Linear transformation and change of basis

[negation]

Homework Statement

Let B = {(1, -2),(2, -3)} and S be the standard basis of R2
and [-8,-4;9,4]
be a linear transformation expressed in terms of the standard basis?

The Attempt at a Solution

1) What is the change of basis matrix PSB ?
1,2
-2,-3

2)What is the change of basis matrix PBS ?
-3,-2
2,1

What is the linear transform expressed in terms of the basis B?
i spent 4 hours going through the proof for this part but unable to bridge a connection. Something is lacking in my understanding.
Give me a step by step guidance on this one.

 

[pondzo]

I won't give you a step by step guide but i will provide a link that gives a better one than i ever could!
https://www.khanacademy.org/math/li...transformation-matrix-with-respect-to-a-basis

 

[negation]

I won't give you a step by step guide but i will provide a link that gives a better one than i ever could!
https://www.khanacademy.org/math/li...transformation-matrix-with-respect-to-a-basis

Khan is good but his proof and subscript can at times be sloppy. I enjoy Mathispower4u (discovered it before Physicforum did) but it seems there are no videos on change on basis on Mathispower4u.

But anyway, I have a much rigorous proof on hand that I wish to understand fully.

The transformation machine that maps the vector v with respect to the standard basis, s, is given as [f(v)]_s.
[f(v)]_s = A_ss [v]_s
A_ss = [ [f(e₁)]_s [f(e₂)]_s...[f(e_n)]_s ] (I don't get this part)

There's more to come but I hope to understand this part first before proceeding...

 

[HallsofIvy]

The basic point is this: if we have basis {u, v} then every vector can be written in the form au+ bv and represented as "\begin{bmatrix}a \\ b\end{bmatrix}". In particular, u= 1u+ 0v and is represented as \begin{bmatrix}1 \\ 0\end{bmatrix} while v= 0u+ 1v and is represented as \begin{bmatrix}0 \\ 1 \end{bmatrix}. Now suppose linear transformation, A, is represented by the matrix
\begin{bmatrix}a & b \\ c & d\end{bmatrix}
Then Au is
\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}
and Av is
\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}b \\ d\end{bmatrix}

That is, applying the linear transformation to the basis vectors, and then writing the result as a linear combination of the basis vectors, the coefficents give the columns.

In this particular case,
\begin{bmatrix}-8 & -4\\ 9 & 4\end{bmatrix}\begin{bmatrix}1 \\ -2\end{bmatrix}= \begin{bmatrix}0 \\ 1\end{bmatrix}
and we can write \begin{bmatrix}0 \\ 1\end{bmatrix}= -2\begin{bmatrix}1 \\ -2\end{bmatrix}+ 1\begin{bmatrix}2 \\ -3\end{bmatrix}

so the first column of the matrix representation is \begin{bmatrix}-2 \\ 1\end{bmatrix}

Similarly,
\begin{bmatrix}-8 & -4\\ 9 & 4\end{bmatrix}\begin{bmatrix}2 \\ -3\end{bmatrix}= \begin{bmatrix}-4 \\ 6\end{bmatrix}
and \begin{bmatrix}-4 \\ 6\end{bmatrix}= 0\begin{bmatrix}1 \\-2\end{bmatrix}- 2\begin{bmatrix}2 \\ -3\end{bmatrix}
so the second column is \begin{bmatrix}0 \\ -2\end{bmatrix}

The matrix representation in this (ordered) basis is
\begin{bmatrix}-2 & 0 \\1 & -2\end{bmatrix}

 

[negation]

The basic point is this: if we have basis {u, v} then every vector can be written in the form au+ bv and represented as "\begin{bmatrix}a \\ b\end{bmatrix}". In particular, u= 1u+ 0v and is represented as \begin{bmatrix}1 \\ 0\end{bmatrix} while v= 0u+ 1v and is represented as \begin{bmatrix}0 \\ 1 \end{bmatrix}. Now suppose linear transformation, A, is represented by the matrix
\begin{bmatrix}a & b \\ c & d\end{bmatrix}
Then Au is
\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}
and Av is
\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}b \\ d\end{bmatrix}

That is, applying the linear transformation to the basis vectors, and then writing the result as a linear combination of the basis vectors, the coefficents give the columns.

In this particular case,
\begin{bmatrix}-8 & -4\\ 9 & 4\end{bmatrix}\begin{bmatrix}1 \\ -2\end{bmatrix}= \begin{bmatrix}0 \\ 1\end{bmatrix}
and we can write \begin{bmatrix}0 \\ 1\end{bmatrix}= -2\begin{bmatrix}1 \\ -2\end{bmatrix}+ 1\begin{bmatrix}2 \\ -3\end{bmatrix}

so the first column of the matrix representation is \begin{bmatrix}-2 \\ 1\end{bmatrix}

Similarly,
\begin{bmatrix}-8 & -4\\ 9 & 4\end{bmatrix}\begin{bmatrix}2 \\ -3\end{bmatrix}= \begin{bmatrix}-4 \\ 6\end{bmatrix}
and \begin{bmatrix}-4 \\ 6\end{bmatrix}= 0\begin{bmatrix}1 \\-2\end{bmatrix}- 2\begin{bmatrix}2 \\ -3\end{bmatrix}
so the second column is \begin{bmatrix}0 \\ -2\end{bmatrix}

The matrix representation in this (ordered) basis is
\begin{bmatrix}-2 & 0 \\1 & -2\end{bmatrix}

Would the outcome be any different if I first went down the route of

[-8;-9] = α₁[1;0] + α₂[0;1]
[-4;4] = β₁[1;0] + β₁[0;1]

that is, I express each column of the matrix A as a linear combination of the vectors in S.

 

[negation]

The basic point is this: if we have basis {u, v} then every vector can be written in the form au+ bv and represented as "\begin{bmatrix}a \\ b\end{bmatrix}". In particular, u= 1u+ 0v and is represented as \begin{bmatrix}1 \\ 0\end{bmatrix} while v= 0u+ 1v and is represented as \begin{bmatrix}0 \\ 1 \end{bmatrix}.


Now suppose linear transformation, A, is represented by the matrix
\begin{bmatrix}a & b \\ c & d\end{bmatrix}
Then Au is
\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}
and Av is
\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}b \\ d\end{bmatrix}

That is, applying the linear transformation to the basis vectors, and then writing the result as a linear combination of the basis vectors, the coefficents give the columns.

In this particular case,
\begin{bmatrix}-8 & -4\\ 9 & 4\end{bmatrix}\begin{bmatrix}1 \\ -2\end{bmatrix}= \begin{bmatrix}0 \\ 1\end{bmatrix}
and we can write \begin{bmatrix}0 \\ 1\end{bmatrix}= -2\begin{bmatrix}1 \\ -2\end{bmatrix}+ 1\begin{bmatrix}2 \\ -3\end{bmatrix}

so the first column of the matrix representation is \begin{bmatrix}-2 \\ 1\end{bmatrix}

Similarly,
\begin{bmatrix}-8 & -4\\ 9 & 4\end{bmatrix}\begin{bmatrix}2 \\ -3\end{bmatrix}= \begin{bmatrix}-4 \\ 6\end{bmatrix}
and \begin{bmatrix}-4 \\ 6\end{bmatrix}= 0\begin{bmatrix}1 \\-2\end{bmatrix}- 2\begin{bmatrix}2 \\ -3\end{bmatrix}
so the second column is \begin{bmatrix}0 \\ -2\end{bmatrix}

The matrix representation in this (ordered) basis is
\begin{bmatrix}-2 & 0 \\1 & -2\end{bmatrix}

Ok I got this. Basically, all I had to find was [T(x)_s]_B.