Hi, pearlyc!

Originally Posted by

**pearlyc** Let S denote the standard basis for $\displaystyle \mathbb{R}^2$ and B = { $\displaystyle [\begin{matrix} 1 \\ 2\end{matrix}]$ , $\displaystyle [\begin{matrix} -1 \\ 0\end{matrix}]$ } be another basis.

(a) write down the change of basis matrix P B->S, from the basis B to the basis S.

P B->S = $\displaystyle [\begin{matrix} 1 & -1 \\ 2 & 0\end{matrix}]$

Correct!

Originally Posted by

**pearlyc** (b) Hence find the change of basis matrix P S->B, from the basis S to the basis B.

P S-> B = inverse of (P B->S)

= inverse of $\displaystyle [\begin{matrix} 1 & -1 \\ 2 & 0\end{matrix}]$

= 1/2 $\displaystyle [\begin{matrix} 0 & 1 \\ -2 & 1\end{matrix}]$

Correct!

Originally Posted by

**pearlyc** (c) Find $\displaystyle [x]_B$ if x = (4,-1)

I don't know how to do this one, can you guys help me out?

What's the trouble? You have a vector with coordinates relative to the standard basis, and need to find the coordinates relative to the basis $\displaystyle B$. Use your transition matrix!

Originally Posted by

**pearlyc** (d) Show that $\displaystyle T : \mathbb{R}^2 -> \mathbb{R}^2$ defined by :

$\displaystyle T(x,y) = (-x, + 2y, 3y)$

Let x = [ x1, x2], y = y1, y2]

T(x+y)= T(x1 + y1, x2 + y2) = (-x1 + 2y1, 3y1)

= ( -x1, 0.x2) + (2y1, 3y2)

= T(x1,x2) + T(y1,y2)

T(ax) = T(ax1, ax2)

= (-ax1, a.0)

= a (-x1, 0)

= aT(x)

Are you being asked to prove that $\displaystyle T$ is a linear transformation?

Your work is completely wrong. I have highlighted your errors in red.

Follow the definition of $\displaystyle T$, and don't confuse the different sets of coordinates that you have in use. For example, here is how you could do the first part:

$\displaystyle T(x,\ y) = (-x + 2y,\ 3y)$

$\displaystyle \text{Let }\textbf{u} = (u_1,\ u_2)\text{ and }\textbf{v} = (v_1,\ v_2)$

$\displaystyle \text{So, }T(\textbf{u} + \textbf{v})$

$\displaystyle =T(u_1 + v_1,\ u_2 + v_2)$

$\displaystyle =\left(

-(u_1 + v_1) + 2(u_2 + v_2),\ 3(u_2 + v_2)

\right)$

$\displaystyle =\left(

-u_1 - v_1 + 2u_2 + 2v_2,\ 3u_2 + 3v_2

\right)$

$\displaystyle =\left(

-u_1 + 2u_2 - v_1 + 2v_2,\ 3u_2 + 3v_2

\right)$

$\displaystyle =

(-u_1 + 2u_2,\ 3u_2) + (-v_1 + 2v_2,\ 3v_2)

$

$\displaystyle =

T(u_1,\ u_2) + T(v_1,\ v_2)

$

$\displaystyle =

T(\textbf{u}) + T(\textbf{v})

$

See if you can prove the second property now.

Originally Posted by

**pearlyc** (e) Find the matrix representation of T with respect to the standard basis $\displaystyle [T]_s$

$\displaystyle A_T$ = $\displaystyle [\begin{matrix} -1 & 2 \\ 0 & 3\end{matrix}]$

Correct!

Originally Posted by

**pearlyc** (f) Use your answers from above to find the matrix representation of T with respect to the basis B, $\displaystyle [T]_B$

$\displaystyle T[b] = P S->B . [T]_S . P B-> S$

= $\displaystyle [\begin{matrix} 3 & 0 \\ 0 & -1\end{matrix}]$

Looks good to me.

Originally Posted by

**pearlyc** (g) Find $\displaystyle [T(x)]_B$ where $\displaystyle [x]_B$ = $\displaystyle [\begin{matrix} -3 \\ 5\end{matrix}]$

$\displaystyle [Tx]_B$ = [tex][T]_B[x]_B = $\displaystyle [\begin{matrix} -9 \\ 5\end{matrix}]$

Check your multiplication here.