# Using coordinates in B to find coordinates in B'

Part 1:
$B=\{v_1,v_2,v_3\}$ is a basis of the real vector space $V$.
${B'}=\{v_1+v_2,av_1+v_3, bv_1-v_3\},\quad a,b\in\mathbb R$
What conditions should $a$ and $b$ satisfy for $B'$ to be a basis of $V$?

Part 2:
If $a=1, b=2$. Given $u\in V$ such that $[u]_B=\begin{pmatrix}-1\\2\\1\end{pmatrix}$, find $[u]_{B'}$

There are two parts to the problem, I am unsure how to do both. I tried reverse engineering the math behind finding coordinates but it didn't work out for me. The problem may be worded oddly, I had to translate it from spanish to english.

## 1 Answer

1) You can apply the definition. Let $\mu_1,\mu_2,\mu_3$ be such that $$0=\mu_1(v_1+v_2)+\mu_2(av_1+v_3)+\mu_3(bv_1-v_3)=(\mu_1+a\mu_2+b\mu_3)v_1+\mu_1v_2+ (\mu_2-\mu_3)v_3\ \ \ (*)$$ Since $v_1,v_2,v_3$ is a basis, we must have $0= \mu_1=\mu_2 - \mu_3=\mu_1+a\mu_2+b\mu_3$. Thus $\mu_2 = \mu_3$ and $\mu_2(a+b)=0$. $B'$ is linearly independent if and only if equation $(*)$ imply $0=\mu_1=\mu_2=\mu_3$. It is now clear that it is true if and only if $a \neq -b$. From the above discussion we know that $a \neq -b$ imply $0=\mu_1=\mu_2 = \mu_3$. Now suppose $a = -b$, if $a=0$, then clearly $B'$ is not linearly independent and if $a \neq 0$, choose $\mu_1 = 0,\mu_2 = \mu_3= 1$. Note that it is sufficient to check if $B'$ is linearly independent to know if it is a basis since it contains $3$ vectors.

2) So you have $u = -v_1+2v_2+v_3$ and you want to find $\mu_1,\mu_2,\mu_3$ such that $u = \mu_1v_1'+\mu_2v_2'+\mu_3v_3'$ whith $B'=\{v_1',v_2',v_3'\}$ (then $[u]_{B'}=(\mu_1,\mu_2,\mu_3)^T$). By identification with equation $(*)$ you get the system $$\left\{\begin{array}{rcl}-1&=&\mu_1+\mu_2+2\mu_3\\2 &=& \mu_1\\ 1&=&\mu_2-\mu_3 \end{array}\right. \iff \left\{\begin{array}{rcl}\mu_1 &=& 2\\ \mu_2&=& -\frac{1}{3}\\ \mu_3&=& -\frac{4}{3}\end{array}\right.$$

• Upon attempting part 2 again I got u=v1(r+2t)+v2(r+s)+v3(s-t) which I then took the r,s,t components and set them equal to the other coordinates components (r+2t=-1, ...), however, the solution to the system of equations does not exist. I believe I am still doing something wrong. – Stephensson Jun 29 '14 at 15:11
• @Stephensson Please check that the OP has been correctly copied to know if the typo comes from me or the passage from image to Latex. – Surb Jun 29 '14 at 15:47
• The OP is in line with what the problem asks. No mistake from latex side. – Stephensson Jun 29 '14 at 16:23
• @Stephensson Ok so there is a mistake in my post, because I took $B' = \{v_1+v2,av_{\color{red}2}+v_3,bv_1-v_3\}$. Gimme little time to edit that and answer your question about 2). – Surb Jun 29 '14 at 16:31
• @Stephensson I edited my answer. – Surb Jun 29 '14 at 17:39