- #1

Kernul

- 211

- 7

## Homework Statement

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##.

## Homework Equations

## 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##.