Recent content by Ninty64

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    Fermat's Last Theorem related question

    Homework Statement Show that x^{n}+y^{n}=z^{n} has a nontrivial solution if and only if the equation \frac{1}{x^{n}}+\frac{1}{y^{n}}=\frac{1}{z^{n}} has a nontrivial solution. Homework Equations By nontrivial solutions, it is implied that they are integer solutions. The Attempt at a Solution...
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    Prove that every non-zero vector in V is a maximal vector

    Homework Statement Let V be a finite dimensional vector space and T is an operator on V. Assume μ_{T}(x) is an irreducible polynomial. Prove that every non-zero vector in V is a maximal vector. Homework Equations μ_{T}(x) is the minimal polynomial on V with respect to T. The Attempt at...
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    Vector Spaces and Correspondence

    I think I get it now. The function maps the coset of V mod U to the same coset of V mod U by mapping each individual element to another element in that coset. I'm sorry if it seemed I brushed over your post and didn't completely read it. I did. I just didn't understand it. I'm having trouble...
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    Vector Spaces and Correspondence

    This is where I confuse myself. T is a linear transformation from V to V. If it maps the set v+U to the set v+U, then wouldn't that be mapping cosets of V mod U to cosets of V mod U instead of mapping V to V?
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    Vector Spaces and Correspondence

    Homework Statement This question came out of a section on Correspondence and Isomorphism Theorems Let V be a vector space and U \neq V, \left\{ \vec{0} \right\} be a subspace of V. Assume T \in L(V,V) satisfies the following: a) T(\vec{u} ) = \vec{u} for all \vec{u} \in U b) T(\vec{v} + U) =...
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    Isomorphic Vector Spaces Proof

    Thank you so much for your help!
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    Isomorphic Vector Spaces Proof

    What I take away from the First Isomorphism Theorem is that if two vectors \vec{v}, \vec{u} \in V, then for any Linear transformation T:V\rightarrow W, if T(\vec{u}) = T(\vec{v}) then \vec{u} \equiv \vec{v} modulus Ker(T) So if you only take the cosets of V mod Ker(T), then it follows that...
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    Isomorphic Vector Spaces Proof

    We went over all three of those. I don't fully understand them, though. I think that's my main problem. I went back to basis because I was familiar with that. It is "the collection of cosets of V modulo Ker(f)". I quoted that from the book because I would have gotten that wrong. I understand...
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    Isomorphic Vector Spaces Proof

    Homework Statement Let V be a vector space over the field F and consider F to be a vector space over F in dimension one. Let f \in L(V,F), f \neq \vec{0}_{V\rightarrow F}. Prove that V/Ker(f) is isomorphic to F as a vector space. Homework Equations L(V,F) is the set of all linear maps...
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    Sum of this geometric sequence doesn't make sense!

    The sum of a geometric series is defined as: a+ar+ar^2+ar^3+...+ar^{n-1} = a\frac{1-r^n}{1-r} If n started at 0, then a would be 2. Since n starts at 1, in order to form a geometric series we must group it as following: \frac{8}{3} + \frac{8}{3}(\frac{4}{3}) + \frac{8}{3}(\frac{4}{3})^2 +...
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    Sum of this geometric sequence doesn't make sense!

    I'm assuming that second 2 is a typo and should be an n. \sum ^{14}_{n=1} 2(\frac{4}{3})^n I believe the equation is working. a represents the first term in the series. In this case, what is a?
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    Proof involving Rank Nullity Theorem

    I just fixed a pretty bad typo my self. Mine should be correct now.
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    Proof involving Rank Nullity Theorem

    If Kernel(T)=Kernel(T^2), then Range(T)=Range(T^2) First I started by saying that Kernel(T) = Kernel(T^2) \Rightarrow Nullity(T) = Nullity(T^2) By the Rank Nullity Theorem we have the following: Dim(V) = Rank(T) + Nullity(T) Dim(V) = Rank(T^2) + Nullity(T^2) \Rightarrow Rank(T) +...
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    Proof involving Rank Nullity Theorem

    Thank makes sense! Then it follows that since \text{ker}(T) \subset \text{ker}(T^2) and Nullity(T)=Nullity(T^2) then \text{ker}(T) = \text{ker}(T^2) Since the Kernel is, by definition, a subspace of V. And if a Vector Space with dimension n is contained in another Vector Space with...
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    Proof involving Rank Nullity Theorem

    Ah, ok. I think that makes sense. Thanks! Ooh. I didn't understand that operator on V implied T:V->V. Does being an operator imply that it is a linear transformation? Or is that part just assumed in the problem? Ok. So trying that I get T^2(c_{1}\vec{v_{1}}+c_{2}\vec{v_{2}}) =...
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