Matrices as Linear Maps

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November 15th 2011, 07:22 PM
AllanCuz

Matrices as Linear Maps

Hey Team,

Quote:

Let A be an M by N matrix, B be an N by P matrix

a) Show that AB is the sum of n matrices each of rank at most 1
b) If the rank of A is n, what is the rank of AB?

I'm not sure where to start on part a, but part b I think I have a decent start.

$Rank(AB) = dim(R[AB])$

We know that A maps from Fn to Fm, and B maps from Fp to Fn. So we define the left multiplication of these guys as such,

$L_A: F^N \to F^M$

$L_B: F^P \to F^N$

So to find the range of AB we can let Z be some arbitrary vector in $F^P$ and see for what values is it good for

$dim(R[AB])=dim(L_A L_B (Z))$

We can note that $L_B(Z)$ is a subset of $F^N$ so the following inequality results,

$dim(L_A L_B (Z)) \le dim(L_A (F^N)) = dim(A(F^N)) = Rank(A) = N$

We can further note that if $L_B$ is a surjective map then its range is actually equal to $F^N$ turning the inequality into equality.

Was the above process correct for b, and if so, are there any hints for a?

Thanks!
November 15th 2011, 09:00 PM
Drexel28

Re: Matrices as Linear Maps

Quote:

Originally Posted by AllanCuz

Hey Team,

I'm not sure where to start on part a, but part b I think I have a decent start.

$Rank(AB) = dim(R[AB])$

We know that A maps from Fn to Fm, and B maps from Fp to Fn. So we define the left multiplication of these guys as such,

$L_A: F^N \to F^M$

$L_B: F^P \to F^N$

So to find the range of AB we can let Z be some arbitrary vector in $F^P$ and see for what values is it good for

$dim(R[AB])=dim(L_A L_B (Z))$

We can note that $L_B(Z)$ is a subset of $F^N$ so the following inequality results,

$dim(L_A L_B (Z)) \le dim(L_A (F^N)) = dim(A(F^N)) = Rank(A) = N$

We can further note that if $L_B$ is a surjective map then its range is actually equal to $F^N$ turning the inequality into equality.

Was the above process correct for b, and if so, are there any hints for a?

Thanks!

That looks right or me. For the second part, consider writing the matrix as a sum of column matrices.
November 15th 2011, 09:52 PM
Deveno

Re: Matrices as Linear Maps

strictly speaking, such a sum does not exist, perhaps you meant a sum of matrices that are all 0's except for one column?
November 16th 2011, 02:52 PM
Drexel28

Re: Matrices as Linear Maps

Quote:

Originally Posted by Deveno

strictly speaking, such a sum does not exist, perhaps you meant a sum of matrices that are all 0's except for one column?

Obviously, I was letting the OP fill things in for themselves, you know?
November 16th 2011, 10:37 PM
Deveno

Re: Matrices as Linear Maps

well you know me, i'm sorta slow-witted....