Homework Statement
Let T: Rn -> Rn be a linear transformation, and let B be an orthonormal basis for R^n. Prove that [ the length of T(x) ] = [ the length of x ] if and only if [T]B (the B-matrix for T) is an orthogonal matrix.
Homework Equations
None I don't think.
The Attempt at...
In the problem it says that A and B are real matrices such that A = PBP^-1 for some invertible 2 x 2 complex matrix P. Since P is invertible, so is P^-1, and this can be rewritten as AP = PB. We can do the same thing with the real matrix Q to get AQ = PQ. So in the equation AX = XB (I think...
Hmm, not entirely sure. If we consider the equation XA = BX, we must be able to show that it has a complex solution, which would account for the A = PBP^-1 and that it also has a real solution, which would account for the A = QBQ^-1. Maybe if we start with one and then work to the other?
Ah! Makes so much more sense! Thank you SO much for your help :)
I just finished 4a. Find a basis for nul((C-lambda I)^2) and relate this to nul(C - lambda I).. if you haven't already.
Ok, so for your case 1, we have an eigenvalue that gave us some eigenvector v. I define V to be the span of v. Because it is just the one vector, it is linearly independent (which we know is true since an eigenvector is not the zero vector). Thus, the dimension is equal to 1. And if v is in...
Let A and B be 2 x 2 real matrices such that A = PBP^-1 for some invertible 2 x 2 complex matrix P. Prove that A = QBQ^-1 for some invertible 2 x 2 real matrix Q.
Gah! I still don't get it. I don't even think I have any sort of argument going... I'm just that confused. I just don't see how we can say that the dimension of V is either 1 or 2 when we don't have much information about A.
Ok, so I know that if A has distinct eigenvalues λ1, ..., λp, then for 1 <= k <= p the dimension of the eigenspace for λk is less than or equal to the multiplicity of the eigenvalue λi. So for this problem, that would imply that there has to be an eigenvalue with multiplicity of 1 or an...
Ok, I'm still confused how to do this problem...
So you have a matrix A. I solve det(A-λI) = 0 and find the eigenvalues. Suppose I take an eigenvalue and plug it into A-λI and solve the equation (A-λI)x = 0. The the set of all solutions to this equation is the eigenspace. I know that the...
I guess I was just trying to stick with the eigenspace argument. So Av = λv and the set of all solutions v that satisfies this is in the eigenspace. And (A-λI)v = 0. So how do I prove that the dimension of the eigenspace is between or equal to 1 and 2?
Homework Statement
Let A be a real n x n matrix. Prove that we can find a subspace V in R^N such that 1 <= dim V < = 2 and A(V) is a subset of V.
Homework Equations
None I don't think.
The Attempt at a Solution
I know that the eigenspace of a matrix satisfies the condition that...
Oh sorry about the confusion with the title. When I initially posted a question, it had to do with rank, but then I figured that one out so I just edited the question rather than post a new thread and must have forgotten to change the title.
I suppose my confusion was just because it's hard...
Prove that the function [ ]B: L(V) -> Mnxn(R) given by T -> [T]B is an isomorphism. [T]B is the B-matrix for T, where T is in the vector space of all linear transformations.
I don't quite understand this...