0
$\begingroup$

Let $u_1=(1,2,0,-1)$, $u_2=(0,2,-1,1)$, $u_3=(3,4,1,-4)$ and $v_1=(-2,-2,1,3)$, $v_2=(2,3,2,-6)$, $v_3=(-1,4,6,-2)$. Let $H =span\{u_1,u_2,u_3\}$ and $K = span\{v_1,v_2,v_3\}$.

In here I have to find bases for $H$, $K$ and $H+K$. I can't understand how to do it. I wrote vectors in $H$ and $K$ as linear combinations. Then I think I have to prove that those vectors are linearly independent. But I don't know how to do it. Can you help me to find an answer for this question?

$\endgroup$
0
$\begingroup$

Note that $u_3 = 3u_1-u_2$, then $\{u_1,u_2,u_3\}$ is linearly dependent and $u_3\in span\{u_1,u_2\}$.

Hint. In general, to see that vectors $w_1,\cdots,w_k$ are linearly independent, write $$\alpha_1 w_1 + \cdots + \alpha_k w_k = 0$$ and try to prove that $\alpha_i = 0$ for all $i\leq k$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

So $$\alpha \cdot v_1 +\alpha_2\cdot v_2= v_3\implies \alpha_1\cdot (-2,-2,1,3)+\alpha_2 \cdot (2,3,2,-6)=(-1,4,6,-2) \implies -2\alpha _1+2\alpha _2=-1,-2\alpha_1+3\alpha_2=4,\alpha _1+2\alpha_2=6 \text { and }3\alpha_1-6\alpha _2=-2 \implies -1-2\alpha_2=4-3\alpha_2 \implies \alpha_2=5 \text { and } \alpha_1= \frac {11}2 \text { and } \alpha_1+2\alpha_2=6 $$, a contradiction. So $v_1, v_2 \text { and }v_3 $ are linearly independent. .. Now we know $\{u_1, u_2\} $ is a basis for H (by @ridias), and $\{v_1,v_2, v_3\} $ a basis for K... Now for H+K, check if any $v_i , i=1,2,3$ is a linear combination of $u_1$ and $u_2$ . This will enable you to determine how many and which of $u_1,u_2,v_1,v_2,\text {and }v_3 $are linearly independent, and get a basis for H +K...

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.