# Linear Algebra II - Change of Basis

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1. Aug 24, 2015

### lemonsare

1. The problem statement, all variables and given/known data

From Linear Algebra with applications 7th Edition by Keith Nicholson.
Chapter 9.2 Example 2.

Let T: R3 → R3 be defined by T(a,b,c) = (2a-b,b+c,c-3a).
If B0 denotes the standard basis of R3 and B = {(1,1,0),(1,0,1),(0,1,0)}, find an invertible matrix P such that P-1MB0(T)P=MB(T).

2. Relevant equations

3. The attempt at a solution

I know to find P, I have to first find MB0(T) and MB(T).

MB0(T) is easy because I have to just find each a, b, and c as linear combinations of B0 and the coefficients are MB0(T).

However, I'm not sure how to find MB(T). In the textbook, they write:

MB(T) = [CB(1,1,-3) CB(2,1,-2) CB(-1,1,0)]

I cannot figure out where they came up with these numbers (1,1,-3), (2,1,-2), and (-1,1,0).

2. Aug 24, 2015

### RUber

$M_B(T)$ is the 3x 3 matrix where the columns are the result of T on each of the basis vectors.
T(1,1,0) = (1,1,-3).
So your matrix will look like:
$\begin{bmatrix} 1& 2& -1 \\ 1& 1& 1 \\ -3 & -2 & 0 \end{bmatrix}$

3. Aug 24, 2015

### lemonsare

Oh wow thanks! That was simple.

But in the textbook $M_B(T)$ = $\begin{bmatrix} 4&4&-1\\-3&-2&0\\-3&-3&2\end{bmatrix}$

From trying to figure out where (1,1,-3) came from, I figured out that (4,-3,-3) came from the coefficients of the linear combination of each of the basis vectors. But is that wrong?