How to find the basis (and cartesian equation(s)) of a sum of two vectorial subspaces?

Question

I have the following vectorial subspaces :

U = span $\left(\begin{pmatrix} 2 \\ 0 \\ 1 \\ -2 \end{pmatrix},\begin{pmatrix} 3 \\ 6 \\ 9 \\ -12 \end{pmatrix} \right)$ and V = span $\left(\begin{pmatrix} 0 \\ 2 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix} \right)$

To calculate the basis of U + V should have use this method : $\left(\begin{array}{cccc|c} 2 & 3 &0 & -1& 0 \\ 0 & 6 &2&1 & 0 \\ 1 & 9 &1&0&0 \\ -2 & -12&0&1 & 0 \end{array}\right)$ which generate this solution after reduction $\left(\begin{array}{cccc|c} 1 & 0 &0 & -\frac{1}{2}& 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$ $\sim$ $\begin{pmatrix} 1 & 0 & 0 & -\frac{1}{2} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{2} \end{pmatrix}$

Now I don't understand how to have cartesian equation(s) of this hyperplane. Can someone help me understanding this?

Answer 1

We have obtained that a basis is given by the following vectors

$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}= a\begin{pmatrix} 2 \\ 0 \\ 1 \\ -2 \end{pmatrix} +b\begin{pmatrix} 3 \\ 6 \\ 9 \\ -12 \end{pmatrix}+ c\begin{pmatrix} 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}$$

to obtain the cartesian equation we need to eliminate $a$,$b$, and $c$ from the system

• $x_1=2a+3b$
• $x_2=6b+2c$
• $x_3=a+9b+c$
• $x_4=-2a-12b$

and determine the relation between $x_1$, $x_2$, $x_3$ and $x_4$.

We can also proceed by elimination using the augmented matrix

$$\left(\begin{array}{ccc|c} 2 & 3 &0 & x_1 \\ 0 & 6 &2 & x_2 \\ 1 & 9 &1&x_3 \\ -2 & -12&0 & x_4 \end{array}\right)$$

operating in the same way as you already did, you'll obtain the cartesian equation in the last row.

Answer 2

Since it is a subspace, it pass the origin, so has an equation like: $$ax+by+cz+dt=0$$ Now put $4$ points in that and solve a system of linear equations to find $a,b,c,d$.