Hey. Im doing some last minute exam study and came across this question:(adsbygoogle = window.adsbygoogle || []).push({});

let A and B be nxn matrices. Prove that [tex]Null(AB) \geq Null(B)[/tex]

Now i think i did it correctly, but i wasn't sure if i perhaps overlooked something and there is a flaw in my proof.

I began by stating that if any vector is in the nullspace of B, then it MUST also be in the nullspace of AB. Let x be in the nullspace of B, then:

Bx = 0

Then:

ABx = A(Bx) = A(0) = 0

Hence any vector in the nullspace of B is also in the nullspace of AB i.e. the nullspace of B is a subset of the nullspace of AB, hence the dimension (nullility) of the nullspace of AB will be greater (or equal).

That look ok?

The second part of the question was:

When is the Null(AB) = Null(B)?

I said when A is invertible, since:

ABx = 0 imples that Bx = 0 if the inverse of A exists. Hence, any vector in the nullspace of AB will also be in the nullspace of B. Since every element of the nullspace of AB is also in nullspace of B, and every element of the nullspace of B is also in the nullspace of AB (from the first part of the question), the two subspaces are equal and hence have the same dimension.

Do they look alright?

Thanks in advance,

Dan.

PS. I am so sick of typing the word nullspace

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# Homework Help: Nullspace Proof

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