# Math Help - Prove A . In = A = In . A

1. ## Prove A . In = A = In . A

How do you prove this:

The matrix multipication definition I have is:

Given A = (aij) : m x n matrix
B = (bij) : n x p matrix

We define the product C = AB where C has general element cij = n sigma k=1 aikbkj
=ai1bij + ai2b2j + ai3b3 + .. + ainbnj

2. Is that Prove A times the identity is the identity times A?

3. Originally Posted by dwsmith
Is that Prove A times the identity is the identity times A?
No. Mainly prove that A times the Identity is equal to the identity...

Can you help?

4. When you multiple anything times the identity what happens? You will have a sum of a_{ij}s but all but 1 will be zero per location (unless that a_{ij} in A is zero then the sum will add up to zero).

5. That is only true if A is the identity.

The identity matrix is a square matrix that acts like the number 1 from the reals.

6. Originally Posted by dwsmith
When you multiple anything times the identity what happens? You will have a sum of a_{ij}s but all but 1 will be zero per location.
I understand that AI = I through many examples, I am just struggling to prove it using the matrix multiplication definitions! I don't quite understand your comment..

7. This is a sheer exercise in notation.
Do you know about the Kronecker’s delta notation: $\[
\delta _{jk} = \left\{ {\begin{array}{*{20}c} {1,} & {j = k} \\ {0,} & {j \ne k} \\ \end{array} } \right~?$

It is used to define $I_n=(\delta_{jk}):~n\times n$.

We must also assume that $A=(a_{jk}):~n\times n$.

So $A \cdot I_n = \sum\limits_{j = 1}^n {\left( {\sum\limits_{k = 1}^n{a_{jk} \cdot \delta _{kj} } } \right)}$.

Now ask yourself, ‘If $1\le J\le n~\&~1\le K\le n$ is it true that
$a_{JK} = \sum\limits_{j = 1}^n {\left( {\sum\limits_{k = 1}^n {a_{Jk} \cdot \delta _{kK} } } \right)}~?$

If so then you are done.

8. Originally Posted by Plato
This is a sheer exercise in notation.
Do you know about the Kronecker’s delta notation: $\[
\delta _{jk} = \left\{ {\begin{array}{*{20}c} {1,} & {j = k} \\ {0,} & {j \ne k} \\ \end{array} } \right~?$

It is used to define $I_n=(\delta_{jk}):~n\times n$.

We must also assume that $A=(a_{jk}):~n\times n$.

So $A \cdot I_n = \sum\limits_{j = 1}^n {\left( {\sum\limits_{k = 1}^n{a_{jk} \cdot \delta _{kj} } } \right)}$.

Now ask yourself, ‘If $1\le J\le n~\&~1\le K\le n$ is it true that
$a_{JK} = \sum\limits_{j = 1}^n {\left( {\sum\limits_{k = 1}^n {a_{Jk} \cdot \delta _{kK} } } \right)}~?$

If so then you are done.
Thanks for that well written explanation. I have heard of the kronecker delta notation but not actually come across it yet in my course.

I don't quite understand the final line though. I understand everything up to that, but not quite how you get to the final part of the proof ..

9. Originally Posted by chr91
I don't quite understand the final line though. I understand everything up to that, but not quite how you get to the final part of the proof ..
I must tell you that I find that statement hard to believe.
Either the notation is clear or it is not clear.
Note that $\delta_{KK}=1$ and it is zero else where.