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.