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:
[itex] \begin{bmatrix} -1 & 1 & -1 \\ 1 & 2 & -1 \\ 2 & 0 & 1 \end{bmatrix}[/itex]
Which is -1.

Now you have to go the other way around. I mean you know how [itex]v[/itex]'s are given in terms of [itex]u[/itex]'s... you should actually go to the inverse relation and see how [itex]u[/itex]'s are written in terms of [itex]v[/itex]'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 [itex]v_i[/itex].

Try to find [itex]u_1[/itex] in terms of [itex]v_{1,2,3}[/itex].

(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 [itex]u_i[/itex] in terms of [itex]v_j[/itex]:

[itex] \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} [/itex]

To find ##M## here:

[itex] \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix}= M \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} [/itex]

 

[ChrisVer]

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

That means you are taking a particular basis for [itex]u_i[/itex]. Then the [itex]v[/itex]'s are:

[itex] v_1 = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}[/itex], [itex] v_2 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}[/itex], [itex] v_3 = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}[/itex].

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

[itex]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}[/itex].
That's 3 equations with 3 unknown variables (a,b,c) so it's solvable:
[itex] \begin{align*} -a +b& +2c = 1 \\ a+2b&=0 \\ -a-b&+c =0 \end{align*} [/itex]

And similarily work for the rest [itex] u_2 \& u_3[/itex]. 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 [itex]M= \begin{bmatrix} a & b & c \\ d & f & e \\ h & w & r \end{bmatrix}[/itex] 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.