19 Eigenvalues and Eigenvectors
As we come to the end of the course, we will cover some material on eigenvalues, eigenvectors, diagonalizability, and similarity. These topics are mainly from Chapter Five, Section II of the textbook. All of this works best over the complex numbers. That is, we consider complex vector spaces, complex eigenvalues, etc. However, much of it also applies to real eigenvalues and vector spaces over the real numbers. We can define eigenvalues and eigenvectors of homomorphisms:
Definition: Let $h\colon V\to V$ be a vector space homomorphism. An eigenvector for $h$ is a non-zero vector $\vec v\in V$ such that $h(\vec v) = \lambda\cdot\vec v$ for some scalar $\lambda$. The scalar $\lambda$ is then an eigenvalue of $h$, and $\vec v$ is said to be an eigenvector for $h$ with eigenvalue $\lambda$.
However, we will work almost exclusively with eigenvalues and eigenvalues of matrices. An eigenvalue or eigenvector of a square matrix is just an eigenvalue or eigenvector of the homomorphism given by multiplication by that matrix:
Definition: Let $A$ be an $n\times n$ matrix. An eigenvector for $A$ is a non-zero vector $\vec v\in V$ such that $A\vec v = \lambda\cdot\vec v$ for some scalar $\lambda$. The scalar $\lambda$ is then an eigenvalue of $A$, and $\vec v$ is said to be an eigenvector for $A$ with eigenvalue $\lambda$.
$\lambda$ is an eigenvalue of $A$ if and only if there is a non-zero vector $\vec v$ such that $A\vec v = \lambda\vec v = (\lambda I_n)\cdot\vec v$, which is true if and only if there is a non-zero vector $\vec v$ in the null space of the matrix $A-\lambda I_n$. That in turn can happen if and only if the matrix $A-\lambda I_n$ is singular, which is the case if and only if the determinant, $|A-\lambda I_n|$, is zero. So, to find the eigenvalues of $A$, we must find values of the variable $x$ that make $|A-x I_n| = 0$.
Now, the determinant $|A-x I_n|$ is a polynomial in $x.$ It is called the characteristic polynomial of $A.$ The eigenvalues of $A$ are precisely the roots of the characteristic polynomial of $A$.
If $\lambda$ is any eigenvalue of $A$, it must have an eigenvector. In fact, any non-zero multiple of an eigenvector for $\lambda$ is also an eigenvector for $\lambda$. And the sum of two eigenvectors for $\lambda$ is an eigenvector for $\lambda$, as long as it is not $\vec 0.$ This lets us define something called an "eigenspace."
Definition: Let $A$ be an $n\times n$ matrix, and let $\lambda$ be an eigenvalue of $A$. We define the eigenspace of $A$ for eigenvalue $\lambda$ to be $\{\vec v\in\C^n\,|\,A\vec v = \lambda v\}$. This eigenspace consists of all of the eigenvectors for $\lambda$ together with $\vec 0.$
An eigenspace is a subspace of $\C^n$.
Defintion: Two $n\times n$ matrices $A$ and $B$ are said to be similar if there is a non-singular $n\times n$ matrix $Q$ such that $A = QBQ^{-1}$.
It is easy to see that similar matrices have the same characteristic polynomial and therefore the same eigenvalues. They do not have the same eigenvectors, but if $A=QBQ^{-1}$, then $\vec v$ is an eigenvector for $B$ with eigenvalue $\lambda$ if and only if $Q\vec v$ is an eigenvector for $A$ with eigenvalue $\lambda$.
We can think of $Q$ as a change of basis matrix. From that point of view, $A$ and $B$ represent the same homomorphism in different bases. (Multiplication by $A$ represents the homomorphism $h(\vec v) = A\vec v$ in the standard basis; $B$ represents this same homomorphism in the basis whose elements are the column vectors of $A$.)
A nice special case occurs when $A$ has a basis of eigenvectors. That is, there is a basis $B= \langle \vec\beta_1, \vec\beta_2, \dots, \vec\beta_n \rangle$ of $\C^n$ where each basis vector in $B$ is an eigenvector for $A$. That is, for $i = 1,2,\dots,n$, $A\vec\beta_i = \lambda_i\vec\beta_i$ for some scalar $\lambda_i$. (Note that some of the $\lambda_i$ can be the same.) Therefore, in the basis $B$, the homomorphism $h(\vec v) = A\vec v$ is represented by the diagonal matrix $$\begin{pmatrix} \lambda_1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda_2 & 0 & \cdots & 0 \\ 0 & 0 & \lambda_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_n \end{pmatrix}$$ Note that the eigenvalues of a diagonal matrix are the diagonal entries of the matrix, and the eigenvectors are the standard basis vectors.
We say that an $n\times n$ matrix is diagonalizable if it is similar to a diagonal matrix. A matrix is diagonalizable if and only if $\C^n$ has a basis of eigenvectors for that matrix. Not every matrix is diagonalizable. For example, the matrix $\big(\begin{smallmatrix} 1 & 1\\ 0 & 1\end{smallmatrix}\big)$ is not diagonalizable: its only eigenvectors are multiples of $\big(\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\big).$
(If you want to find a class of fairly simply matrices such that every matrix is similar to one of the matrices in the class, you need to use matrices in "Jordan form." Jordan form matrices are diagonal matrices except that some 1's can occur immediately below the diagonal. Assuming that we are working with complex vector spaces, every matrix is similar to a Jordan form matrix. Jordan form matrices are covered in Chapter Five, Sections III and IV in the textbook, but they are not part of the material for this course.)