1. ## Linear Transformations

Determine the matrix [T: B_1, B_2] for the given L.T. T and the basis B_1 and B_2.

B_1={(1,1),(1,0)} and B_2={(2,3),(4,5)}.

I need Solution..

2. One thing to keep in mind here is that you're just trying to find the $B_2$ coordinates of a point that is given to you in the $B_1$ coordinate system, so the point itself does not change.

A point in the $B_1$ coordinate system has the coordinates (x, y) expressed as the linear combination of the $B_1$ basis vectors:

$x\begin{bmatrix}1 \\ 1\end{bmatrix} + y\begin{bmatrix}1 \\ 0\end{bmatrix}$

In the $B_2$ coordinate system, a point (u, v) will have coordinates of the expressed as linear combinations of the $B_2$ basis vectors:

$u\begin{bmatrix}2 \\ 3\end{bmatrix} + v\begin{bmatrix}4 \\ 5\end{bmatrix}$

These are the same point (we are just changing the coordinate system), so:

$x\begin{bmatrix}1 \\ 1\end{bmatrix} + y\begin{bmatrix}1 \\ 0\end{bmatrix} = u\begin{bmatrix}2 \\ 3\end{bmatrix} + v\begin{bmatrix}4 \\ 5\end{bmatrix}$

Thus,

$B_1\begin{bmatrix}x \\ y\end{bmatrix} = B_2\begin{bmatrix}u \\ v\end{bmatrix}$, where $B_1 = \begin{bmatrix}1 & 1\\ 1 & 0\end{bmatrix}$ and $B_2 = \begin{bmatrix}2 & 4\\ 3 & 5\end{bmatrix}$

In order to get $\begin{bmatrix}u \\ v\end{bmatrix}$ by itself, multiply both sides of the equation on the left by $B_2^{-1}$ to get:

$\begin{bmatrix}u \\ v\end{bmatrix} = B_2^{-1} B_1\begin{bmatrix}x \\ y\end{bmatrix}$

So, the matrix that changes base from $B_1$ to $B_2$ is $B_2^{-1} B_1$.