# Checking if intersection is correct

1. Nov 20, 2016

### Kernul

1. The problem statement, all variables and given/known data
The exercise gives me $V = <(1, 0, 0, 1), (-7, 0, 1, 0)>$ and $U = <(0, 4, 1, -1), (1, 12, 0, 26)>$ and I have to find dimension and a base of $V \cap U$ and $V + U$.

2. Relevant equations

3. The attempt at a solution
They both have dimension $2$. I find a vector $\vec v \in V \implies \vec v = \alpha (1, 0, 0, 1) + \beta (-7, 0, 1, 0) = (\alpha - 7 \beta, 0, \beta, \alpha)$
Now I have to see if $\vec v \in U$ and this implies that the vector just found has to be linearly dependent to the other two of U. I can see for what parameters this is true by putting all in a matrix and implying the determinant to be $0$.
$$\begin{pmatrix} \alpha - 7 \beta & 0 & \beta & \alpha \\ 0 & 4 & 1 & -1 \\ 1 & 12 & 0 & 26 \end{pmatrix}$$
I use Kronecker's theorem and put the two determinants in a system.
$$\begin{cases} \begin{vmatrix} \alpha - 7 \beta & 0 & \beta \\ 0 & 4 & 1 \\ 1 & 12 & 0 \end{vmatrix} = 0 \\ \begin{vmatrix} \alpha - 7 \beta & 0 & \alpha \\ 0 & 4 & -1 \\ 1 & 12 & 26 \end{vmatrix} = 0 \end{cases}$$
But doing the calculations I find myself with this:
$$\begin{cases} -4 \beta - 12 \alpha + 84 \beta = 0 \\ 116 \alpha - 812 \beta - 4 \alpha = 0 \end{cases}$$
$$\begin{cases} 12 \alpha = 80 \beta \\ 112 \alpha = 812 \beta \end{cases}$$
And if I keep on going, I'll have $\alpha = \frac{20}{3} \beta$. Should I stop here and go substituting this in the vector and get the intersection? What about the other equation in the system? Is it okay to leave it like that? Because if I substitute the $\alpha$ just found in there, I would get $\beta = 0$.

2. Nov 20, 2016

### ehild

Are the four vectors (1, 0, 0, 1), (-7, 0, 1, 0),(0, 4, 1, -1), (1, 12, 0, 26) independent?

3. Nov 22, 2016

### Kernul

I put them in a matrix and did the determinant. I have a determinant different from $0$ ($-196$), so they are independent.

4. Nov 22, 2016

### ehild

So what is the intersection of the subspaces U and V? Can be any linear combination of the base vectors of U element of V?

Can you consider the four independent vectors as base of a 4-dimensional space? So what is the union of U and V ?

5. Nov 22, 2016

### Kernul

Oh, so the intersection is void? This means that the sum will be a direct sum, right?

6. Nov 22, 2016

### ehild

Yes, the intersection contains only the zero vector.
I do not know what you mean on U+V, but the "Union" of the two subspaces would be a sensible question.

7. Nov 22, 2016

### Kernul

The exercise asks me about the intersection and the sum of the two subspaces. I know that when the intersection is the zero vector, then the sum of those two subspaces is a sum called "direct sum".
By the way, the union wouldn't simply have the dimension as the sum of the two dimension's subspaces and a base as the union of the two bases?

8. Nov 22, 2016

### ehild

Yes, I think so. All the four vectors have four components, so they are elements of a four-dimensional space. As they are independent, they can be considered as an other base of that space. You can consider it as the direct sum of the two two-dimensional spaces U and V.

Last edited: Nov 22, 2016