Notes on diagonalizable linear operators

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Notes on diagonalizable linear operators

May 11, 2011 ivanky Leave a comment Go to comments

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 $A$ 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.
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.

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Categories: Justincase, Teachings Tags: linear algebra, mathematics, teaching

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Jitse Niesen

May 12, 2011 at 9:07 am | #1

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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 $d_k$.
ivanky

May 12, 2011 at 11:23 am | #2

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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.
toasty redhead

May 14, 2011 at 5:57 am | #3

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I didn’t know that.