# Linear transformation - matrices

1. Nov 7, 2005

### Benny

Linear transformation - matrices *edit* Question resolved (y)

I'm not sure what to do in the following question.
A linear transformation has matrix
$$P = \left[ T \right]_B = \left[ {\begin{array}{*{20}c} 3 & { - 4} \\ 1 & { - 1} \\ \end{array}} \right]$$
with respect to the standard basis B = {(0,1),(0,1)} for R^2.
a) Give the matrix $$Q = \left[ T \right]_{B'}$$ with respect to the basis B' = {(3,1),(2,1)}.
b) Determine whether or not the transformation is diagonalizable.
c) Find a general expression for Q^n. Hence deduce the formula for $$P^n = \left[ {\begin{array}{*{20}c} 3 & { - 4} \\ 1 & { - 1} \\ \end{array}} \right]^n$$.
Here is what I've thought about.
a) The format of this question is quite new to me, I haven't seen any questions set out in this way before so I'm just going to have a guess...
$$T\left( {\left[ {\begin{array}{*{20}c} x \\ y \\ \end{array}} \right]} \right) = \left[ T \right]_b \left[ {\begin{array}{*{20}c} x \\ y \\ \end{array}} \right] = \left( {3x - 4y,x - y} \right)$$
Oh wait...is this just an application of transitition matrices? As in, I have [T]_B and I want [T]_B'. I also have the two basis sets, so I can use transition matricies to find [T]_B'?
b) I don't understand this question. I've only done questions on determining whether or not a matrix is diagonalizable. I'm not sure how this relates to the transformation itself. That is, how a matrix of a transformation relates to whether or not the transformation is diagonalizable
c) I don't get the point of this question. How does having Q^n allow me to deduce the formula for P^n? Perhaps if someone tells me what I need to do in part 'a' then it should make this part more clear.
Any help with either of the 3 parts of the question would be good thanks.

Last edited by a moderator: Nov 7, 2005
2. Nov 7, 2005

### HallsofIvy

Staff Emeritus
Benny, I accidently hit "edit" when I meant to hit "quote". It think I put it back right! Sorry about that.
I think you are heading the right way. Remember that any basis vectors, written in that basis, look like (1, 0), (0, 1). Multiplying a matrix by (1,0) just gives the first column, multiplying a matrix by (0,1) just gives the second column.
What is P(3,1)? How would that be written as a linear combination of (3,1) and (2,1)? Those coefficients form the first column of Q. What is P(2,1)? How would that be written as a linear combination of (3,1) and (2,1)? Those coefficients form the second columnm of Q.
Yes, you can use the transition matrix. In fact, you will need it for (c).
Yes, a transformation is diagonalizable if and only if its matrix representation is diagonalizable. In fact the word "diagonalizable" really applies to the transformation, not the matrix. When you "diagonalize" a matrix you are finding another matrix that would represent the same transformation in another basis.
Find the eigenvalues and eigenvectors of Q (in fact, find them of P also!)
What does that tell you? (note: for a diagonal matrix, obviously (1,0) and (0,1) are eigenvectors!)
Because Q is a particularly simple matrix, it is easy to calculate a few powers and conjecture what Qn looks like (and, of course, use induction on n to prove it). Since Q and P represent the same transformation in different bases, there exist, as you said above, a transition matrix, T, such that P= TQT-1. Then P2= (TQT-1)(TQT-1)= (TQ)(T-1T)(QT-1)= TQ2T-1. You try P3! Do you see the point?

3. Nov 7, 2005

### Benny

Thanks for the help HallsofIvy. In most of the questions I've seen, the matrices equivalent to Q has been a diagonal matrix(this time I found it be diagonal but with another non-zero off diagonal entry). The formula for Q^n was rather easy to deduce but the nature of the question was suprising. I checked my final answer for P^n and it's correct.

I got P^n by using the transition matrices. In most of the questions I've done, when Q was a diagonal matrix, one of the transition matrices had columns consisting of eigenvectors of what would be equivalent to the matrix Q. In this case, the formula for P^n (AQ^nA^-1) works simply because A and A^-1 are inverses. So I'm thinking that A only has eigenvectors as its columns when Q is diagonal.

4. Nov 8, 2005

### HallsofIvy

Staff Emeritus
A linear transformation is "diagonalizable" if and only if there exist a basis for the vector space made entirely of eigenvectors of the transformation. Written in that basis, the transformation is represented by a diagonal matrix with eigenvalues on the diagonal.
If a linear transformation does not have a "full set of eigenvectors", then it is not diagonalizable but can be put in "Jordan normal form". That is a matrix with 0s below the diagonal, eigenvalues on the diagonal and, where there are eigenvalues of multiplicity greater than 1, "blocks" with 1 just above the diagonal.
Example:
$$\left[ {\begin{array}{*{20}c}2 & 1 & 0 & 0 & 0 \\0 & 2 & 0 & 0 & 0 \\0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 3\end{array}} \right]$$
is a matrix with double eigenvalue 2, triple eigenvalue 3, in Jordan normal form.

5. Nov 8, 2005

### Benny

Ok, I haven't thought of a transformation that way before. Jordan normal form will be one of the things I study if I end up continuing with linear algebra in later years.