The characteristic polynomial
Let’s start with $T: V \to V$ a linear transformation between finite dimensional vector spaces.
Lemma. For a linear transformation $T : V \to V$ with $\dim V < \infty$, we have $\mathcal Z(T) = 0$ if and only if $T$ is an isomorphism.
Proof. (Expand to view)
To check that $T$ is an isomorphism it suffices to check $T$ is a bijection and it suffices to show that $\mathcal Z(T) = 0$ and $\mathcal R(T) = V$ to check it is bijection.
So clearly if $T$ is an isomorphism, we must have $\mathcal Z(T) = 0$.
In the other direction, we use the Rank Nullity Theorem \(\dim \mathcal Z(T) + \dim \mathcal R(T) = \dim V\)
Assume that $\mathcal Z(T) = 0$. Then, \(\dim \mathcal R(T) = \dim V\) so $\mathcal R(T) = V$ ■
We have seen that $\lambda$ is eigenvalue for $T$ if and only if \(\mathcal Z(T - \lambda \operatorname{Id}) \neq 0\)
Let’s pick a basis and get a matrix representation $A$ of $T$. With the matrix, we have the power of the determinant.
Proposition. $\lambda$ is an eigenvalue of $A$ if and only \(\det (A - \lambda I_n) = 0\)
Proof. (Expand to view)
From the discussion above, we know that $\lambda$ is an eigenvalue if and only $\mathcal Z(A-\lambda I_n) \neq 0$.
We know that $\mathcal Z(A -\lambda I_n) \neq 0$ if and only if $A - \lambda I_n$ invertible from the previous lemma. Finally, we saw that $A - \lambda I_n$ is invertible if and only \(\det(A - \lambda I_n) \neq 0.\) ■
This proposition tells us we really want to understand the zeroes of a function of a single variable.
Definition. The characteristic polynomial $\chi_A(x)$ of a $n \times n$ matrix $A$ is the function \(x \mapsto \det(A - xI_n)\)
We should probably check $\chi_A(x)$ is actually a polynomial in $x$. Before giving a general proof, let’s look at some low dimensions.
For $A = (a)$, we just have \(\chi_A(x) = x-a\)
For \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) we have \(A - xI_2 = \begin{pmatrix} a-x & b \\ c & d-x \end{pmatrix}\) and so \(\chi_A(x) = (a-x)(d-x) - bc = x^2 - (a+d)x + (ad-bc)\)
For \(A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\) we have \(A - xI = \begin{pmatrix} a-x & b & c \\ d & e-x & f \\ g & h & i-x \end{pmatrix}\) We compute $\chi_A(x)$ using a cofactor expansion along the top row: \(\begin{aligned} \chi_A(x) & = (a-x) \det\begin{pmatrix} e-x & f \\ h & i-x\end{pmatrix} - b \det \begin{pmatrix} d & f \\ g & i-x\end{pmatrix} + c \det \begin{pmatrix} d & e-x \\ g & h \end{pmatrix} \\ & = (a-x)(e-x)(i-x) - (a-x)fh - bd(i-x)+bfg + cdh - cg(e-x) \\ & = -x^3 + (a+e+i)x^2 - (ae+ai+ie-fh-bd-cg)x \\ & + aei - afh - bdi + bfg + cdh - ceg \end{aligned}\)
Proposition. For any $n \times n$ matrix $A$, $\chi_A(x)$ is a polynomial of degree $n$ with constant coefficient equal to $\det A$ and whose $x^n$ coefficient is $(-1)^n$.
Proof. (Expand to view)
We first prove the more general claim.
Claim. Suppose that we have a $n \times n$ matrix $B$ satisfying the condition: for some $1 \leq j \leq n$ we have exactly $j$ columns where one entry is of the form $c-x$ with $c \in k$ and the others being elements of the field $k$. Then, $\det B$ is a polynomial of degree at most $j$.
We prove this claim using induction on $n$. The case of $n=1$ is clear.
Assume that the statement is true for $n \times n$ matrices and let $B$ be a $(n+1) \times (n+1)$ satisifying the condition. If no column of $B$ has a $c-x$ then we are done since we have a constant. Otherwise, pick a column $\mathbf{C}_l(A)$ with $c-x$ as an entry and using the cofactor expansion. \(\det B = \sum_{i=1}^{n+1} (-1)^{i-1} B_{iL} \det M_{iL}\) Each $M_{i1}$ is $n \times n$ matrix which has at most $j-1$ columns containing a $c-x$. Applying the induction hypothesis, we know that the degree of $\det M_{il}$ is at most $j-1$. Thus, $ \operatorname{deg} (B_{il} \det M_{il}) \leq j
$$ as is the sum giving $\det B$.
Let’s use this claim to prove the proposition. We expand $\chi_A(x)$ using the first row \(\chi_A(x) = \sum_{j=1}^n (A-xI)_{1j} \det M_{1j}(A-xI)\) If $j \neq 1$, then $M_{1j}$ still has $n-2$ columns with a $c-x$ as we have removed the $(1,1)$ along with a $j$-column. Also $(A-xI)_{1j} = A_{1j}$. So $(A-xI)_{1j} \det M_{1j}$ has degree at most $n-2$ by the claim.
The term $(A-xI)_{11} \det M_{11}$ has degree at most $n$ by the claim also. Furthermore, if the degree $n$ term is nonzero it must come from this term since the others are degree $n-2$ at most. Using induction, we see that the degree $n$ term comes from \((A-xI)_{11}(A-xI)_{22} \cdots (A-xI)_{nn}\) which expands to \((-1)^nx^n + \cdots\)
Finally, for the constant term of a polynomial we just need to evaluate it at $x=0$. We have \(\chi_A(0) = \det (A-0I) = \det A.\) ■
A scalar $a \in k$ is a root of a polynomial $p(x)$ if \(p(a) = 0.\)
Thus, to find the eigenvalues of $A$, we need to find the roots of $\chi_A(x)$.
Example. Let’s compute the eigenspaces of \(A = \begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix}\) We do this by first finding the eigenvalues using the characteristic polynomial.
We have \(A - xI = \begin{pmatrix} -x & -1 \\ 1 & 2-x \end{pmatrix}\) so \(\det(A -xI) = -x(2-x)-1 = -x^2 - 2x -1 = -(x+1)^2\) We see that the only eigenvalue is $-1$.
Now, we want to find the null space of $A - \lambda I$ for each eigenvalue. The null space of \(A + I = \begin{pmatrix} -1 & -1 \\ 1 & 1 \end{pmatrix}\) is \(E_{-1}(A) = \left\lbrace \begin{pmatrix} a \\ - a \end{pmatrix} \mid a \in k \right\rbrace\) This is one dimensional so we do not have a basis of eigenvectors for $A$.