Home > Justincase, Teachings > Notes on diagonalizable linear operators

Notes on diagonalizable linear operators

Okay, I have a few things in my mind about this topic and I don’t want to lose them tomorrow or the day after. Therefore, I created this note. Hope it is useful to other people as well. Background of this note are as follows:

  1. Suppose A is an n \times n matrix and it is diagonalizable. Therefore, by definition, there is an invertible matrix P and a diagonal matrix D. However, I always forget the relationship between A and D, whether it is D=PAP^{-1} or D=P^{-1}AP. Generally, those two are no different, but we can choose P such that the column vectors of matrix P are the eigenvectors of matrix A, hence we have to know precisely which one it is, otherwise it gets wrong. Until now, when I forget about this, I always derive it and it takes some of my precious time (haha) but now after I have this, I can just open the Internet and look for this note.
  2. Point 1 above talks about diagonalizability of a matrix. But generally, we have a linear operator acting on a general vector field instead. A linear operator is called diagonalizable if there is a basis B such that the matrix representation of this linear operator is a diagonal matrix. It seems for me, diagonalizability of a linear operator and that of a matrix are talking about two different things. One is talking about how to find a basis, while the other is talking about how to find an invertible matrix. However, these two concepts turn out to be the same.

We begin with the following theorem.

Theorem 1 (Coordinate-change matrix)
Let E be a standard basis and B be another basis for an n-dimensional linear space V. Then there is an invertible matrix P such that for every v \in V

P (v)_B=(v)_E.                                                                                                          (1)

Matrix P is called the transition or coordinate-change matrix from the basis B to the basis E. Note that (v)_B and (v)_E are the coordinates of vector v with respect to basis B and E respectively. In fact, the column vectors of matrix P are the coordinates of the basis vectors of B with respect to the standard bases E (i.e. if B=\{ v_1,v_2,\dots,v_n \} then P=[(v_1)_E (v_2)_E \dots (v_n)_E] ).

We won’t prove the theorem as it can be found in any linear algebra textbook. However, using this theorem we can relate the diagonalizability of a linear operator and that of a matrix.Suppose T is a diagonalizable linear operator on an n-dimensional linear space V. Take x \in V then let y be y=T(x). Suppose E is a standard basis in V and B is a basis in V that consists of independent eigenvectors of T.

Then we have the following.

(i) (y)_E = [T(x)]_E = [T]_E (x)_E = A (x)_E, where A is a representation matrix of T with respect to standard basis E.
(ii) (y)_B = [T(x)]_B = [T]_B (x)_B = D (x)_B, where D is a representation matrix of T with respect to basis B. We also know that D is a diagonal matrix.

We want to show that using the theorem 1, D=PAP^{-1} or D=P^{-1}AP. From equation (1) we get:

P (y)_B = (y)_E = A (x)_E = A P (x)_B.

Hence, we have (y)_B = P^{-1}AP (x)_B. While from point (ii) we have (y)_B = D (x)_B. As a conclusion, we have,

D = P^{-1} A P.

We have derived that in fact, D = P^{-1} A P is the right one. We also derived that the invertible matrix P has such a close relationship with the basis B as the column vectors of P are the basis vectors of B, which are the eigenvectors of the linear operator T. This conclude my note.

  1. Jitse Niesen
    May 12, 2011 at 9:07 am

    Hi Ivanky, good to see you’re still doing maths. 🙂

    I have the same problem that I cannot remember whether it’s $A = P^{-1} D P$ or $A = P D P^{-1}$. This is how I derive that it’s the latter: $A = P D P^{-1}$ implies $A P = P D$. Take the k-th column: $(AP)_k = (PD)_k$ or $AP_k = PD_k$ where $X_k$ denotes the k-th column of a matrix $X$. Now, the k-th column of the diagonal matrix $D$ is $d_k e_k$ where $d_k$ is the k-th value on the diagonal and $e_k$ is the k-th vector in the standard basis. So we get $AP_k = d_k P e_k = d_k P_k$. This says that the k-th column of $P$ is an eigenvector of $A$ with eigenvalue $d_k$.

    If instead you start with $A = Q^{-1} D Q$ (I use $Q$ here because it’s another matrix; in fact $Q = P^{-1}$), then you get $Q A = D Q$ and taking the k-th column will not get you anywhere. Instead, you can take the k-th row. Using Matlab notation $X(k,:)$ to denote the k-th row of a matrix $X$, we find $Q(k,:) A = D(k,:) Q = d_k e_k^T Q = d_k Q(k,:)$. Thus, the k-th row of $Q = P^{-1}$ is a left eigenvector of $A$ with eigenvalue $d_k$.

  2. ivanky
    May 12, 2011 at 11:23 am

    Hi Jitse, thanks for your comment. It is very useful, I never thought this way. I need diagonalization for my normal form computation. I am going to Italy this June but I don’t know if I can go to England to meet you guys.

  3. May 14, 2011 at 5:57 am

    I didn’t know that.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: