Is There an Inner Product on R^2 Matching the Norm |x1| + |x2|?

[jimmypoopins]

1. Suppose $$V = U \bigoplus W$$ where U and W are nonzero subspaces of V. Find all eigenvalues and eigenvectors of $$P_{U,W}$$.

As long as $$\lambda$$=0 is an eigenvalue of $$P_{U,W}$$, i can prove that $$\lambda$$=0 and $$\lambda$$=1 are the only eigenvalues and then find the corresponding eigenvectors. can anyone help me to show how $$\lambda$$=0 is an eigenvalue of $$P_{U,W}$$? does it have to do with the fact that Null($$P_{U,W}$$)=W?

2. Prove or disprove: there is an inner product on $$R^{2}$$ such that the associated norm is given by $$\|(x_1,x_2)\| = |x_1| + |x_2|$$ for all $$(x_1,x_2)$$ in $$R^{2}$$.

we just started inner product spaces and the wording on this problem confuses me a lot. i can find an example of two vectors that disprove the statement for one inner product induced by the norm, but not for all inner products. the question wants an example that works for all inner products, right? (if i am to disprove it, i mean.) can someone push me in the right direction?

thanks for the help.

edit: nevermind, I think i figured out 1. since W = null P, all nonzero w in W satisfy (P-lambdaI)w=Pw=0, so lambda=0 is an eigenvalue. i still need help with 2 though

 

[HallsofIvy]

Yes, it has everything to do with nullspace($$P_{U,W}$$)= V!

If V is trivial, that is if V contains only the 0 vector, U= W and, of course, $P_{U,W}$ is just the identity operator. It has only eigenvalue 1 and every vector in W as an eigenvector. So, strictly speaking it is not true that 0 must be an eigenvalue. However, as you say, if V is not trivial then P(v)= 0= 0v for every vector in P and 0 is an eigenvalue with any vector in V an eigenvector.

2. Prove or disprove: there is an inner product on R² such that the associated norm is given by ||x₁,x₂||= |x<sub>1[/sup]|+ |x2[/sup]| for all (x1, x2) in R².

we just started inner product spaces and the wording on this problem confuses me a lot. i can find an example of two vectors that disprove the statement for one inner product induced by the norm, but not for all inner products. the question wants an example that works for all inner products, right?</sub>

<sub>Two vectors that disprove what statement? No, this does want an example that works for all inner products. It specifically says "there is an inner product".

Given an inner product, <u, v> the "associated norm" is defined by $||x||= \sqrt{<x, >}$. If x= (x1, x2), then you want $\sqrt{<(x_1, x_2), (x_1,x_2)>}= |x_1|+ |x_2|$. Can you define an inner product so that is true?
To simplify, let $<x[_1, x_2]), (x_1,x_2)>= a$. You want $\sqrt{a}= |x_1|+ |x_2|$. Does such an a satisfy the conditions for an inner product?</sub>