Question

I have 2 square matrices, A and B, both of size 5x5.
rank(B) > rank(A)
and the solution to Ax=0 equals to Sp{(0,3,-1,2,1),(4,-2,1,4,0)}

I need to prove that AB =/ 0 (AB does not equals 0)

I'm really lost here and would love some help

Answer 1

In order to show AB is not the zero matrix, it is sufficient to show there one vector v such that
\[ ABv \neq 0 \]
We know that the nullity of A is 2 and hence the rank of A is 3, because rank + nullity = 5.

Hence there is a 3-dimensional subspace \( S \subset \mathbb{R}^5 \) such that for every non-zero vector \( v \in S \) we have that \( Av \neq 0 \).

We also know that rank(B) > rank(A). Hence there is a 4 or 5 dimensional sub-space \( T \subset \mathbb{R}^5 \) such that for all non-zero \(w \in T \), we know that \(Bw \neq 0 \). In other words, the image of B, im(B) is also a 4 or 5 dimensional subspace of \( \mathbb{R}^5 \).

Now ...

Answer 2

Now, this diagram will help. We're looking for a vector w such that ABw is not zero.

In other words, we want a vector w such Bw does not lie in the null space of A.

Given that image(B) = image(W), which dimension 4 or 5, you need to convince yourself that it must intersect with the subspace S, which has dimension 3. If that's right, then we can find a nonzero vector
\[ v \in S \cap im(W) \] and thus there is a vector \( w \) such that
\[ Bw = v \]
and
\[ Av \neq 0 \]
therefore
\[ ABw = Av \neq 0 \]

Answer 3

(*correction in picture: T maps to the image(T), not image(W).
Also, for the record, im(B) = im(T) )

Answer 4

Make sense?

Answer 5

I get the 1st part, still working on understanding the second part. BTW, we havn't reached images where I'm learning, so is there a way to explain this without the use of images? I'm really not sure what that is.

Answer 6

You've learnt about functions. The image of a function is the range of a function. So call it range if you like.

Answer 7

is that correct to say that im(B) is like Sp{B}?

Answer 8

The image of B is the set \( \{ Bv \ | v \in \mathbb{R}^5 \} \).