Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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.

Linear Algebra
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

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

So which axioms are you having trouble showing?
All of them. I just do not understand for some reason.
I can walk you through some of them, and hopefully you'll be able to finish the rest.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

ok
Oh, before I start, we're assuming this space of symmetric matrices has real valued entries, and has 2x2 matrices. Correct?
yes
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?
yes.
So then Axiom 2 would be satisfied right?
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.
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.
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.
yes those last two are starting to make sense now.
Sorry, for Axiom 3, I should have said "use the fact that addition of real numbers is associative."
I knew what you meant.
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.
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.
ok starting to make sense now thanks.
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 \[\vec{1}=\text{Id}_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]
Call me back if you need help with any more of the axioms.
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.
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question