# Polynomial subspaces

1. Nov 13, 2011

### gotmejerry

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution
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 $dim(M+N)=dim M + dim N- dim(M \cap N)$. The diemnsion of the subspace generated by p1 and p2 is 2, and so is the dimension of the subspace generated by q1 and q2. Am I right? But then from this $dim(M+N)=dim M + dim N- dim(M \cap N)$ I get that $(M \cap N)$ has a dimension of 1.

Thank you!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 13, 2011

### Deveno

1. how do you conclude that dim(M+N) = 3? clearly it's ≤ 3, but how do you know that we need all 3 basis elements to describe something in M+N? it's not immediately clear that span({p1,p2,q1,q2}) is all of R2[t].

2. before you can conclude dim(M) and dim(N) = 2, you should show that {p1,p2} and {q1,q2} are actually linearly independent. this isn't hard to do.

3. assuming you do (1) and (2) above, then you are in a position to use

dim(M+N) = dim(M) + dim(N) - dim(M∩N).

finding the dimensions isn't really the hard part. finding the basis elements is. what would an element of M∩N look like?

3. Nov 13, 2011

### gotmejerry

1. If I take the linear combinations of p1 p2 q1 q2 as I have written and Im not wrong I think the basis {t^2 t 1} is ok.

2. I found out that {p1,p2} and {q1,q2} are linearly independent, because a*p1+b*p2=0 s only solution is the trivial soulution, same for {q1,q2}. So dim(M)=2 and dim(N)=2.

3. An element in $M \cap N$ is something which is in the span({p1,p2}) and in the span({q1,q2}) at the same time. But how do I find a basis ?

4. Nov 13, 2011

### Deveno

for span({p1,p2,q1,q2}) to equal R2[t], every polynomial must be in the span.

if we have any abitrary polynomial At2 + Bt + C, in the span, this means that we have:

a+b+c+d = A
a+2c-d = B
a+b+3c+d = C

this is equivalent to:

$$\begin{bmatrix}1&1&1&1\\1&0&2&-1\\1&1&3&1 \end{bmatrix}\begin{bmatrix}a\\b\\c\\d \end{bmatrix} = \begin{bmatrix}A\\B\\C \end{bmatrix}$$

now, it isn't obvious that the 4x3 matrix actually has rank 3. i think you might have made a lucky guess.

as for M∩N, suppose we have a(1+t+t2) + b(1-t2) = c(1-t+t2) + d(3+2t+t2). what can we say about a,b,c and d?

5. Nov 13, 2011

### gotmejerry

a(1+t+t2) + b(1-t2) = c(1-t+t2) + d(3+2t+t2) is true if

a-b=c+d
a=2d-c
a+b=c+3d

What do I do now?