Am I understanding this right?

The set of all 2x2 matrices of the form $$\left[\begin{matrix}a & b \ c & 1\end{matrix}\right]$$ would not be a vector space because it is not closed under scalar multiplication.

i.e.

$$2\left[\begin{matrix}a & b \ c & 1\end{matrix}\right]$$

would result in

$$\left[\begin{matrix}2a & 2b \ 2c & 2\end{matrix}\right]$$

which would not be a part of the original matrix form defined.

Question

Am I understanding this right?

The set of all 2x2 matrices of the form $$\left[\begin{matrix}a & b \ c & 1\end{matrix}\right]$$ would not be a vector space because it is not closed under scalar multiplication.

i.e.

$$2\left[\begin{matrix}a & b \ c & 1\end{matrix}\right]$$

would result in

$$\left[\begin{matrix}2a & 2b \ 2c & 2\end{matrix}\right]$$

which would not be a part of the original matrix form defined.

jamesj · Answer

Correct.

chriss · Answer

thank you

jamesj · Answer

It also isn't a closed subset under addition

jamesj · Answer

and finally, the zero matrix isn't a member of this subset.  So in fact, every axiom needed for this set to be a vector sub-space of the 2x2 matrices fails.

chriss · Answer

right, so the idea is that I should be able to perform any standard operation on the matrix and get a result that still fits the pattern of the original matrix

jamesj · Answer

Any vector operation, yes, and I assume the binary operation you're interested in is addition.

There are also matrices where the binary operation you want is multiplication, but that doesn't come into play so much with vector space theory.

chriss · Answer

In this case I only have to find that it either passes or fails.  I'm trying to find it in my notes, but I can't... however, even though there are ten vector space axioms total, the one I cited plus the 2 you added are sufficient to determine if the matrix form passes or fails, correct?

jamesj · Answer

If it fails any axiom, then it's not a vector space.  In order to show it is not a vector space, it is sufficient to show that it fails just one, as you have done.  (And yes, there are others.  For example, the existence of an additive inverse.)

But to show that a set V with operations is a vector space, you must show every axiom is satisfied.

chriss · Answer

Ok.  That's been very helpful.  I might have been misremembering, but I thought he said that there were 3 axioms that if it passed those, it would pass all the others, but again I can't find it in my notes so I am probably not remembering it right.

Thanks again for your help!

jamesj · Answer

Look that up.  It's an important trick to know how to check 3 things vs. 10, particularly in an exam.

chriss · Answer

I found it, and I did somewhat misremember it.  That is in reference to subspaces.  If W is a nonempty subset of V, then W is a subspace of V iff it is closed under addition and scalar multiplication.