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# Linear independence watch

1. 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

2. Determinant is not 0
3. Last two can be done by inspection or general rules about n-dimensional vector spaces.
4. (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?
5. (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
6. (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?
7. (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.
8. (Original post by Glutamic Acid)
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.
Aren't they square matrices? b.) looks like a 3x4 to me. I'm not too sure about the space that would be generated
9. (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.
10. Anybody know how to answer these?
11. (Original post by TwiMaster)
Anybody know how to answer these?
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.

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