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anonymous
 3 years ago
Let V be the set of 2x2 matrices with standard definitions for vector addition and scalar multiplication. Determine whether V is a vector space. If V is not a vector space, show that at least one of the 10 axioms does not hold.
Have to show that the set of all real symmetric matrices, that is, the set of all matrices such that A^t=A is a vector space.
anonymous
 3 years ago
Let V be the set of 2x2 matrices with standard definitions for vector addition and scalar multiplication. Determine whether V is a vector space. If V is not a vector space, show that at least one of the 10 axioms does not hold. Have to show that the set of all real symmetric matrices, that is, the set of all matrices such that A^t=A is a vector space.

This Question is Closed

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1So which axioms are you having trouble showing?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0All of them. I just do not understand for some reason.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I can walk you through some of them, and hopefully you'll be able to finish the rest.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Oh, before I start, we're assuming this space of symmetric matrices has real valued entries, and has 2x2 matrices. Correct?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1So if \(v\in V\), \(v\) looks like \[\begin{bmatrix}a&b\\b&a\end{bmatrix}\]for \(a,b\in\mathbb{R}\). Axiom 1: If \(u,v\in V\), then \(u+v\in V\). Pf: Let \[u=\begin{bmatrix}a&b\\b&a\end{bmatrix},\quad v=\begin{bmatrix}c&d\\d&c\end{bmatrix}.\]Then \[u+v=\begin{bmatrix}a+c&b+d\\b+d&a+c\end{bmatrix}.\]This is still a symmetric matrix, so axiom 1 is satisfied. Make sense?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So then Axiom 2 would be satisfied right?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Almost. There's one more small thing we have to show. Axiom 2: If \(u,v\in V\), then \(u+v=v+u\). Pf: We know\[u+v=\begin{bmatrix}a+c&b+d\\b+d&a+c\end{bmatrix}=\begin{bmatrix}c+a&d+b\\d+b&c+a\end{bmatrix}=v+u.\]So axiom 2 is satisfied because addition of real numbers is commutative.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Axiom 3: If \(u,v,w\in V\), then \((u+v)+w=u+(v+w)\). Pf: I think you can do this yourself. Just let \(u,v\) be the same matrices as above, and let \[w=\begin{bmatrix}e&f\\f&e\end{bmatrix}\]Then add the matrices together, and use the fact that addition of real numbers is commutative.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Axiom 4: There exists \(\vec{0}\in V\) such that \(\vec{0}+v=v+\vec{0}=v\) for all \(v\in V\). Pf: Again, I think you can do this. Let \[\vec{0}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\]and do the necessary operations.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes those last two are starting to make sense now.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry, for Axiom 3, I should have said "use the fact that addition of real numbers is associative."

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I knew what you meant.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Axiom 5: For all \(v\in V\) there exists some \(v\in V\) such that \(v+(v)=v+v=\vec{0}\). Pf: Another very straightforward one. \[v=\begin{bmatrix}a&b\\b&a\end{bmatrix},\quad v=\begin{bmatrix}a&b\\b&a\end{bmatrix}.\]Add them together.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Axiom 6: For all \(c\in\mathbb{R}\) and \(v\in V\), \(cv\in V\). Pf: Let\[v=\begin{bmatrix}a&b\\b&a\end{bmatrix}.\]Then\[cv=c\begin{bmatrix}a&b\\b&a\end{bmatrix}=\begin{bmatrix}ca&cb\\cb&ca\end{bmatrix}.\]This is still in the form we want, so this axiom is satisfied.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok starting to make sense now thanks.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I think you'll be able to handle the rest. Just make sure to take things one step at a time. Also, for the last axiom, use \[\vec{1}=\text{Id}_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Call me back if you need help with any more of the axioms.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have another question for you. Let V=\[\left\{ \left(\begin{matrix}\cos t \\ \sin t\end{matrix}\right) \right\}\] and define \[\left(\begin{matrix}\cos t_1 \\ \sin t_1\end{matrix}\right)⊕ \left(\begin{matrix}\cos t_2 \\ \sin t_2\end{matrix}\right)=\left(\begin{matrix}\cos (t_1+t_2) \\ \sin (t_1+t_2)\end{matrix}\right)\] c ⨀\[\left(\begin{matrix}\cos t \\ \sin t\end{matrix}\right)=\left(\begin{matrix}\cos ct \\ \sin ct\end{matrix}\right)\] Show that if ⊕ and ⨀ are the standard componentwise operations, then V is not a vector space. I'm not quite sure it is not a vector space.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I think your problem will be with axiom 5: For all \(v\in V\), there exists a \(v\) such that \(v+v=v+(v)=\vec{0}\). There is a \(\vec0\) in this space (try to find this), but you may not have the inverses.
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