First Post... Desperately need help before exam.

Hi Everyone, I firstly want to apologize for the long post, but I have a Linear Algebra Exam on Wednesday, and We received a review sheet in class today, but no answers. I've been working on it for a while and would really appreciate some affirmation, and clearing up.

1) Let v be a non zero vector in Rn, and let W={x in Rn such that x^{T}*v=0} Show that W is a subspace of dimension (n-1)

I can show that W is in fact a subspace through the closure properties, however I'm having trouble finding a minimal spanning set with n-1 vectors in it.

2) Let L: Rn to Rm be a linear transformation, and W a subspace of Rm. Let L_{W} = {x in Rn given L(x) is in W} Show that L_{W }is a subspace of Rn.

Here I believe it is just showing the zero vector is contained and that the 2 closure properties follow, but I'm not sure about showing this with a Linear Transformation.

3) Let P be an n x n matrix satisfying the equation P^{m. }= P for some m > 1 Suppose that N(P) = {0}. Show that if P is diagonalizable, then P^{2} = Id (Identity Matrix)

This question really has me puzzled right now.

4) Suppose T: Rm to Rn is a linear transformation, and {v_{1},.....,v_{k} } are k vectors in Rm for which {T(v_{1}),.....,T(v_{k})} are linearly independent in Rn. Show that {v_{1},.....,v_{k}} are linearly independent in Rm.

This makes sense to me that any vectors having undergone a linear transformation, and are still linearly independent, must have been linearly independent beforehand, but I'm lacking on a more mathematically driven way to show it.

Lastly True False, with explanations regarding why it is true or false. I haven't attempted these yet so have no current answers, but am working on them tonight and hoping this can be answered by tomorrow night so I may use it to study.

A) If A and B are n x n matrices, then Det(A+B) = Det(A) + Det(B).

B) If A is nonsingular and diagonalizable, then A^{-1}is also diagonalizable

C) If S = {v_{1},v_{2},.....,v_{n}} is a basis for Rn and W is a subspace of Rn, then SnRn = {v_{1} given v_{1}is in W} is a basis for W.

D) Let W C Rn be a subspace, and let W(Perpendicular sign) = {y in Rn such that x (dot) y =0 for all x in W}. Then W(perpendicular) is a subspace of Rn

If anyone could reply with what the work should look like, or an explanation of the steps for these problems, I would greatly appreciate it so that I can check my work and be more prepared for Wednesday.

Thanks Everyone!

Re: First Post... Desperately need help before exam.

Quote:

Originally Posted by

**AlgebrahOSU** Hi Everyone, I firstly want to apologize for the long post, but I have a Linear Algebra Exam on Wednesday, and We received a review sheet in class today, but no answers. I've been working on it for a while and would really appreciate some affirmation, and clearing up.

1) Let v be a non zero vector in Rn, and let W={x in Rn such that x

^{T}*v=0} Show that W is a subspace of dimension (n-1)

I can show that W is in fact a subspace through the closure properties, however I'm having trouble finding a minimal spanning set with n-1 vectors in it.

You are given that v is non zero so at least one component of v is non-zero. Let vi be a non-zero component. Define u1= <1, 0, ..., ai, 0, ..., 0> such v1+ aivi= 0. That is, the first component is 1, the ith component is v1/vi, and all other components are 0. Define u2= <0, 1, ..., bi, 0, ..., 0> such that v2+ bivi= 0 and all other components are 0. That is, the second component is 1, the ith component is v2/vi and all other components are 0. Continue in that way defining uj to be the vector such the jth component is 1 for all j= 1 to n but j NOT equal to I, the ith component vj/vi, and all other components are 0. That will give n- 1 vectors that, by construction, have dot product with v equal to 0. It needs to be shown that they are independent and so span W.

Quote:

2) Let L: Rn to Rm be a linear transformation, and W a subspace of Rm. Let L

_{W} = {x in Rn given L(x) is in W} Show that L

_{W }is a subspace of Rn.

Here I believe it is just showing the zero vector is contained and that the 2 closure properties follow, but I'm not sure about showing this with a Linear Transformation.

Yes, and it is pretty straight forward. If u and v are in L_{W}, the L(u) is in W and L(v) is in W. All you need to show is that L(u+ v)= L(u)+ L(v) is in W and L(ku)= kL(u), where k is a scalar, is in W.

Quote:

3) Let P be an n x n matrix satisfying the equation P

^{m. }= P for some m > 1 Suppose that N(P) = {0}. Show that if P is diagonalizable, then P

^{2} = Id (Identity Matrix)

This question really has me puzzled right now.

4) Suppose T: Rm to Rn is a linear transformation, and {v

_{1},.....,v

_{k} } are k vectors in Rm for which {T(v

_{1}),.....,T(v

_{k})} are linearly independent in Rn. Show that {v

_{1},.....,v

_{k}} are linearly independent in Rm.

This makes sense to me that any vectors having undergone a linear transformation, and are still linearly independent, must have been linearly independent beforehand, but I'm lacking on a more mathematically driven way to show it.

Prove it by contradiction. If they are NOT linearly independent, then there exist scalars, a1, a2, ..., ak, not all 0, such that a1v1+ a2v2+ ...+ akvk= 0. Then T(a1v1+ a2v2+ ...+ akvk)= a1T(v1)+ a2T(v2)+ ...+ akT(vk)= T(0)= 0.

[quote]Lastly True False, with explanations regarding why it is true or false. I haven't attempted these yet so have no current answers, but am working on them tonight and hoping this can be answered by tomorrow night so I may use it to study.

A) If A and B are n x n matrices, then Det(A+B) = Det(A) + Det(B).[/tex]

Look at and .

Quote:

B) If A is nonsingular and diagonalizable, then A^{-1}is also diagonalizable

A is diagonalizable if and only if there exist a matrix B such that where D is the diagonal matrix having the eigenvalues of A on its main diagonal. Since A is nonsingular, none of those is 0. Show that and is the diagonal matrix having the **reciprocals** of the eigenvalues of A on its diagonal.

Quote:

C) If S = {v_{1},v_{2},.....,v_{n}} is a basis for Rn and W is a subspace of Rn, then SnRn = {v_{1} given v_{1}is in W} is a basis for W.

Are you sure you have written this correctly? What you have written is that, given any basis for Rn and any subspace, W, of Rn, the **first** vector in the basis of Rn is a basis for W. Although it is NOT what you have written, if you replace v_{1} with v_{i}, you would be saying that selecting a subset from the first basis, of all basis vectors that are in W, you have a basis for W. That sounds better but is still NOT true. For example, {<1, 0>, <0, 1>} is a basis for R2 and W= {<x, y>| x+ y= 0} is a subspace of R2 but neither vector in {< 1, 0>, <0, 1> } is in W.

Quote:

D) Let W C Rn be a subspace, and let W(Perpendicular sign) = {y in Rn such that x (dot) y =0 for all x in W}. Then W(perpendicular) is a subspace of Rn

Again, it is just a matter of checking that the sum of two vectors in the set is in the set and that the scalar product of a vector in the set with any scalar is in the set.

Suppose u and v are in W. Then x.u= 0 and x.v= 0 for all x in W. So what is x.(u+ v)? What is x.(ku) for any k?

Quote:

If anyone could reply with what the work should look like, or an explanation of the steps for these problems, I would greatly appreciate it so that I can check my work and be more prepared for Wednesday.

Thanks Everyone!

Re: First Post... Desperately need help before exam.

Thank you so so much Halls. I did see your other response as well, sorry for that, but I was unsure of the speed of the replies being a first time poster. Your answers were extremely well explained and I feel infinitely more confident about this exam after comparing my work with what you provided. You are a life saver, thanks a million.

Re: First Post... Desperately need help before exam.

Also On True false C, It was a typo that is supposed to be S intersected with W not S intersected with Rn

Re: First Post... Desperately need help before exam.

I found this helpful too, thanks!