# Transition Matrix and Ordered Bases

1. Apr 14, 2013

### LosTacos

Let B and C be ordered bases for ℝn. Let P be the matrix whose columns are the vectors in B and let Q be the matrix whose columns are the vectors in C. Prove that the transition matrix from B to C equals Q-1P.

I am stuck. Here is what I have.

I know that if B is the standard basis in ℝn, then the transition matrix from B to C is given by [1st vector in C 2nd vector in C ........... nth vector in C]-1.

Also, if C is a standard basis in ℝn, then the transition matrix from B to C is given by [1st vector in B 2 vector in B ......... nth vector in B].

Since I konw what the transition matrix is from B to C given different standard bases, I am having a difficult time relating this to teh columns of each.

2. Apr 14, 2013

### Fredrik

Staff Emeritus
You're going to need a notation for the basis vectors. I suggest
\begin{align} B&=\{e_1,\dots,e_n\}\\
C &=\{f_1,\dots,f_n\}
\end{align} How do you define the transition matrix from B to C? Is it the M defined by
$$f_i=\sum_j M_{ij} e_j$$ or the M defined by
$$f_i=Me_i=\sum_j (Me_i)_j e_j=\sum_j M_{ji} e_j?$$ (The latter M is the transpose of the former). You want to prove that (with one of these choices of M), we have $M=Q^{-1}P$. This is equivalent to $QM=P$, which is equivalent to $P_{ij}=(QM)_{ij}=$what? Use the definition of matrix multiplication to rewrite $(QM)_{ij}$. Then you can start thinking about rows and columns.

3. Apr 15, 2013

### LosTacos

Im confused with the notation of the matrix. How do you rewrite the rows and columns

4. Apr 15, 2013

### Fredrik

Staff Emeritus

One notation that can be useful is to denote the number on row i, column j of a matrix A by $A^i_j$ instead of $A_{ij}$. Then you can just denote the ith row by $A^i$.

So for example, we have $P_i=e_i$ for all i.

5. Apr 16, 2013

### HallsofIvy

Staff Emeritus
If we have basis $\{u_1, u_2, \cdot\cdot\cdot, u_n\}$ for vector space U, then we can represent a vector $u= a_1u_1+ a_2u_2+ \cdot\cdot\cdot+ a_nu_n$ as the array $\left<a_1, a_2, \cdot\cdot\cdot, a_n\right>$.
In particular the basis vectors themselves are very easy:
$u_1= \left< 1, 0, \cdot\cdot\cdot, 0\right>$
$u_2= \left<0, 1, \cdot\cdot\cdot, 0\right>$
... $u_n= \left<0, 0, \cdot\cdot\cdot, 1\right>$
Now, look at what when you multiply each of those, written as a column by a matrix:
Multiplying $u_1$ gives just the first column, multiplying $u_2$ gives the second column, etc.
Example:
$$\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}= \begin{bmatrix}a_{12} \\ a_{22} \\ a_{32}\end{bmatrix}$$
which will then be the coefficients of the expansion of Au in whatever basis we are using for the range space. That is, to represent linear transformation A from U to V, using a given ordered basis for each, apply A to each basis vector for U in turn, writing the result as a linear combination of the basis vectors for V. The coefficients of that linear combination will be the columns of the matrix representation.

Last edited by a moderator: Apr 16, 2013