Characteristic Polynomial and Characteristic Equation

The characteristic polynomial

For an $n \times n$ matrix $A$, the characteristic polynomial is $p(\lambda) = \det(A - \lambda I)$

It is a polynomial of degree exactly $n$ in $\lambda$, and its roots are the eigenvalues of $A$. Setting $p(\lambda) = 0$ gives the characteristic equation.

Key facts about $p(\lambda)$

For any $n \times n$ matrix: $p(\lambda) = (-1)^n \lambda^n + (-1)^{n-1} \operatorname{tr}(A) \lambda^{n-1} + \cdots + \det(A)$

The leading coefficient alternates sign with $n$. The constant term (value at $\lambda = 0$) is $\det A$.

For $2 \times 2$: $p(\lambda) = \lambda^2 - \operatorname{tr}(A) \lambda + \det(A)$

Memorizing this short form saves time for $2 \times 2$ eigenvalue problems.

Step-by-step

$A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$ (an upper triangular matrix).

Step 1 — Compute $A - \lambda I$. $A - \lambda I = \begin{pmatrix} 3 - \lambda & 1 \\ 0 & 2 - \lambda \end{pmatrix}$

Step 2 — Compute the determinant. $p(\lambda) = \det(A - \lambda I) = (3 - \lambda)(2 - \lambda) - (1)(0) = (3 - \lambda)(2 - \lambda)$

Step 3 — Expand and verify. $p(\lambda) = \lambda^2 - 5\lambda + 6$

Quick sanity check via short form: $\operatorname{tr}(A) = 3 + 2 = 5$ and $\det(A) = 3 \cdot 2 - 0 = 6$. So $p(\lambda) = \lambda^2 - 5\lambda + 6$ ✓

Step 4 — Find the roots. $\lambda^2 - 5\lambda + 6 = 0 \implies (\lambda - 3)(\lambda - 2) = 0$ $\lambda_1 = 3, \quad \lambda_2 = 2$

These are the eigenvalues.

Triangular shortcut

For a triangular matrix (upper or lower), the eigenvalues are simply the diagonal entries. You can read them off without any calculation: $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} \implies \lambda = 3, 2$

This works because $\det(A - \lambda I)$ for a triangular matrix is the product of diagonal entries $(3 - \lambda)(2 - \lambda)$.

Algebraic vs. geometric multiplicity

If $(\lambda - \lambda_i)^k$ divides the characteristic polynomial, $\lambda_i$ has algebraic multiplicity $k$. Its geometric multiplicity is $\dim \operatorname{Null}(A - \lambda_i I)$.

• Geometric multiplicity $\le$ algebraic multiplicity always.
• If equal for every eigenvalue, $A$ is diagonalizable.
• If strictly less for some eigenvalue, $A$ is defective (not diagonalizable).

For our matrix, both eigenvalues have algebraic multiplicity 1 → diagonalizable.

Cayley-Hamilton theorem

Every matrix satisfies its own characteristic equation: $p(A) = 0$.

For our example: $p(\lambda) = \lambda^2 - 5\lambda + 6$, so $A^2 - 5A + 6 I = 0$.

Check: $A^2 = \begin{pmatrix} 9 & 5 \\ 0 & 4 \end{pmatrix}$, $5 A = \begin{pmatrix} 15 & 5 \\ 0 & 10 \end{pmatrix}$, $6 I = \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}$. $A^2 - 5A + 6I = \begin{pmatrix} 9 - 15 + 6 & 5 - 5 + 0 \\ 0 & 4 - 10 + 6 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} ✓$

This gives an easy way to compute $A^{-1}$: $A \cdot (5I - A) = 6 I$, so $A^{-1} = (5 I - A)/6$. (Works when $\det A \ne 0$, i.e. the constant term is non-zero.)

Common mistakes

• Sign slip in $A - \lambda I$. The diagonal gets $-\lambda$, nothing else.
• Missing the $(-1)^n$ coefficient. For large matrices, the leading sign matters.
• Assuming real roots. Real matrices can have complex eigenvalues (e.g. rotations), in conjugate pairs.
• Using $\det(\lambda I - A)$ instead of $\det(A - \lambda I)$. These differ only by a sign of $(-1)^n$ — same roots, same polynomial up to an overall sign.

Try it in the visualization

Plot $p(\lambda) = \det(A - \lambda I)$ against $\lambda$. Zero crossings = eigenvalues. Adjust $A$'s entries and watch the polynomial warp; the roots slide accordingly.