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I am aware that this might be a trivial question, but is there an identity matrix I which has dimensions $m \times m$ for which IA=A for all matrices A with dimensions $m \times n$? I know that for square matrices, it is 1 along a diagonal and 0 for the other entries, but for non-square matrices this will be different. Thanks.

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    $\begingroup$ If $m \neq n$ then you cannot even multiply two $m \times n$ matrices. $\endgroup$
    – badjohn
    Commented May 19 at 5:47
  • $\begingroup$ but then is there a matrix I for which IA=A, where A is a m by n matrix $\endgroup$
    – GSmith
    Commented May 19 at 6:18
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    $\begingroup$ The $m \times m$ identity matrix works here. $\endgroup$ Commented May 19 at 6:33

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The identity matrix that is multiplication-compatible with the given matrix will work. The reason is simple, and I will illustrate this with an example. Say $m =2$ and $n = 3$. Consider the product $AB$ where $$ A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}, \quad B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{bmatrix}.$$ The product $AB$ can be viewed as $A$ acting on each column of $B$, as follows: $$AB = A \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{bmatrix} = \begin{bmatrix} A \begin{bmatrix} b_{11} \\ b_{21} \end{bmatrix} & A \begin{bmatrix} b_{12} \\ b_{22} \end{bmatrix} & A \begin{bmatrix} b_{13} \\ b_{23} \end{bmatrix} \end{bmatrix}.$$ Then, it is clear that the matrix $A$ that sends each vector to itself (and hence $B$ to itself), is the identity $\mathbf{I}_{m}$.

The key is to realize that the left-multiplication of $A$ with $B$ is the same as $A$ acting separately on the columns on $B$.

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