Linear independance (1 Viewer)

[Librah]

Need help with c, and maybe a better explanation for b. Is there a shorter way of doing c then the way i'm thinking, which is to do u_2 dot u_3. Since you already proved the other cases in a, you just need to do the last possible combination, which is that. Part d is pretty simple.

 

[InteGrand]

Need help with c, and maybe a better explanation for b. Is there a shorter way of doing c then the way i'm thinking, which is to do u_2 dot u_3. Since you already proved the other cases in a, you just need to do the last possible combination, which is that. Part d is pretty simple.

Are you sure we can use the previous parts to do (c)? Because I think (c) is asking for the case where the vectors v_i are any general vectors, whereas the previous parts were for specific cases?

 

[Librah]

Are you sure we can use the previous parts to do (c)? Because I think (c) is asking for the case where the vectors v_i are any general vectors, whereas the previous parts were for specific cases?

Well it does ask in general any of the two vectors u_1, u_2, u_3. But if you can prove it in general, better.

 

[InteGrand]

And for (b), can't we just say that since the determinant is nonzero, and the columns of it are the components of the vectors v_i, then the vectors are linearly independent? This is a well-known fact, and since it just says "explain", it doesn't look like they want a proof of it?

 

[Librah]

Also if possible would also like to know why if you have an invertible matrix 2x2 matrix P, where the columns of P are eigenvectors, then PD_1D_2P(inverse)= PD_2D_1P(Inverse). D_1,D_2 are diagonal matrices.

 

[Librah]

And for (b), can't we just say that since the determinant is nonzero, and the columns of it are the components of the vectors v_i, then the vectors are linearly independent? This is a well-known fact, and since it just says "explain", it doesn't look like they want a proof of it?

That's what i wrote, but i didn't think that would be worth 4 marks.

 

[InteGrand]

That's what i wrote, but i didn't think that would be worth 4 marks.

OK, then maybe we could say that since the det(A) ≠ 0 (calling that matrix A), the equation Ax = 0 must have a unique solution (i.e. the zero vector), and since Ax is a linear combination of the columns of A, it follows that the only way to write 0 as a linear combination of the columns of A is to take 0 of each (i.e. the columns of A are linearly independent, i.e. the vectors v_i are linearly independent).

But surely we'd need to use some result of det(A) ≠ 0. I doubt we'd have to derive everything from scratch.

 

[awesome-0_4000]

Also if possible would also like to know why if you have an invertible matrix 2x2 matrix P, where the columns of P are eigenvectors, then PD_1D_2P(inverse)= PD_2D_1P(Inverse). D_1,D_2 are diagonal matrices.

The answer to this question has nothing to do with the properties of P, multiplication of diagonal matrices is commutative.

 

[RenegadeMx]

isnt this gram schmidt pretty much? and c) is the projection forces the vectors to be perpendicular