## 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:

- Suppose is an matrix and it is diagonalizable. Therefore, by definition, there is an invertible matrix and a diagonal matrix . However, I always forget the relationship between and , whether it is or . Generally, those two are no different, but we can choose such that the column vectors of matrix are the eigenvectors of matrix , 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.
- 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 be a standard basis and be another basis for an -dimensional linear space . Then there is an invertible matrix such that for every *

*. (1)*

*Matrix is called the transition or coordinate-change matrix from the basis to the basis . Note that and are the coordinates of vector with respect to basis and respectively. In fact, the column vectors of matrix are the coordinates of the basis vectors of with respect to the standard bases (i.e. if then ). *

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 is a diagonalizable linear operator on an -dimensional linear space . Take then let be . Suppose is a standard basis in and is a basis in that consists of independent eigenvectors of .

Then we have the following.

(i) , where is a representation matrix of with respect to standard basis .

(ii) , where is a representation matrix of with respect to basis . We also know that is a diagonal matrix.

We want to show that using the theorem 1, or . From equation (1) we get:

.

Hence, we have . While from point (ii) we have . As a conclusion, we have,

.

We have derived that in fact, is the right one. We also derived that the invertible matrix has such a close relationship with the basis as the column vectors of are the basis vectors of , which are the eigenvectors of the linear operator . This conclude my note.

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 $d_k$.

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.

I didn’t know that.