Today we finish 4.3. **Read** Section 4.4 for Monday
and also read **Appendix D** on polynomials (**self-study**).
Work through recommended homework questions.

**Tutorials:** Quiz 5 is next week.

**Office hour:** Monday, 1:30-2:30, MC103B.

**Help Centers:** Monday-Friday 2:30-6:30 in MC 106.

**Midterm 2 Solutions** are available from the course home page.
The average was 27/40 = 68%.

**Question:**
If $P$ is invertible, how do $\det A$ and $\det(P^{-1}AP)$ compare?

They are equal:
$\det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \frac{1}{\det (P)} \det(A) \det(P) = \det A$.

**Definition:** Let $A$ be an $n \times n$ matrix.
A scalar $\lambda$ (lambda) is called an **eigenvalue** of $A$ if
there is a nonzero vector $\vx$ such that $A \vx = \lambda \vx$,
i.e. $(A - \lambda I) \vx = \vec 0$.
Such a vector $\vx$ is called an **eigenvector** of $A$ corresponding to $\lambda$.

**Definition:** The collection of **all** solutions to
$(A - \lambda I) \vx = \vec 0$ is a subspace called the
**eigenspace** of $\lambda$ and is denoted $E_\lambda$. In other words,
$$ E_\lambda = \null(A - \lambda I) . $$
It consists of the eigenvectors plus the zero vector.

**Definition:**
If $A$ is $n \times n$, $\det (A - \lambda I)$ will be a degree $n$
polynomial in $\lambda$.
It is called the **characteristic polynomial** of $A$, and
$\det (A - \lambda I) = 0$ is called the **characteristic equation**.

By the fundamental theorem of invertible matrices, the solutions to the characteristic equation are exactly the eigenvalues.

1. Compute the characteristic polynomial $\det(A - \lambda I)$.

2. Find the eigenvalues of $A$ by solving the characteristic equation
$\det(A - \lambda I) = 0$.

3. For each eigenvalue $\lambda$, find a basis for $E_\lambda = \null (A - \lambda I)$
by solving the system $(A - \lambda I) \vx = \vec 0$.

**Theorem D.4 (The Fundamental Theorem of Algebra):**
A polynomial of degree $n$ has at most $n$ distinct roots.

Therefore:

**Theorem:**
An $n \times n$ matrix $A$ has at most $n$ distinct eigenvalues.

Also:

**Theorem D.2 (The Factor Theorem):** Let $f$ be a polynomial and
let $a$ be a constant. Then $a$ is a zero of $f(x)$ (i.e. $f(a) = 0$)
if and only if $x - a$ is a factor of $f(x)$ (i.e. $f(x) = (x - a) g(x)$
for some polynomial $g$).

The largest $k$ such that $(x-a)^k$
is a factor of $f$ is called the **multiplicity** of the root $a$ in $f$.

In the case of an eigenvalue, we call its multiplicity in the characteristic
polynomial the **algebraic multiplicity** of this eigenvalue.

We also define the **geometric multiplicity** of an eigenvalue $\lambda$
to be the dimension of the corresponding eigenspace $E_\lambda$.

**Theorem 4.15:** The eigenvalues of a triangular matrix
are the entries on its main diagonal (repeated according to
their algebraic multiplicity).

**Example:** If $A = \bmat{rrr} 1 & 0 & 0 \\ 2 & 3 & 0 \\ 4 & 5 & 1 \emat$,
then
$$
\kern-6ex
\det(A - \lambda I) = \bdmat{ccc} 1-\lambda & 0 & 0 \\ 2 & 3-\lambda & 0 \\ 4 & 5 & 1-\lambda \edmat
= (1 - \lambda)^2 (3 - \lambda) ,
$$
so the eigenvalues are $\lambda = 1$ (with algebraic multiplicity 2)
and $\lambda = 3$ (with algebraic multiplicity 1).

**Question:**
What are the eigenvalues of a diagonal matrix?

The eigenvalues are the diagonal entries.

**Question:**
What are the eigenvalues of $\bmat{cc} 0 & 4 \\ 1 & 0 \emat$?

The characteristic polynomial is
$$
\bdmat{rr} -\lambda & 4 \\ 1 & -\lambda \edmat = \lambda^2 - 4 = (\lambda-2)(\lambda+2),
$$
so the eigenvalues are 2 and -2. Trick question.

**Question:**
How can we tell whether a matrix $A$ is invertible using eigenvalues?

$A$ is invertible if and only if 0 is not an eigenvalue,
because 0 being an eigenvalue is equivalent to $\null(A)$ being
non-trivial, which is equivalent to $A$ not being invertible,
by the fundamental theorem.

So we can extend the fundamental theorem with two new entries:

**Theorem 4.17:**
Let $A$ be an $n \times n$ matrix. The following are equivalent:

a. $A$ is invertible.

b. $A \vx = \vb$ has a unique solution for every $\vb \in \R^n$.

c. $A \vx = \vec 0$ has only the trivial (zero) solution.

d. The reduced row echelon form of $A$ is $I_n$.

f. $\rank(A) = n$

g. $\nullity(A) = 0$

h. The columns of $A$ are linearly independent.

i. The columns of $A$ span $\R^n$.

j. The columns of $A$ are a basis for $\R^n$.

k. The rows of $A$ are linearly independent.

l. The rows of $A$ span $\R^n$.

m. The rows of $A$ are a basis for $\R^n$.

n. $\det A \neq 0$

o. $0$ is not an eigenvalue of $A$

**Theorem 4.18:**
If $\vx$ is an eigenvector of $A$ with eigenvalue $\lambda$,
then $\vx$ is an eigenvector of $A^k$ with eigenvalue $\lambda^k$.
This holds for each integer $k \geq 0$, and also for $k < 0$ if
$A$ is invertible.

In contrast to some other recent results, this one is very useful computationally:

**Example 4.21:** Compute $\bmat{rr} 0 & 1 \\ 2 & 1 \emat^{10} \coll 5 1$.

**Solution:**
By finding the eigenspaces of the matrix, we can show that
$$
\bmat{rr} 0 & 1 \\ 2 & 1 \emat \coll 1 {-1} = - \coll 1 {-1}
\qtext{and}
\bmat{rr} 0 & 1 \\ 2 & 1 \emat \coll 1 2 = 2 \coll 1 2
$$
Write $A = \bmat{rr} 0 & 1 \\ 2 & 1 \emat$, $\vx = \coll 5 1$,
$\vv_1 = \coll 1 {-1}$ and $\vv_2 = \coll 1 2$.
Since $\vx = 3 \vv_1 + 2 \vv_2$ we have
$$
\begin{aligned}
A^{10} \vx &= A^{10} (3 \vv_1 + 2 \vv_2)
= 3 A^{10} \vv_1 + 2 A^{10} \vv_2 \\
&= 3 (-1)^{10} \vv_1 + 2(2^{10}) \vv_2
= \coll {3+2^{11}}{-3+2^{12}}
\end{aligned}
$$
**Much faster** than repeated matrix multiplication, especially if $10$ is replaced with $100$.

This raises an interesting question. In the example, the eigenvectors were a basis for $\R^2$, so we could use this method to compute $A^k \vx$ for any $\vx$. However, last class we saw a $3 \times 3$ matrix with two one-dimensional eigenspaces, so the eigenvectors didn't span $\R^3$. We will study this further in Section 4.4, but right now we can answer a related question about linear independence.

**Theorem:** If $\vv_1, \vv_2, \ldots, \vv_m$ are eigenvectors of $A$
corresponding to distinct eigenvalues
$\lambda_1, \lambda_2, \ldots, \lambda_m$, then
$\vv_1, \vv_2, \ldots, \vv_m$ are linearly independent.

**Proof in case $m = 2$:**
If $\vv_1$ and $\vv_2$ are linearly dependent, then
$\vv_1 = c \vv_2$ for some $c$.
Therefore
$$
A \vv_1 = A \, c \vv_2 = c A \vv_2
$$
so
$$
\lambda_1 \vv_1 = c \lambda_2 \vv_2 = \lambda_2 \vv_1
$$
Since $\vv_1 \neq \vec 0$, this forces $\lambda_1 = \lambda_2$, a contradiction.$\quad\Box$

The general case is very similar; see text.

Next: how to become a Billionaire using the material from this course.