Matrix Base and Transition matrices

Let $B_1 = \{v_1, v_2, v_3\}$ be a basis for a vector space $V$, and let \left\{\begin{align} u_1 &= v_1, \\ u_2 &= v_1 + v_2, \\ u_3 &= v_1 + v_2 + v_3. \end{align}\right.

(a) Show that $B_2 = \{u_1, u_2, u_3\}$ is also a basis for $V$.

(b) Find the transition matrices $P_{B_1 \to B_2}$ and $P_{B_2 \to B_1}$

I'm not sure how to prove part (a) besides saying that $B_2$ is a basis since it follows the additive property of vector spaces. Are there any other ways to prove $B_2$ is a basis?

On part (b) I'm not sure how to find transition matrices for general matrices. I've only done them for matrices with concrete values.

• A collection of vectors $\{u_1, \dots, u_n\}$ is a basis for a vector space $V$ if it satisfies two conditions. What are they? Commented Oct 26, 2015 at 20:33
• I edited your question to make the $\LaTeX$ mark-up more readable. Have a look at the source (click "edit" under the question) to learn how some of the commands/syntax for future questions. Here's a tutorial, too: meta.math.stackexchange.com/q/5020/6509 Commented Oct 26, 2015 at 21:12

a) I recall you that a list $w_1,w_2,w_3$ of vectors are linearly independent in $V$ if $$\forall a,b,c \qquad aw_1+bw_2+cw_3 = 0 \quad \implies \quad a=b=c=0.$$
So let $a,b,c$ be such that $au_1+bu_2+cu_3 = 0$, then we have $$0=au_1+bu_2+cu_3=(a+b+c)v_1+(b+c)v_2+cv_3.$$ Now, $v_1,v_2,v_3$ is a basis, so we must have $$c=b+c=a+b+c=0.$$ I leave you to show that this implies that $a=b=c = 0$.
b) For this part, it is not clear what you mean exactly by transition matrix. Moreover, note that the coefficients of a matrix depends on the basis in which you express them. I will show you how to find a matrix $P$ such that $Pu_i = v_i$ for $i=1,2,3$ and express this matrix with respect to the basis $B_1$. You should then be able to reuse this example to answer the question.
We have $$Pu_1 = Pv_1, \quad Pu_2 = Pv_1+Pv_2, \quad Pv_3=Pv_1+Pv_2+Pv_3,$$ and we want $Pu_1 = v_1,Pu_2=v_2$ and $Pu_3=v_3$. This is implies $$Pv_1 = v_1, \quad Pv_2 = v_2-Pv_1=v_2-v_1, \quad Pv_3 = v_3-Pv_2-Pv_1=v_3-v_2.$$ Hence, the matrix $P$ has to satisfy, $$P\begin{pmatrix}1\\ 0 \\ 0 \end{pmatrix}_{B_1}={\color{red}{\begin{pmatrix}1\\ 0 \\ 0 \end{pmatrix}_{B_1}}},\quad P\begin{pmatrix}0\\ 1 \\ 0 \end{pmatrix}_{B_1}={\color{blue}{\begin{pmatrix}-1\\ 1 \\ 0 \end{pmatrix}}}, \quad P\begin{pmatrix}0\\ 0 \\ 1 \end{pmatrix}_{B_1}={\color{green}{\begin{pmatrix}0\\ -1 \\ 1 \end{pmatrix}_{B_1}}}.$$ It follows that $$[P]_{B_1,B_1}=\begin{pmatrix} \color{red}1 & \color{blue}{-1} & \color{green}0 \\ \color{red}0 & \color{blue}1 & \color{green}{-1} \\ \color{red}0 & \color{blue}0 & \color{green}1 \end{pmatrix}.$$ It is now easily checked that $$Pu_1 = P\begin{pmatrix}1\\ 0 \\ 0 \end{pmatrix}_{B_1}=\begin{pmatrix}1\\ 0 \\ 0 \end{pmatrix}_{B_1}=v_1, \quad Pu_2 = P\begin{pmatrix}1\\ 1 \\ 0 \end{pmatrix}_{B_1}=\begin{pmatrix}0\\ 1 \\ 0 \end{pmatrix}_{B_1}=v_2, \quad Pu_3 = P\begin{pmatrix}1\\ 1 \\ 1 \end{pmatrix}_{B_1}=\begin{pmatrix}0\\ 0 \\ 1 \end{pmatrix}_{B_1} =v_3,$$ i.e. $Pu_i=v_i$ for $i=1,2,3$.