Hello everyone,(adsbygoogle = window.adsbygoogle || []).push({});

it's final's time next week , so I will be posting here more often than usual

Here is one problem I came across when doing review:

The nullspace of non-zero 4x4 matrix cannot contain a set of 4 lin. indep. vectors. (T/F)

The way I was thinking is that if I solve a homogeneous s-m with this matrix, and if the dimension of nullspace is 4, that means that there have to be 4 free variables in the homogeneous s-m, but matrix is just 4x4.

And then there is this rank-nullity theorem thatn = rank(A) + nullity(A), so in this case rank(A) = 0, is it ever possible? My guess is not.

Does the same hold for dim of nullspace (nullity): it has to have at least one solution (trivial, where everyting = 0, but that does not mean that the nullity is an empty set!) ?

Is it correct?

Thanks in advance!

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# Nullspace of non-zero 4x4 matrix

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