Math 1600 Lecture 33, Section 002, 29 Nov 2024

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Announcements:

Today we continue Section 4.4. Read the part Section 4.6 on linear recurrence relations for next class. (We are not covering the rest of 4.6.)

Work through suggested exercises.

Homework 10 is on WeBWorK and is due on today. (Homework 11 will be on Gradescope next week.)

Math Help Centre: M-F 12:30-5:30 in PAB48/49 and online 6pm-8pm.

My next office hour is today 2:30-3:20 in MC130.

Questions to discuss with your neighbour: What does it mean for $A$ and $B$ to be similar? What properties do similar matrices have in common? What does it mean for $A$ to be diagonalizable? How do we tell if it is, and how do we diagonalize $A$?

Review of Lecture 32: Section 4.4:

Definition: Let $A$ and $B$ be $n \times n$ matrices. We say that $A$ is similar to $B$ ($A \sim B$) if there is an invertible matrix $P$ such that $P^{-1} A P = B$.

Theorem 4.22: Let $A$ and $B$ be similar matrices. Then $A$ and $B$ have the same determinant, rank, characteristic polynomial and eigenvalues.

Definition: $A$ is diagonalizable if it is similar to some diagonal matrix.

If $A$ is similar to a diagonal matrix $D$, then $D$ must have the eigenvalues of $A$ on the diagonal. But how to find $P$?

Theorem 4.23: Let $A$ be an $n \times n$ matrix. If $P$ is an $n \times n$ matrix whose columns are linearly independent eigenvectors of $A$, then $P^{-1} A P$ is a diagonal matrix $D$ with the corresponding eigenvalues of $A$ on the diagonal.

On the other hand, if $P$ is any invertible matrix such that $P^{-1} A P$ is diagonal, then the columns of $P$ are linearly independent eigenvectors of $A$.

It follows that $A$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors.

This theorem is one of the main reasons we want to be able to find eigenvectors of a matrix. Moreover, the more eigenvectors the better, so this motivates allowing complex eigenvectors.

Theorem 4.24: If $\lambda_1, \ldots, \lambda_k$ are distinct eigenvalues of $A$ and, for each $i$, $\cB_i$ is a basis for the eigenspace $E_{\lambda_i}$, then the union of the $\cB_i$'s is a linearly independent set.

Combining Theorems 4.23 and 4.24 gives the following important consequence:

Theorem: An $n \times n$ matrix is diagonalizable if and only if the sum of the geometric multiplicities of the eigenvalues is $n$.

In particular:

Theorem 4.25: If $A$ is an $n \times n$ matrix with $n$ distinct eigenvalues, then $A$ is diagonalizable.

So it is important to understand the geometric multiplicities better. Here is a helpful result:

Lemma 4.26: If $\lambda_1$ is an eigenvalue of an $n \times n$ matrix $A$, then $$ \kern-6ex \text{geometric multiplicity of } \lambda_1 \leqslant \text{algebraic multiplicity of } \lambda_1 $$

It follows that the only way for the geometric multiplicities to add to $n$ is if they are equal to the algebraic multiplicities and the algebraic multiplicities add to $n$:

Theorem 4.27 (The Diagonalization Theorem): Let $A$ be an $n \times n$ matrix with distinct eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_k$. Let their geometric multiplicities be $g_1, g_2, \ldots, g_k$ and their algebraic multiplicities be $a_1, a_2, \ldots, a_k$. Then the following are equivalent:
a. $A$ is diagonalizable.
b. $g_1 + \cdots + g_k = n$.
c. $g_i = a_i$ for each $i$ and $a_1 + \cdots + a_k = n$.

Note: This is stated incorrectly in the text. The red part must be added unless you are working over $\C$, in which case it is automatic that $a_1 + \cdots + a_k = n$. With the way I have stated it, it is correct over $\R$ or over $\C$.

New material:

Example: Is $A = \bmat{rr} 0 & -1 \\ 1 & 0 \emat$ diagonalizable?

Summary of diagonalization: Given an $n \times n$ matrix $A$, we would like to determine whether $A$ is diagonalizable, and if it is, find the invertible matrix $P$ and the diagonal matrix $D$ such that $P^{-1} A P = D.$ The result may depend upon whether you are working over $\R$ or $\C$.

Steps:

1. Compute the characteristic polynomial $\det(A - \lambda I)$ of $A$.
2. Find the roots of the characteristic polynomial and their algebraic multiplicities by factoring.
3. If the algebraic multiplicities don't add up to $n$, then $A$ is not diagonalizable, and you can stop. (If you are working over $\C$, this can't happen.)
4. For each eigenvalue $\lambda$, compute the dimension of the eigenspace $E_\lambda$. This is the geometric multiplicity of $\lambda$, and if it is less than the algebraic multiplicity, then $A$ is not diagonalizable, and you can stop.
5. Compute a basis for the eigenspace $E_\lambda$.
6. If for each eigenvalue the geometric multiplicity equals the algebraic multiplicity, then you take the $n$ eigenvectors you found and put them in the columns of a matrix $P$. Put the eigenvalues in the same order on the diagonal of a matrix $D$.
7. Check that $AP=PD$.

Note that step 4 only requires you to find the row echelon form of $A - \lambda I$, as the number of free variables here is the geometric multiplicity. In step 5, you solve the system.

Exercise 4.4 12: Find eigenvalues of $\bmat{rrr} 1 & 0 & 0 \\ 2 & 2 & 1 \\ 3 & 0 & 1 \emat$ and their geometric multiplicities. Is it diagonalizable? If so, find $P$ and $D$. On board.

Exercise similar to 4.4 41: Find eigenvalues of $\bmat{rrr} -3 & -2 & 0 \\ 6 & 5 & 0 \\ 4 & 4 & -1 \emat$ and their geometric multiplicities. Is it diagonalizable? If so, find $P$ and $D$. On board.

Powers:

Suppose $P^{-1} A P = D$, where $D$ is diagonal. Then $A = P D P^{-1}$. We can use this to compute powers of $A$. For example, $$ \kern-8ex \begin{aligned} A^5 &= (P D P^{-1}) (P D P^{-1}) (P D P^{-1}) (P D P^{-1}) (P D P^{-1}) \\ &= P D^5 P^{-1} \end{aligned} $$ and $D^5$ is easy to compute since $D$ is diagonal: you just raise the diagonal entries to the fifth power.

More generally, $A^k = P D^k P^{-1}$. This is an efficient way to compute powers! Note that we need to know $P$, not just $D$, to do this. Also note that even if $A$ is real, it would work to diagonalize $A$ over $\C$. The answer would be real, but the intermediate calculations would be complex.

Also note that if $A$ is invertible, then $D$ will be as well (why?), and $$ A^{-1} = (P D P^{-1})^{-1} = (P^{-1})^{-1} D^{-1} P^{-1} = P D^{-1} P^{-1}. $$ More generally, $A^k = P D^k P^{-1}$, for any integer $k$.