix is the original matrix, just as the multiplicative identity for $\mathbb{R}$ is $1$.

For an $m\times n$ matrix $A$, we have $$(AI)_{ij} = A_{i1}I_{1j}+A_{i2}I_{2j}+\dots+A_{in}I_{nj}$$But $I_{ij}=0$ when $i \neq j$ and 1 otherwise. So $(AI)_{ij} = A_{ij}$. We get $$\begin{align}(AI)_{ij} &= A_{i1}I_{1j}+A_{i2}I_{2j}+\dots+A_{ij}I_{jj}+\dots+A_{in}I_{nj} \\ &= A_{i1}\cdot0+A_{i2}\cdot0+\dots+A_{ij}\cdot1+\dots+A_{in}\cdot0\\ &=A_{ij}\end{align}$$

Showing $(IA)_{ij} = A_{ij}$ is similar.

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