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I'm trying to solve the problem below.

$U$ and $W$ are subspaces of polynomials over $\mathbb{R}$.

$U = Span(t^3 + 4t^2 - t + 3, t^3 + 5t^2 + 5, 3t^3 + 10t^2 -5t + 5)$ $W = Span(t^3 + 4t^2 + 6, t^3 + 2t^2 - t + 5, 2t^3 + 2t^2 -3t + 9)$

What is $dim(U \cap W)$?

I have solved it using the fact that $dim(U) + dim(W) - dim(U \cap W) = dim(U \cup W)$, but was wondering how to solve it without using this fact.

In order to find $dim(U \cap W)$, I first try and find $U \cap W$.

Clearly if $v \in U \cap W$, then $$\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5) = \beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$ for some $\alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3 \in \mathbb{R}$.

Using this fact, you can reduce a system of linear equations to work out that:

$\alpha_1 + 5\alpha_3 - \beta_2 - 3\beta_3 = 0$

$\alpha_2 -2 \alpha_3 + 2\beta_2 + 6\beta_3 = 0$

$\beta_1 + 2\beta_2 + 5\beta_3 = 0$

But I don't know where to go from here.

Any help would be greatly appreciated.

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2 Answers 2

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Consider a polynomial $p(t)=at^{3}+bt^{2}+ct+d$ belonging to $U$ and $V$ at the same time and solve two separated systems to find the conditions for $p(t)$ to be in $U$ and beside find the coditions for $p(t)$ to be in $V$ and once you find those two list of conditions consider a system on the unkowns $a,b,c$.

$$p(t)=\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5)$$

$$ p(t)=\beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$

then

$$at^{3}+bt^{2}+ct+d=\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5)$$

$$at^{3}+bt^{2}+ct+d=\beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$

this implies that

For $U$ you need to find the conditions on $a,b$ and $c$ for the following system to have solution:

$$at^{3}+bt^{2}+ct+d=\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5)$$

$$A_{U}=\left(\begin{array}{ccc|c} 1&1&3&a\\ 4&5&10&b\\ -1&0&-5&c\\ 3&5&5&d \end{array}\right)$$

For $V$ you need to find the conditions on $a,b$ and $c$ for the following system to have solution:

$$at^{3}+bt^{2}+ct+d=\beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$

$$A_{V}=\left(\begin{array}{ccc|c} 1&1&2&a\\ 4&2&2&b\\ 0&-1&-3&c\\ 6&5&9&d \end{array}\right)$$

The row reduced echelon form of $A_{U}$ will give you the conditions for a polynomial to be in $U$. The same for $A_{V}$. These conditions are linear equations in $a,b,c$ that you can put in a linear system of equations in the variables $a,b,c$ to finally find the conditions for $p(t)$ to be in the intersection of $U$ and $V$.

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  • $\begingroup$ Thanks for your reply Hector. How can I use this to find $U \cap W$? $\endgroup$
    – Jack
    Commented Nov 12, 2017 at 15:51
  • $\begingroup$ Find the conditions for a polynomial $p(t)$ to be in $U$ and separated from that find also the conditions for the same $p(t)$ to be in $V$. Then, you put in a system of equations together both sets of conditions. $\endgroup$ Commented Nov 12, 2017 at 16:00
  • $\begingroup$ Hector, I've reduced the system of equations as far as they go at the bottom of my question. Where do I go from here? $\endgroup$
    – Jack
    Commented Nov 12, 2017 at 16:03
  • $\begingroup$ Reduce the two separated systems I showed in my answer to clarify the idea :) $\endgroup$ Commented Nov 12, 2017 at 16:08
  • $\begingroup$ @Jack: Find the conditions for a polynomial $p(t)$ to be in $U$ and separated from that find also the conditions for the same $p(t)$ to be in $V$. Then, you put in a system of equations together both sets of conditions. I show the idea in my answer. $\endgroup$ Commented Nov 12, 2017 at 16:35
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Consider the matrix:

\begin{bmatrix} 1 & 4 & -1 & 3 \\ 1 & 5 & 0 & 5 \\ 3 & 10 & -5 & 5 \\ 1 & 4 & 0 & 6 \\ 2 & 2 & -3 & 9 \\ \end{bmatrix} Row reduce the matrix to its Row reduced echelon form to get the basis of $U + W = \text{Span}\{v_1,v_2,v_3,v_4,v_5,v_6 \}$.

Then use the formula:

$$\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)$$

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  • $\begingroup$ Ah, I specified in my question not to use this approach. Is there another way in which you could solve this problem? $\endgroup$
    – Jack
    Commented Nov 12, 2017 at 16:02

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