So that part makes sense now.
How do we complete the proof if |H| = 1?
The questions were
For the first one I claimed it was true by using a uniqueness result
and by pairing T(v) with 0. But I can't figure out why this argument does not work for the second one?
Why do you think that "uniqueness result" is true? Add anything orthogonal to x to y and you won't change the inner product of x with it.
(Also the truth of b) is (perhaps surprisingly) dependent on the field your vector space is over.)
Last edited by seanieg89; 6 Apr 2017 at 4:44 PM.
Is there an intuitive explanation for this?
Let T be a linear map on a finite-dimensional inner product space V
Then T is an isometry iff T is unitary
So the question is obviously easy first year stuff. I'd prove linear independence and then use dim(V) = B to deduce that it's a basis.
However, for the linearly independence step
I just want a validity check because I'm having second doubts. Mostly because no solutions made a remark on this.
I differentiated w.r.t t and then subbed in t=2 to prove c1 = 0. (And then repeated this to show c2 = c3 = 0.) Is this ok?
In an inner product space, the norm is defined in terms of the inner product. This means that any operator that preserves the inner product will preserve the norm.
Less obvious is the fact that in an inner product space, the inner product can be written in terms of the induced norm (*). Consequently anything that preserves the norm will preserve the inner product.
Of course, you don't need to prove (*) in order to answer this particular question, but it kind of hits at the heart of the relationship between inner products and induced norms on a real/complex inner product space and is a good exercise.
Note also that finite dimensionality is not required in any of these arguments.
The question is an extension on c. How can I explain this answer from first-year calculus?
I don't know much about Graph Theory and Group Theory but are they two different topics? Or Different names but same subjects?
I can't be bothered Googling it
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I was wondering if given the trace and the determinant of a matrix could you write down a unique matrix satisfying these conditions, or a simple formula for the family of matrices satisfying it?
Mostly asking for the 2x2 case
Last edited by InteGrand; 8 May 2017 at 11:21 PM.
The intuition of the question was this
I then realised that B may share the same eigenvectors as A, and have eigenvalues equal to the square root of those of A. But I'm not sure where to proceed from there.
No it need not be. Say V = R^2 and W1 = R^2 (= V) and W2 be the line (t, 0) (the x-axis). Take T to be a rotation map by 90 degrees counter-clockwise about the origin say. Then T is a linear map from V to V, so T(V) = T(W1) is a subspace of W1 = V = R^2, and W2 is a subspace of W1 (which is a subspace of V), but clearly W2 is not invariant under T (e.g. the point (1, 0) in W1 does not get mapped to a point in W2 by T; it gets mapped to (0, 1)).
Last edited by InteGrand; 22 May 2017 at 11:20 PM.
My approach thus far: Write
Is this a dead end? Because I don't see how I can use what I know about T here
Last edited by leehuan; 7 Jun 2017 at 9:13 PM.
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