# Thread: Identity Matrix - Show That AI=IA=A For All Square Matrices

1. ## Identity Matrix - Show That AI=IA=A For All Square Matrices

Hey all,

I've got a question that I'm having a little trouble with. I understand what the identity matrix is and how it works, but am unsure how to clearly show this. The question is as follows.

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Let I be a n x n matrix whose entry in row i and column j is

{ 1 if i = j
{ 0 if i != j
*This is an identity matrix

Show that AI = IA = A for every n x n matrix A.
--

Once again I know that this statement is true as I know what the identity matrix is and when I use the identity matrix I understand why this works, but I am unsure how I would clearly show this. Any help is greatly appreciated. Thank you very much!

2. Originally Posted by BraydenW
Hey all,

I've got a question that I'm having a little trouble with. I understand what the identity matrix is and how it works, but am unsure how to clearly show this. The question is as follows.

--
Let I be a n x n matrix whose entry in row i and column j is

{ 1 if i = j
{ 0 if i != j
*This is an identity matrix
We call this $\displaystyle \delta_{ij}$ for short, the so called "Kronecker delta".

Show that AI = IA = A for every n x n matrix A.

Once again I know that this statement is true as I know what the identity matrix is and when I use the identity matrix I understand why this works, but I am unsure how I would clearly show this. Any help is greatly appreciated. Thank you very much!
Let's just calculate what the element in the i-th row and k-th column will be for the two cases:

$\displaystyle (AI)_{ik} = \sum_j a_{ij}\delta_{jk}= a_{ik}$ by definition of $\displaystyle \delta_{jk}$, since for all $\displaystyle j\neq k$ we have $\displaystyle \delta_{jk}=0$.

and, similarly,

$\displaystyle (IA)_{ik} = \sum_j \delta_{ij} a_{jk}= a_{ik}$, by definition of $\displaystyle \delta_{ij}$

So what we have shown that $\displaystyle AI=A$ and $\displaystyle IA=A$, and thus$\displaystyle AI=IA$.

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