Need Help with upcoming exam review problems

AlgebrahOSU

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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 xT*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 LW = {x in Rn given L(x) is in W} Show that LW 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 Pm. = P for some m > 1 Suppose that N(P) = {0}. Show that if P is diagonalizable, then P2 = Id (Identity Matrix)

This question really has me puzzled right now.

4) Suppose T: Rm to Rn is a linear transformation, and {v1,.....,vk } are k vectors in Rm for which {T(v1),.....,T(vk)} are linearly independent in Rn. Show that {v1,.....,vk} 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-1is also diagonalizable
C) If S = {v1,v2,.....,vn} is a basis for Rn and W is a subspace of Rn, then SnRn = {v1 given v1is 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!
 
I believe you posted this on a different website and I responded to all except
3) Let P be an n x n matrix satisfying the equation Pm. = P for some m > 1 Suppose that N(P) = {0}. Show that if P is diagonalizable, then P2 = Id (Identity Matrix)
there. This is probably the hardest problem in the lot. The fact that N(P)= {0} means that P is invertible and that none of its eigenvalues is 0. The fact that P^m= P (I assume that "m" was meant to be a power) means that P^mv= Pv for all v. In particular if v is an eigenvector corresponding to eigenvalue \(\displaystyle \lambda\) then \(\displaystyle Pv= \lambda v\), \(\displaystyle P^2 v= P(Pv)= P(\lambda v)= \lambda Pv= \lambda(\lambda v)= \lambda^2 v\) and, generally, \(\displaystyle P^mv= \lambda^m v\) (prove that by induction). So for any eigenvector v, we have \(\displaystyle \lambda^m v= \lambda v\) whence \(\displaystyle \lambda^m= \lambda\). \(\displaystyle \lambda^m- \lambda= \lambda(\lambda^{m-1}- 1)= 0\) so that \(\displaystyle \lambda= 0\) or \(\displaystyle \lambda= 1\). Since P is invertible, all eigenvalues are 1.

If P is diagonalizable, then there exist B such that \(\displaystyle P= BDB^{-1}\) where D is a diagonal matrix having the eigenvalues of P on its diagonal. But if the eigenvalues are all 1 then D is the identity matrix and we have \(\displaystyle P= BIB^{-1}= BB^{-1}= I\). So all that remains is to prove that P is diagonalizable. `
 
I have the same review sheet, so if you could link to where you responded that would be very helpful :)
 
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