Let $U_1, U_2 $ be sub spaces in $R_4[x]$, such that:

$$U_1 = Sp\{x^3+2x^2+3x+6, 4x^3-x^2+3x+6, 5x^3+x^2+6x+12\}$$ $$U_2 = Sp\{x^3-x^2+x+1,2x^3-x^2+4x+5\}$$

Find $U_1 \cap U_2$

My idea (i need help how to proceed): compare linear combination of both spanning sets, get the general solution for the homogeneous system.

In the end im not sure, its becoming too complicated technically.

Let $a_1,a_2,a_3,a_4,a_5 \in \Bbb R$:

$$a_1(x^3+2x^2+3x+6)+a_2(4x^3-x^2+3x+6)+a_3(5x^3+x^2+6x+12) = a_4(x^3-x^2+x+1)+a_5(2x^3-x^2+4x+5)$$

$$\Downarrow$$

$$a_1(x^3+2x^2+3x+6)+a_2(4x^3-x^2+3x+6)+a_3(5x^3+x^2+6x+12) - a_4(x^3-x^2+x+1) - a_5(2x^3-x^2+4x+5) = 0$$

$$\Downarrow$$

$$\begin{bmatrix}1&4&5&-1&-2&|0 \\ 2&-1&1&1&1&|0 \\ 3&3&6&-1&-4&|0 \\ 6&6&12&-1&-5&|0\end{bmatrix} \xrightarrow{rank} \begin{bmatrix}1&4&5&-1&-2&|0 \\ 0&-9&-9&3&5&|0 \\ 0&0&0&-1&-3&|0 \\ 0&0&0&0&0&|0\end{bmatrix}$$

$$a_5 = t$$ $$a_4 = -3t$$ $$a_3 = s$$ $$a_2 = 9s+4t$$ $$a_1 = -41s-17t$$

Therefore, the general solution:

$$\Rightarrow t(-17,4,0,-3,1) + s(-41,9,1,0,0)$$

Take t = 1: $$a_5 = 1$$ $$a_4 = -3$$ $$\Rightarrow a_4(x^3-x^2+x+1)+a_5(2x^3-x^2+4x+5)$$ $$\Rightarrow -3(x^3-x^2+x+1)+1(2x^3-x^2+4x+5)$$ $$\Downarrow$$ $$-x^3+2x^2+x+2$$

Is it the only vector in the intersection?

$$dim(U_1 \cup U_2) = dim(U_1) + dim(U_2) - dim(U_1 \cap U_2)$$

I calculated in previous question so: $$dim(U_1 \cap U_2) = 2 + 2 - 3 = 1$$

So there is only 1 vector in the base of $(U_1 \cap U_2)$.

$$B_{(U_1 \cap U_2)} = \{-x^3+2x^2+x+2\}$$