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.

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

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,

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

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

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

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

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

$\displaystyle 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 $\displaystyle L_B $ is a surjective map then its range is actually equal to $\displaystyle F^N$ 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.

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?

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?

Re: Matrices as Linear Maps

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