the identity matrix is unique

Question

let A be an $m\times n$ matrix. Prove that there are unique matrices $I_m$ and $I_n$ such that :
$$I_mA=AI_n=A$$

Actually I can't prove the uniqueness here,any help is appreciated.

Thanks

Take $A$ to be the matrix with only non-zero entry : 1 in the $i,j^{th}$ spot — Prahlad Vaidyanathan, Sep 22 '13 at 14:37
mxn is the dimension of the matrix and A is different from zero matrix,you have right — dimitris, Sep 22 '13 at 14:52
This is badly put. "Let $A$..." means that $A$ is given, and in that case the identity is not necessarily unique. What it should say is that $I_mA = AI_n = A$ for all matrices $A$ of order $m \times n$. — Vedran Šego, Sep 22 '13 at 14:52
@PrahladVaidyanathan : you should expand your comment into an answer. — Stefan Smith, Sep 22 '13 at 15:47

score 2 · Answer 1 · answered Sep 22 '13 at 15:39

Let us first address the question as it was asked: for some (i.e., for a given) $A$,

$$I_m A = A I_n = A. \tag{*}$$

Let $X_k$ denote any matrix of order $k$, and let

$$A = \operatorname{diag}(1, 0, \dots, 0) = \begin{bmatrix} 1 & 0 & \dots & 0 \\ 0 & 0 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 0 \end{bmatrix}.$$

Then any

$$I_m = \operatorname{diag}(1, X_{m-1}) = \begin{bmatrix} 1 & 0 \\ 0 & X_{m-1} \end{bmatrix}, \quad I_n = \operatorname{diag}(1, X_{n-1}) = \begin{bmatrix} 1 & 0 \\ 0 & X_{n-1} \end{bmatrix}$$

will satisfy $(*)$. This means that for some $A$, the statement in the question is not true.

Let us now address what the question should have been:

If $(*)$ is true for any (complex or real) matrix $A$ of order $m \times n$, then $I_m$ and $I_n$ are unique.

We observe only $I_m$, as the proof for $I_n$ is equivalent.

Let us define a sequence of matrices:

$$A_k := \operatorname{diag}(\underbrace{0, \dots, 0}_{k-1}, 1, \underbrace{0, \dots, 0}_{m-k}) \in \mathbb{F}^{m \times m},$$

where $\mathbb{F} = \mathbb{C}$ or $\mathbb{F} = \mathbb{R}$. Descriptively, $A_k$ is constructed form a zero matrix of order $m \times m$ be replacing its $k$-th diagonal element with $1$.

Then $I_m A_k$ differs from a zero matrix only in the $k$-th column, which is equal to the $k$-th column of $I_m$. Given that $I_m A_k = A_k$, we see that the $k$-th column of $I_m$ is equal to the $k$-th column of $A$, i.e., it is equal to

$$\begin{bmatrix} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}\!\! \begin{array}{l} \left.{ \vphantom{ \begin{array}{c} 0 \\ \vdots \\ 0 \end{array} }}\right\}k-1 \\ \vphantom{0} \\ \left.{ \vphantom{ \begin{array}{c} 0 \\ \vdots \\ 0 \end{array} }}\right\}m-k \end{array}.$$

Going through $k=1,2,\dots,m$, we get that $I_m = \operatorname{diag}(1,1,\dots,1)$, so it is unique.

We do the same with $A I_n = A$, only this time we're working with the rows of $I_n$.

the identity matrix is unique

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