5.2 The characteristic equation

Big idea

Last section left one job undone: producing the eigenvalues. A scalar $\lambda$ is an eigenvalue of $A$ when $(A - \lambda I)\mathbf{x} = \mathbf{0}$ has a nonzero solution, which means $A - \lambda I$ is not invertible, which means its determinant is zero. That single chain turns a search over all scalars into one equation in one unknown.

Expanding $\det(A - \lambda I)$ gives a polynomial in $\lambda$ of degree $n$. Its roots are the eigenvalues, all of them, with nothing else mixed in. So eigenvalue hunting becomes polynomial root finding, and the algebra you own from precalculus does the rest.

The polynomial carries more than a list of roots. A root can repeat, and how many times it repeats is a number worth tracking, because it will not always match the dimension of the corresponding eigenspace. That mismatch is the whole obstacle to diagonalization in the next section.

Decoder

The characteristic polynomial of $A$ is $\det(A - \lambda I)$, and its roots, with multiplicity, are the spectrum of $A$.

Unpacked: subtract $\lambda$ from each diagonal entry, take the determinant of what is left, and treat the result as a polynomial in $\lambda$. Setting it to zero gives the characteristic equation. “Spectrum” is just the collection of eigenvalues. “With multiplicity” means a root counted as many times as it appears as a factor, so a double root is listed twice.

Definitions and results

The characteristic equation. $\lambda$ is an eigenvalue of the $n \times n$ matrix $A$ exactly when

$$ \det(A - \lambda I) = 0 $$

Each direction of that statement is one of the standing facts about invertibility: a square matrix has a nonzero null space exactly when it is singular, and it is singular exactly when its determinant is zero.

The characteristic polynomial. The left side, expanded, is a polynomial in $\lambda$ of degree exactly $n$, with leading term $(-1)^n\lambda^n$. An $n \times n$ matrix therefore has at most $n$ distinct eigenvalues, and exactly $n$ counted with multiplicity if you allow complex roots.

The $2 \times 2$ shortcut. For

$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$

the characteristic polynomial is $\lambda^2 - (a+d)\lambda + (ad - bc)$. The coefficient of $\lambda$ is minus the trace, and the constant term is the determinant. Memorize this one: it turns most $2 \times 2$ eigenvalue problems into a quadratic you can factor in your head.

Trace and determinant in general. For any size, the sum of the eigenvalues (with multiplicity) is the trace of $A$, and their product is $\det A$. These are free checks on an answer, and they cost one addition and one multiplication.

Algebraic multiplicity. Factor the characteristic polynomial. The number of times $(\lambda - \lambda_0)$ appears is the algebraic multiplicity of the eigenvalue $\lambda_0$. Multiplicities across all eigenvalues add up to $n$.

Geometric multiplicity. The dimension of the eigenspace for $\lambda_0$, that is, the number of free variables in $(A - \lambda_0 I)\mathbf{x} = \mathbf{0}$.

How the two compare. The geometric multiplicity is at least $1$ and never exceeds the algebraic multiplicity. It can be strictly smaller. When it is, the matrix has fewer eigenvector directions than its size, and the next section will call it not diagonalizable. When every eigenvalue has matching multiplicities, the eigenvectors fill the space.

Triangular case again. For a triangular matrix, $A - \lambda I$ is triangular, so its determinant is the product of the shifted diagonal entries. The characteristic polynomial is already factored, and the eigenvalues are the diagonal entries with their repetitions as algebraic multiplicities.

Similar matrices. $A$ and $B$ are similar when $B = P^{-1}AP$ for some invertible $P$. Similar matrices have the same characteristic polynomial, hence the same eigenvalues with the same algebraic multiplicities. The eigenvectors differ: if $A\mathbf{v} = \lambda\mathbf{v}$ then $B(P^{-1}\mathbf{v}) = \lambda(P^{-1}\mathbf{v})$, so the eigenvectors are carried across by $P^{-1}$.

Similarity is not row equivalence. Row operations change eigenvalues. Swapping two rows of the identity gives a matrix with eigenvalue $-1$, which the identity does not have. Never row reduce a matrix and then take the eigenvalues of the result.

Why nobody computes this way at scale. Expanding a determinant symbolically costs too much for large $n$, and polynomial roots are sensitive to tiny changes in coefficients. The characteristic equation is the definition and the hand-computation tool. Production software iterates instead, which is the subject of a later section.

Worked examples

A two by two from the shortcut

Let

$$ A = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix} $$

The trace is $3$ and the determinant is $2 - 30 = -28$, so the characteristic polynomial is

$$ \lambda^2 - 3\lambda - 28 = (\lambda - 7)(\lambda + 4) $$

The eigenvalues are $7$ and $-4$, each with algebraic multiplicity $1$. Check against trace and determinant: $7 + (-4) = 3$ and $7 \cdot (-4) = -28$. Both match.

Confirm one of them directly. $A - 7I$ has rows $(-6, 6)$ and $(5, -5)$, whose determinant is $30 - 30 = 0$, so $7$ really is a root. The eigenvector follows from $-6x_1 + 6x_2 = 0$, giving $(1,1)$, and $A(1,1) = (7,7)$.

A three by three with a repeated root

Let

$$ B = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} $$

Expand $\det(B - \lambda I)$ along the first row:

$$ (1-\lambda)\left[(1-\lambda)^2 - 4\right] - 2\left[2(1-\lambda) - 4\right] + 2\left[4 - 2(1-\lambda)\right] $$

The first bracket is $\lambda^2 - 2\lambda - 3$, the second is $-2 - 2\lambda$, the third is $2 + 2\lambda$. Collecting:

$$ (1-\lambda)(\lambda^2 - 2\lambda - 3) + (4 + 4\lambda) + (4 + 4\lambda) = -\lambda^3 + 3\lambda^2 + 9\lambda + 5 $$

Factor out $-1$ and factor the cubic:

$$ \det(B - \lambda I) = -(\lambda - 5)(\lambda + 1)^2 $$

So $5$ has algebraic multiplicity $1$ and $-1$ has algebraic multiplicity $2$. Check: the eigenvalues sum to $5 - 1 - 1 = 3$, which is the trace, and multiply to $5 \cdot (-1) \cdot (-1) = 5$, which is $\det B$ (expand it and you get $1(1-4) - 2(2-4) + 2(4-2) = -3 + 4 + 4 = 5$).

Now the geometric multiplicity of $-1$. The shifted matrix $B + I$ has every entry equal to $2$, so it has rank $1$ and its null space has dimension $2$. A basis is $(1,-1,0)$ and $(1,0,-1)$. Check the first: $B(1,-1,0) = (1 - 2, \, 2 - 1, \, 2 - 2) = (-1,1,0)$, which is $-1$ times the vector. Here the two multiplicities agree.

When the multiplicities disagree

Let

$$ C = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} $$

The trace is $6$ and the determinant is $9$, so the characteristic polynomial is $\lambda^2 - 6\lambda + 9 = (\lambda - 3)^2$. The only eigenvalue is $3$, with algebraic multiplicity $2$.

Its eigenspace comes from

$$ C - 3I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $$

One pivot, one free variable, so the eigenspace is the line spanned by $(1,0)$ and the geometric multiplicity is $1$. Check: $C(1,0) = (3,0) = 3(1,0)$, and any vector with a nonzero second entry, say $(0,1)$, maps to $(1,3)$, which is off its own line.

Two for the algebraic count, one for the geometric count. A $2 \times 2$ matrix with only one eigenvector direction cannot be rebuilt from a basis of eigenvectors, and the next section names that failure.

Practice

Build the characteristic polynomial and solve it. Start with matrices small enough to factor by hand.

The determinant is the engine here, so drill it on its own until $2 \times 2$ and $3 \times 3$ cases are automatic.

Then separate the two counts: how many times a root repeats, and how many free variables the shifted matrix leaves.

Videos

The first video is the geometric side: watch for the moment the determinant of the shifted matrix is set to zero. The second works the algebra through by hand at board speed, including what repeated roots do.

Quiz

Six items on characteristic polynomials, determinants of small matrices, and the two multiplicities.