Recent content by wurth_skidder_23

so basically I'm trying to prove that for some nonzero x, l_j(x)=0 ?

Assume that m<n and l_1,l_2,...,l_m are linear functionals on an n-dimensional vector space X . Prove there exists a nonzero vector x \epsilon X such that < x,l_j >=0 for 1 \leq j \leq m. What does this say about the solution of systems of linear equations?This implies l_j(x)...

For the second one, which is basically just an addition to the first, is this correct? Property 1 of a linear functional is satisfied as follows: \ y (x+z)=x(-2)+z(-2)+\int_0^1\ (x(t^2)+z(t^2))\, dt \ y (x+z)=x(-2)+\int_0^1\ x(t^2)\, dt + z(-2)+\int_0^1\ z(t^2)\, dt \ y (x+z)=y(x)+y(z)...

I am studying for a final I have tomorrow in linear algebra, and I am still having trouble understanding linear functionals. Can someone help me out with this example problem, walk me through it so I can understand exactly what a linear functional is? Is the following a linear functional? \ y...

Okay, maybe I should state the whole problem: Let X and Y be inner product spaces in R with inner products < , >_X and < , >_Y Suppose that T is in L(X,Y). Show that <x,y> = <Tx,Ty>_Y is an inner product on X if and only if T is one to one.

distance between two vectors x, y in an inner product space is ||x-y||

Is there a way to show that mathematically using the adjoint or something similar?

How would I prove that an isometry is one to one? General definition of isometry, A: <Ax,Ay> = <x,y> Where < , > is an inner product (scalar product, dot product, etc.) How do I prove A has to be one to one for this to work?

That is also true for the characteristic polynomial. I know that the characteristic polynomial can be divided by a minimal polynomial and I want to show that the minimal polynomial is equal to (s-a_1)^(d_1)*...*(s-a_k)^(d_k)

The book defines the index as: "N_m = N_m(a) the nullspace of (A-a*I)^m. The subspaces N_m consist of generalized eigenvectors; they are indexed increasingly, that is N_1 is included in N_2 is included in... Since these are subspaces of a finite-dimensional space, they must be equal from a...

Try integration by parts with u = (ln(x))^3 and dv = x dx

Let A be an n x n matrix; denote its distinct eigenvalues by a_1,...,a_k and denote the index of a_i by d_i. How do I prove that the minimal polynomial is then: m_A(s) = (s-a_1)^d_1*...*(s-a_k)^d_k ? The characterstic polynomial is defined as: p_A(s) = (s-a_1)*...*(s-a_n);

It would have to be 1. I see now, thanks. I should've thought of that.

I know about the sign, but I don't see where you're going. We call it the signature in our book. It would be great to know how to prove it without brute force. I'm an engineering major, so I'm not great at proofs.

If p1 = 2, then p2 has to equal 3 for p1 to go to 3 the second time. That means p3 = 4 for 2 to go to 4 the second time. For those to hold true, p4 = 2 so 3 goes to 2, but then 4 goes to 2 which goes to 3, so it doesn't hold. Similarly, if p1 = 4, then p4 = 3. 3 has to go to 1 for p4 to go...