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So a matrix, A, is linearly independent if Ax = 0 has only the trivial solution (x=0)

so let A = AB, then if the columns of AB are linearly independent, the equation (AB)x = 0 has only the trivial solution

by the associative law of multiplication, (AB)C = A(BC), so (AB)x = A(Bx) = 0

I'm not sure where to go from here, or even if this is a proper way to go about proving that the columns of B are linearly independent

Satan Lucifer
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1 Answers1

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If $Bx=0$ has a nonzero solution $x$, then it also solves $ABx=0$.

Berci
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  • ohhh, that makes sense. So if B is linearly dependent, x can be a non-zero vector and ABx=0 would have a non-trivial solution (which can't be the case if the columns of AB are linearly independent). So B's columns must be linearly independent. Thank you for the help! – Satan Lucifer Nov 01 '20 at 09:06