Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Okay, here's the problem:

Suppose that {$u_1,u_2,u_3$} forms a basis for a vector space V. Show that if $$v_1=u_1+2u_3$$$$v_2=u_1+2u_2+3u_3$$$$v_3=u_2-u_3$$ then {$v_1,v_2,v_3$} forms a basis for V.

So what I've decided so far is that if $M=\begin{bmatrix} 1&1&0\\ 0&2&1\\ 2&3&-1 \end{bmatrix}$ and $U=\begin{bmatrix} u_1&u_2&u_3 \end{bmatrix}$ then the vectors $v_1,v_2,v_3$ are the columns of $ UM$. So I just need to show, I guess, that the columns of $UM$ are linearly independent. $M$ is invertible; I'm pretty sure that's important. If I had a theorem that said that the product of two matrices with linearly independent columns has linearly independent columns then I'd be set, but I don't believe I do (I would if I knew that U was square). Am I going in a good direction here, or is there a smarter way to do this?

share|cite|improve this question
up vote 0 down vote accepted

Your reasoning is also correct. Here's the argument.

If $\{v_1,v_2,v_3\}$ is linearly independent then this forms a basis.

So, let us suppose that $\alpha_1 v_1+\alpha_2 v_2+\alpha_3 v_3=0$.

Then $(\alpha_1+\alpha_2)u_1+(2\alpha_2+\alpha_3)u_2+(2\alpha_1+3\alpha_2-\alpha_3)u_3=0$.

Since $\{u_1, u_2, u_3\}$ forms a basis, we have

$\begin{eqnarray*}\alpha_1+\alpha_2=0\\ 2 \alpha_2+\alpha_3=0\\ \text{ and }\\ 2\alpha_1+3\alpha_2-\alpha_3=0\end{eqnarray*}$

The above system has the only solution $\alpha_1=\alpha_2=\alpha_3=0$ iff the coefficient matrix $M$ is invertible.

share|cite|improve this answer
Thanks, I'm starting to get this stuff I think. – crf Nov 1 '11 at 21:00

Let $v \in V$ be arbitrary. Because $u_1,u_2,u_3$ are a basis you know that $v=a_1 u_1+a_2 u_2+a_3 u_3$, but by solving the linear system you have that

$$u_1=\frac{1}{3} (5 v_1 - 2 v_2 + 4 v_3)$$

$$u_2=\frac{1}{3} (-v_1 + v_2 + v_3)$$

$$u_3=\frac{1}{3} (-v_1 + v_2 - 2 v_3)$$

Therefore you know that $v \in \text{span}\{v_1,v_2,v_3\}$, because $v$ was arbitrary it follows that $\{v_1,v_2,v_3\}$ spans your vector space, and because all bases of a vector space have the same size you found a basis of $V$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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