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Linear independence

I think I found a way to solve this but it seems unnecessarily long to solve in echelon form given that it's just for just 1 mark each, surely there's a better way to answer these?
Help would be much appreciated,

Thanks

Linear algebra hw5 Q1.jpg

Determinant is not 0

Last two can be done by inspection or general rules about n-dimensional vector spaces.

Reply 3

TwiMaster

Original post by ubisoft

Determinant is not 0

what does it mean when the determinant is not equal to zero? what's the significance of it?

Reply 4

ubisoft

Original post by TwiMaster

what does it mean when the determinant is not equal to zero? what's the significance of it?

Det NOT = 0 => independent

Det = 0 => dependant

Reply 5

TwiMaster

Original post by morgan8002

Last two can be done by inspection or general rules about n-dimensional vector spaces.

Is that because they're not square matrices?

Reply 6

Glutamic Acid

Original post by TwiMaster

Is that because they're not square matrices?

Not necessarily that they're not square matrices when you put them together*. But what would it mean for four 3-d vectors to be not linearly dependent? What sort of space would they generate?

* 2 3-d vectors could, of course, be linearly independent.

Reply 7

TwiMaster

Original post by Glutamic Acid

Aren't they square matrices? b.) looks like a 3x4 to me. I'm not too sure about the space that would be generated

Reply 8

poorform

Original post by TwiMaster

Aren't they square matrices? b.) looks like a 3x4 to me. I'm not too sure about the space that would be generated

A square matrix is one with the same number of columns as rows. A nxn matrix is a matrix with n columns and n rows. Apply this to the ones in your question you should see only the first one is square.

Reply 9

TwiMaster

Anybody know how to answer these? Linear algebra hw5 Q3.jpg

Reply 10

poorform

Original post by TwiMaster

Anybody know how to answer these? Linear algebra hw5 Q3.jpg

Well to start with it would be useful to consider the definition of a nullspace. Here the nullspace would be the set of all vectors x such that Ax=0. Consider such a multiplication and you will get x1+2x2+3x3+x4=0. Now you have 1 equation in 4 unknowns so you can pick 3 of the unknowns arbitrarily and then solve the system. (It will have infinitely many solutions.)

Can you find the set that spans the nullspace (if you understand the definitions this should follow without effort from the first part. are they linearly indep? Then you have a basis if you can prove this). Then the last part notice that dim(N(A))=|A|=#of vectors in the basis so that is just stating the number which is easy once you have the basis.

Hopefully this gives you enough to work with.

Good luck.

(edited 8 years ago)