Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- schrodinger

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- KingGeorge

So which axioms are you having trouble showing?

- anonymous

All of them. I just do not understand for some reason.

- KingGeorge

I can walk you through some of them, and hopefully you'll be able to finish the rest.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

ok

- KingGeorge

Oh, before I start, we're assuming this space of symmetric matrices has real valued entries, and has 2x2 matrices. Correct?

- anonymous

yes

- KingGeorge

So 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

yes.

- anonymous

So then Axiom 2 would be satisfied right?

- KingGeorge

Almost. 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

Axiom 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

Axiom 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

yes those last two are starting to make sense now.

- KingGeorge

Sorry, for Axiom 3, I should have said "use the fact that addition of real numbers is associative."

- anonymous

I knew what you meant.

- KingGeorge

Axiom 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

Axiom 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

ok starting to make sense now thanks.

- KingGeorge

I 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

Call me back if you need help with any more of the axioms.

- anonymous

I 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

I 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.