Basis and Components (Vectors)

[RyanH42]

Homework Statement

Let $u_1,u_2,u_3$ be a basis and let $v_1=-u_1+u_2-u_3$ , $v_2=u_1+2u_2-u_3$ , $v_3=2u_1+u_3$ show that $v_1,v_2,v_3$ is a basis and find the components of $a=2u_1-u_3$ in terms of $v_1,v_2,v_3$

Homework Equations

For basis vecor we need to clarify these vector are lineerly independet.We can understand linearly independent as make a matrix 3x3 and see the take the det of matrx. If det of matrix is not zero means linearly independent.

The Attempt at a Solution

I take the determinant of $v_1,v_2,v_3$ and I found -1.It means linearly independent.It means they are basis vectors.Then I tried to show $u_1=(1,0,0)$ then I tried write it in terms of $v$ but I thought that there a lot way to do that or maybe wrong.

The answer is $u_1=-2v_1+v_2-v_3$ and $u_2=3v_1-v_2+2v_3$ and $u_3=4v_1-2v_2+3v_3$ and $a=-8v_!+4v_2-5v_3$
Thanks

 

[ChrisVer]

The matrix you are looking the determinant of is:
$\begin{bmatrix} -1 & 1 & -1 \\ 1 & 2 & -1 \\ 2 & 0 & 1 \end{bmatrix}$
Which is -1.

Now you have to go the other way around. I mean you know how $v$'s are given in terms of $u$'s... you should actually go to the inverse relation and see how $u$'s are written in terms of $v$'s (so you can make the substitution in alpha)...

If you are lucky you can see what you have to do right away by a few tries of adding/substracting and multiplying with factors the $v_i$.

Try to find $u_1$ in terms of $v_{1,2,3}$.

(do you know about inverse matrices?)

 

[RyanH42]

(do you know about inverse matrices?)

Yeah,I know.

 

[ChrisVer]

you can use the inverse matrix to get the $u_i$ in terms of $v_j$:

$\begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 2 & -1 \\ 2 & 0 & 1 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix}$

To find $M$ here:

$\begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix}= M \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}$

 

[ChrisVer]

You can try also your way $u_1 = \begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix}$, $u_2 = \begin{pmatrix} 0 \\ 1 \\0 \end{pmatrix}$, $u_3 = \begin{pmatrix} 0\\ 0 \\1 \end{pmatrix}$.

That means you are taking a particular basis for $u_i$. Then the $v$'s are:

$v_1 = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}$, $v_2 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$, $v_3 = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$.

Now you would like to see how you can add $v_1,v_2,v_3$ in order to get $u_1$. You can try that by solving the equation for $a,b,c$:

$u_1 = av_1+ bv_2 +cv_3 \Rightarrow \begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix}= a \begin{pmatrix} -1 \\ 1 \\-1 \end{pmatrix}+ b\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} +c \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$.
That's 3 equations with 3 unknown variables (a,b,c) so it's solvable:
$\begin{align*} -a +b& +2c = 1 \\ a+2b&=0 \\ -a-b&+c =0 \end{align*}$

And similarily work for the rest $u_2 \& u_3$. That will be in total 9 equations with 9 unknown parameters.

That's an alternative way of saying that you are taking the inverse matrix.. you just wrote for $M= \begin{bmatrix} a & b & c \\ d & f & e \\ h & w & r \end{bmatrix}$ and you go to determine its elements one by one.

 

[RyanH42]

$1=-a+b+2c$
$0=a+2b$
$0=-a-b+c$
then #### $a=-2b$ then If I put this in third equation equation I found $b=-c$ .If I put these info on first equation I get $1=2b+b-2b$, $b=1$ then $a=-2$ and $c=-1$ Which İt fits the answer.
I understand the matrix way and this way.Matrix way is look like simpler but Its hard to find inverse matrix of 3x3.I used online calculator and I found exact solution.
Thanks my friend.