1
$\begingroup$

I've been doing all of these proofs the same basically, I just want to make sure I'm doing them right, I didn't include all the details but I have the outlines of my proofs here.

1) U and W are subspace of V s.t. V = $U \oplus W$. Prove $u_1, \dots, u_m, w_1, \dots, w_m$ is a basis of V where the u's are a basis for U and the w's a basis for W.

I said that since the basis for U is lin ind in V then you can extend it to be a basis by adding the vectors from W. Since they are linearly independent as well, then you add every vector. This shows that $u_1, \dots, u_m, w_1, \dots, w_m$ is linearly independent. It obviously spans V because V = $U \oplus W$, there fore it is a basis

2) Prove or give counter example. $v_1, \dots, v_n$ and $v_1, \dots, v_n$ are linearly independent in V. Then:

$\lambda v_1, \dots, \lambda v_n$ with $\lambda \in F$ is linearly indp.

$\lambda (a_1 v_1 + \dots + a_n v_n) = 0$ (pulling out $\lambda$ from the span, we know the span is linearly indp). Therefore $\lambda 0 = 0$.

$v_1 + w_1, \dots, v_m + w_m$ is linearly indp.

Rearranging the span of this you get: $a_1 v_1 + \dots + a_m v_m + a_1 w_1 +\dots a_m + w_m = 0$ Split this into span(U) = $0$ and span(V) = $0$ implies all the coefficients must be zero.

$\endgroup$
  • 1
    $\begingroup$ Both are incorrect. In the first part you have to use the fact that $U\cap W=\{0\}$; in the second part, what happens if $\lambda=0$? $\endgroup$ – egreg Jul 14 '14 at 21:50
2
$\begingroup$

There are some glitches in your reasoning. In the first part you must use the hypothesis that $U\cap W=\{0\}$. In the second part, you're forgetting that $\lambda v=0$ when $\lambda=0$.

Part 1.

Let $\{u_1,u_2,\dots,u_m\}$ and $\{w_1,w_2,\dots,w_n\}$ be bases of $U$ and $W$ respectively. We want to prove that $$\{u_1,u_2,\dots,u_m,w_1,w_2,\dots,w_n\}$$ is linearly independent. So, suppose $$ a_1u_1+a_2u_2+\dots+a_mu_m+b_1w_1+b_2w_2+\dots+b_nw_n=0 $$ We must obviously use the fact that $U\cap W=\{0\}$ (which is part of the hypothesis $V=U\oplus W$), so we try to get a vector in $U\cap W$: set $$ v=a_1u_1+a_2u_2+\dots+a_mu_m. $$ By definition, $v\in U$. But we also have $$ v=-(b_1w_1+b_2w_2+\dots+b_nw_n) $$ so $v\in W$.

Then …

The part about $\{u_1,u_2,\dots,u_m,w_1,w_2,\dots,w_n\}$ being a spanning set is easy: if $v\in V$, then from $V=U\oplus W$ we know that $v=u+w$, with $u\in U$ and $w\in W$. Then we can write $u=\dots$ and $w=\dots$, so …

Part 2

If $\{v_1,v_2,\dots,v_n\}$ is linearly independent and $\lambda\in F$, we can consider $$ a_1(\lambda v_1)+a_2(\lambda v_2)+\dots+a_n(\lambda v_n)=0 $$ from which $$ \lambda(a_1v_1+a_2v_2+\dots+a_nv_n)=0 $$ that gives $$ \lambda=0\qquad\text{or}\qquad a_1v_1+a_2v_2+\dots+a_nv_n=0 $$ Thus …

$\endgroup$

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.