Transition matrix between two bases?

[concon]

Homework Statement

Let So = {v1,v2,v3,v4} be basis of vector space V.
And S = {u1,u2,u3,u4} be set of vectors defined as follows:
u1 = 20v1 + 46v2 + 116v3 + 170v4
u2 = 20v1 + 86v2 + 147v3 + 174v4
u3 = 30v1 + 89v2 + 59v3 + 81v4
u4 = 15v1 + 27v2 + 12v3 + 9v4

Find transition matrix A from So to S which is transition matrix from S coordinates to So coordinates.

Homework Equations

I know how to find transition matrix, it's hard to explain put you want to but one basis down and the other basis next to it and turn one into identity matrix.

The Attempt at a Solution

So first off we know that S is a basis of vector space V since it just scalar multiples of So.

I have calculated transition for small basis like {(1,2),(2,1)} but am unsure how to set up for this problem

If someone can set it up for me or give advice on what to do I can solve it. Thanks in advance!

 

[electricspit]

You're given $S=\vec{u}$ in terms of the basis $S_0=\vec{v}$. Or:

$\begin{pmatrix}u_1\\u_2\\u_3\\u_4\end{pmatrix}=\begin{pmatrix}20 && 46 && 116 && 170\\20 && 86 && 147 && 174\\30 && 89 && 59 && 81\\15 && 27 && 12 && 9\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\\v_4\end{pmatrix}$

Which can be written more compactly:

$\vec{u}=T\vec{v}$

Where $T$ is the transition matrix from $S_0$ coordinates to $S$ coordinates. I hope this helps!

 

[concon]

You're given $S=\vec{u}$ in terms of the basis $S_0=\vec{v}$. Or:

$\begin{pmatrix}u_1\\u_2\\u_3\\u_4\end{pmatrix}=\begin{pmatrix}20 && 46 && 116 && 170\\20 && 86 && 147 && 174\\30 && 89 && 59 && 81\\15 && 27 && 12 && 9\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\\v_4\end{pmatrix}$

Which can be written more compactly:

$\vec{u}=T\vec{v}$

Where $T$ is the transition matrix from $S_0$ coordinates to $S$ coordinates. I hope this helps!

Okay so then I should take inverse of the T to find transition matrix?

 

[electricspit]

Yes. You should be able to use:

$\vec{u}=T\vec{v}$

To show that. It should take 1 line.

 

[concon]

Yes. You should be able to use:

$\vec{u}=T\vec{v}$

To show that. It should take 1 line.

i just did thst, but apparently my answer was wrong. can you explain exactly the process? Do I use the numbers in the exact order you set up?

 

[electricspit]

Make the augmented matrix:

$A = [T|I]$

and do elementary row operations until you get:

$B = [I|T^{-1}]$

Where $I$ is the identity matrix. Does that help?