Finding $\dim(U\cap V)$ of two different vector subspaces in $\mathbb R^4$

Question

$U=sp\{(1,2,1,0),(-1,1,1,1)\}$ and $V=sp\{(2,-1,0,1),(-5,6,3,0)\}$

My try is $U\cap V$ = $sp\{(-5/4,2/4,3/4,1)\}$. Then $\dim(U\cap V)$ = $1$

Do I have the solution well?. Help, please. I'm very confused.

Refer to How to find basis for intersection of two vector spaces in $\mathbb{R}^n$ — ChoF, Mar 18 '18 at 05:35

Siong Thye Goh · Accepted Answer · 2018-03-18T05:48:06.133

1

Guide:

Let $U = sp \{ u_1, u_2\}$.

Let $V = sp \{ v_1, v_2\}$.

First verify that the $U \ne V$, hence note that $\operatorname{dim}(U \cap V) < 2$.
Solve $\alpha u_1 + \beta u_2 = \gamma v_1 + \delta v_2$

That is $$\begin{bmatrix} u_1 & u_2 & -v_1 & -v_2\end{bmatrix}\begin{bmatrix} \alpha \\ \beta \\ \gamma \\ \delta\end{bmatrix}=0$$

To find a basis, i.e. $\{ \alpha u_1 +\beta u_2\}.$

The claim that the rank is $1$ is correct, but currently the matrix that you proposed doesn't lie in $V$.

edited Mar 18 '18 at 05:48

answered Mar 18 '18 at 05:38

Siong Thye Goh

Thank you, you're very kind :). – Jean Paul Mar 18 '18 at 05:47
How exactly do you select your basis vectors for $U \cap V$ based on the $\alpha,\beta,\gamma,\delta$ you got from solving the system of equations? – mdcq Mar 23 '18 at 19:01
1

After you solve the linear system, you find that the nullity is $1$, that is the nullspace is $1$, That is the solution to $\alpha u_1 + \beta u_2 = \gamma v_1 + \delta v_2$ is unique up to a scalar multiplication. Hence, we pick one of them say $(\alpha_0, \beta_0, \gamma_0, \delta_0)$ is a non-zero solution and ${\alpha_0 u_1 +\beta_0 u_2}$ is a basis. – Siong Thye Goh Mar 23 '18 at 21:00

1 Answers1