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

Question

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 $U_1 \cap U_2$

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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 $(U_1 \cap U_2)$.

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

Answer 1

Note that the sum of the third vector spanning $U_1$ is the sum of the first two. Taking (4 x 1st vector - 2nd vector)/9 we get $u_1=x^2+x+2$. Subtracting $4u_1$ from the 1st vector gives $u_2=x^3-2x^2-x-2$, so $u_1,u_2$ span $U_1$.

For $U_2$ subtracting twice the 1st vector from the second gives $v_1=x^2+2x+3$. Subtracting that from the first vector gives $u_2$, so $U_2$ is spanned by $v_1,u_2$.

Clearly, multiples of $u_2$ belong to $U_1\cap U_2$. So $U_1\cap U_2$ has dimension 1 or 2. But $u_2+u_1=x^3-x^2\in U_1$ and $x^3-x^2\notin U_2$, so $U_1\cap U_2$ must have dimension 1 and hence be just the multiples of $u_2$.

Answer 2

I’d word a few things differently, but your solution looks correct to me.

Note that you could’ve found bases for both $U_1\cap U_2$ and $U_1+U_2$ (which you asked about in a previous question) at the same time. Forming the combined matrix of all of the spanning vectors and fully reducing that matrix, we have $$\left[\begin{array}{ccc|cc}1&4&5&1&2\\2&-1&1&-1&-1\\3&3&6&1&4\\6&6&12&1&5\end{array}\right] \to \left[\begin{array}{ccc|cc}1&0&1&0&\frac79\\0&1&1&0&-\frac49\\0&0&0&1&3\\0&0&0&0&0\end{array}\right].$$ Examining the pivots of the reduced matrix as a whole, we can see that a basis for $U_1+U_2$ is $\{(1,2,3,6)^T,(4,-1,3,6)^T,(1,-1,1,1)^T\}$. This is no doubt a different basis than the one that you obtained by row-reducing the transpose of this matrix. Note, too, that this reduced matrix tells us that $\dim(U_1)=\dim(U_2)=2$, so just as you’ve calculated, we must have $\dim(U_1\cap U_2)=1$.

For $U_1\cap U_2$, we examine the null space, which is spanned by $(1,1,-1,0,0)^T$ and $(7/9,-4/9,0,3,-1)^T$. The first of these only tells us that the generating vectors of $U_1$ aren’t linearly independent, which doesn’t really help. On the other hand, the second one tells us that $$\frac79\begin{bmatrix}1\\2\\3\\6\end{bmatrix}-\frac49\begin{bmatrix}4\\-1\\3\\6\end{bmatrix} = -3\begin{bmatrix}1\\-1\\1\\1\end{bmatrix}+1\begin{bmatrix}2\\-1\\4\\5\end{bmatrix}.$$ This vector, which corresponds to $-x^3+2x^2+x+2$, lies in the intersection of the two spaces, which we know from above is one-dimensional, so that’s our basis for the intersection.