# Dimension and basis

• Nov 12th 2011, 10:05 AM
noricka
Dimension and basis
Let $M$ be a subspace of the vector space $\mathbb{R}_2[t]$ generated by $p_1(T)=t^2+t+1$ and $p_2(T)=1-t^2$, and $N$ be a subspace generated by $q_1(T)=t^2+2t+3$ and $q_2(T)=t^2-t+1$. Show the dimension of the following subspaces: $M+N$, $M \cap N$, and give a basis for each.
• Nov 12th 2011, 10:44 AM
Drexel28
Re: Dimension and basis
Quote:

Originally Posted by noricka
Let $M$ be a subspace of the vector space $\mathbb{R}_2[t]$ generated by $p_1(T)=t^2+t+1$ and $p_2(T)=1-t^2$, and $N$ be a subspace generated by $q_1(T)=t^2+2t+3$ and $q_2(T)=t^2-t+1$. Show the dimension of the following subspaces: $M+N$, $M \cap N$, and give a basis for each.

So, what are your thoughts?
• Nov 12th 2011, 11:23 AM
noricka
Re: Dimension and basis
Quote:

Originally Posted by Drexel28
So, what are your thoughts?

I have tried the following: if I take the linear combination of $p_1$ $p_2$ $q_1$ $q_2$, I get $(a+b+c+d)t^2 + (a+2c-d)t +(a+b+3c+d).$ And a basis of this polynomial is $\{1,t,t^2\}$, which means the dimension of M+N is 3.

And if M and N are finite dimension subspaces then $dim(M+N)=dim M + dim N- dim(M \cap N)$. But what is the dimension of M and N?
• Nov 12th 2011, 12:18 PM
Deveno
Re: Dimension and basis
how many basis elements do M and N have?
• Nov 12th 2011, 12:40 PM
noricka
Re: Dimension and basis
Both of them have 3? (Happy)
• Nov 12th 2011, 01:01 PM
Deveno
Re: Dimension and basis
how can a set generated by 2 elements have 3 basis vectors???

to be fair, p1 and p2 are only listed as generating elements, its not explicitly stated whether or not {p1,p2} forms a basis.

but, by the very definition of "generate" they are elements of a span-set. are they linearly independent? (if you decide they are, how can you be sure?

and what does this mean dim(M) is?)
• Nov 12th 2011, 01:13 PM
noricka
Re: Dimension and basis
p1 and p2 are linearly independent, so their dimension is 2?
• Nov 12th 2011, 01:16 PM
Deveno
Re: Dimension and basis
elements do not have dimension, vector spaces (and their subspaces) have dimension.

the dimension of a vector space, V, is defined to be the number of elements in ANY basis.

if S is a subset of V, then dim(span(S)) ≤ |S|.

if S is a linearly independent set, then dim(span(S)) = |S|.
• Nov 12th 2011, 01:29 PM
noricka
Re: Dimension and basis
So the dimension of the subspace M generated by p1, and p2 is 2, then. And that's the same with N as well.
• Nov 12th 2011, 01:37 PM
noricka
Re: Dimension and basis
Was my idea of the dimension of M+N right? And what about the bases?