Why do we subtract \lambda I instead of just subtracting \lambda from each entry of A?

Subtracting \lambda from each entry would not be matrix arithmetic; we need to subtract the scalar \lambda only from the diagonal, which is exactly what subtracting the matrix \lambda I does. If we subtracted \lambda from every entry, we would lose information and get the wrong eigenvalues.

Can a matrix have repeated eigenvalues?

Yes. For example, the identity matrix I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} has characteristic polynomial (\lambda - 1)^2 = 0, so \lambda = 1 is a repeated eigenvalue with multiplicity 2. Repeated eigenvalues are common in applications like vibration analysis and differential equations.

Do eigenvalues always have to be real numbers?

No. Rotation matrices and other real matrices often have complex eigenvalues. For instance, the 90° rotation matrix \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} has characteristic polynomial \lambda^2 + 1 = 0 with solutions \lambda = \pm i. Complex eigenvalues encode information about oscillation and rotation.

What is the characteristic polynomial actually used for in real applications?

Eigenvalues determine stability in differential equations (medicine, engineering), natural frequencies in vibrating systems (buildings, bridges), ranking in search algorithms (Google PageRank), compression in data analysis (principal component analysis), and long-term behavior in Markov chains (finance, genetics). The characteristic polynomial is the gateway to all these applications.

Characteristic Polynomial

Solve the characteristic polynomial equation to find a matrix's eigenvalues—the scaling factors that reveal how the matrix transforms vectors.

\det(A - \lambda I) = 0

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What each symbol means

What Characteristic Polynomial takes

Symbol	Meaning
$A$	The square matrix whose eigenvalues you want to find; if $A$ is not square, the characteristic polynomial does not exist.
$\lambda$	The eigenvalue variable you are solving for; a scalar (real or complex) satisfying $\det(A - \lambda I) = 0$ .
$I$	The identity matrix of the same size as $A$ , with 1s on the diagonal and 0s elsewhere; using any other matrix here will produce the wrong eigenvalues.

When to use it

Use the characteristic polynomial when you need to find the eigenvalues of a matrix.

Level

Usually taught in: Linear Algebra

Worked examples

1. Find eigenvalues of an upper triangular matrix

Problem

Find all eigenvalues of the matrix

A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}

$A - \lambda I = \begin{pmatrix} 2 - \lambda & 1 \\ 0 & 3 - \lambda \end{pmatrix}$
We form $A - \lambda I$ by subtracting $\lambda$ from each diagonal entry.
$\det(A - \lambda I) = (2 - \lambda)(3 - \lambda) - (1)(0)$
For a 2×2 matrix, the determinant is (top-left)(bottom-right) minus (top-right)(bottom-left).
$(2 - \lambda)(3 - \lambda) = 6 - 5\lambda + \lambda^2$
Expanding: $2 \cdot 3 = 6$ , $2 \cdot (-\lambda) + (-\lambda) \cdot 3 = -5\lambda$ , $(-\lambda)(-\lambda) = \lambda^2$ .
$\lambda^2 - 5\lambda + 6 = 0$
Set the characteristic polynomial equal to zero.
$(\lambda - 2)(\lambda - 3) = 0$
Factor by finding two numbers that multiply to 6 and add to $-5$ : these are $-2$ and $-3$ .
$\lambda = 2 \text{ or } \lambda = 3$
Each factor gives one eigenvalue.

Answer: $\lambda = 2 \text{ or } \lambda = 3$

For upper triangular matrices (all entries below the diagonal are zero), the eigenvalues are exactly the diagonal entries—this is a special property that makes computation fast. Here we see it confirmed: the diagonal holds 2 and 3, matching our eigenvalues.

2. Find eigenvalues when entries are negative

Problem

Find all eigenvalues of

A = \begin{pmatrix} 1 & 2 \\ 2 & -2 \end{pmatrix}

$A - \lambda I = \begin{pmatrix} 1 - \lambda & 2 \\ 2 & -2 - \lambda \end{pmatrix}$
Subtract $\lambda$ from each diagonal entry.
$\det(A - \lambda I) = (1 - \lambda)(-2 - \lambda) - (2)(2)$
Apply the 2×2 determinant formula.
$(1 - \lambda)(-2 - \lambda) = -2 - \lambda + 2\lambda + \lambda^2 = \lambda^2 + \lambda - 2$
Expand: $1 \cdot (-2) = -2$ , $1 \cdot (-\lambda) + (-\lambda) \cdot (-2) = -\lambda + 2\lambda = \lambda$ , $(-\lambda)(-\lambda) = \lambda^2$ .
$\lambda^2 + \lambda - 2 - 4 = \lambda^2 + \lambda - 6$
Subtract the off-diagonal product $(2)(2) = 4$ .
$\lambda^2 + \lambda - 6 = 0$
Set the characteristic equation to zero.
$(\lambda + 3)(\lambda - 2) = 0$
Factor: find two numbers multiplying to $-6$ and adding to 1; these are 3 and $-2$ .
$\lambda = -3 \text{ or } \lambda = 2$
Solve each factor.

Answer: $\lambda = -3 \text{ or } \lambda = 2$

This example shows that matrices with negative entries produce real eigenvalues just as easily—the characteristic polynomial correctly handles mixed signs. One eigenvalue is negative, which in applications means the matrix stretches vectors in one direction while flipping them in another.

3. Model concert ticket sales growth rates

Problem

A concert venue tracks monthly sales of regular and VIP tickets, represented by

R

and

V

(in tens of tickets). Monthly changes follow the coupled system

R_{\text{new}} = 9R + V

and

V_{\text{new}} = R + 9V

. This gives the transition matrix

A = \begin{pmatrix} 9 & 1 \\ 1 & 9 \end{pmatrix}

. Find the growth rates (eigenvalues) that determine long-term sales behavior.

$A - \lambda I = \begin{pmatrix} 9 - \lambda & 1 \\ 1 & 9 - \lambda \end{pmatrix}$
Form $A - \lambda I$ .
$\det(A - \lambda I) = (9 - \lambda)^2 - (1)(1)$
Use the 2×2 determinant formula.
$(9 - \lambda)^2 = 81 - 18\lambda + \lambda^2$
Expand using $(a-b)^2 = a^2 - 2ab + b^2$ with $a = 9$ and $b = \lambda$ .
$81 - 18\lambda + \lambda^2 - 1 = \lambda^2 - 18\lambda + 80$
Subtract 1 from the expanded square.
$\lambda^2 - 18\lambda + 80 = 0$
Set the characteristic polynomial equal to zero.
$\lambda = \frac{18 \pm \sqrt{(-18)^2 - 4(1)(80)}}{2(1)} = \frac{18 \pm \sqrt{324 - 320}}{2}$
Apply the quadratic formula $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 1$ , $b = -18$ , $c = 80$ .
$\lambda = \frac{18 \pm \sqrt{4}}{2} = \frac{18 \pm 2}{2}$
Simplify the discriminant: $\sqrt{4} = 2$ .
$\lambda = \frac{20}{2} = 10 \text{ or } \lambda = \frac{16}{2} = 8$
Compute both solutions.

Answer: $\lambda = 10 \text{ or } \lambda = 8$

The eigenvalues $\lambda = 10$ and $\lambda = 8$ represent two independent growth modes: the dominant mode ( $\lambda = 10$ ) means sales in both categories grow tenfold per month when they move together, while the secondary mode ( $\lambda = 8$ ) captures balanced oscillations between regular and VIP. This analysis is crucial for predicting inventory needs and marketing strategy.

Common mistakes

Where Characteristic Polynomial usually goes wrong

The mistake	Why it is wrong	The fix
Computing $\det(A) = 0$ instead of $\det(A - \lambda I) = 0$	The determinant of $A$ alone is just a number—it has no dependence on $\lambda$ , so it cannot produce the eigenvalues.	Always subtract $\lambda I$ from $A$ before taking the determinant: the equation must be $\det(A - \lambda I) = 0$ , not $\det(A) = 0$ .
Assuming the eigenvalues are the diagonal entries of $A$	Diagonal entries equal eigenvalues only when $A$ is already diagonal; for general matrices, eigenvalues come from solving the polynomial, not reading them off.	Solve $\det(A - \lambda I) = 0$ fully by expanding and factoring or using the quadratic formula; do not assume any entry is an eigenvalue just because it is on the diagonal.
Expanding $\det(A - \lambda I)$ incorrectly, such as writing $\det(A) - n\lambda$ (where $n$ is the matrix size)	The determinant $\det(A - \lambda I)$ is a polynomial of degree $n$ in $\lambda$ , not a linear expression; this mistake loses the coupling terms.	Compute the full determinant carefully: expand products and collect all powers of $\lambda$ from $\lambda^0$ up to $\lambda^n$ , then solve the resulting polynomial equation.

Tips and when to use something else

The sum of all eigenvalues equals the trace of $A$ (sum of diagonal entries), and their product equals $\det(A)$ —use these as quick sanity checks on your answer.
An $n \times n$ matrix always has exactly $n$ eigenvalues when counted with multiplicity and including complex numbers; if you found fewer, check your algebra.
The characteristic polynomial only finds eigenvalues; to find eigenvectors, solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$ separately for each eigenvalue using Gaussian elimination.
If you only need to determine whether a matrix is invertible, skip the characteristic polynomial and just compute $\det(A) \neq 0$ directly—a matrix is invertible if and only if zero is not an eigenvalue.

Related formulas

Eigenvalues and EigenvectorsEigenvalues and eigenvectors show which vectors a matrix scales without rotating, revealing fundamental properties of linear transformations and systems.Determinant of a 2x2 MatrixA single number computed from a 2x2 matrix that tells whether the matrix is invertible and helps solve systems of linear equations.Identity MatrixThe identity matrix leaves any matrix unchanged when multiplied—the matrix equivalent of 1, essential for solving linear systems and finding inverses.Matrix MultiplicationMatrix multiplication combines rows from the first matrix with columns of the second to produce a new matrix, useful for solving systems and transformations.Rank of a MatrixThe rank of a matrix counts its pivot columns, telling you the dimension of its column space and whether a system has a solution.

Frequently asked questions

Why do we subtract $\lambda I$ instead of just subtracting $\lambda$ from each entry of $A$ ?: Subtracting $\lambda$ from each entry would not be matrix arithmetic; we need to subtract the scalar $\lambda$ only from the diagonal, which is exactly what subtracting the matrix $\lambda I$ does. If we subtracted $\lambda$ from every entry, we would lose information and get the wrong eigenvalues.
Can a matrix have repeated eigenvalues?: Yes. For example, the identity matrix $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ has characteristic polynomial $(\lambda - 1)^2 = 0$ , so $\lambda = 1$ is a repeated eigenvalue with multiplicity 2. Repeated eigenvalues are common in applications like vibration analysis and differential equations.
Do eigenvalues always have to be real numbers?: No. Rotation matrices and other real matrices often have complex eigenvalues. For instance, the 90° rotation matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ has characteristic polynomial $\lambda^2 + 1 = 0$ with solutions $\lambda = \pm i$ . Complex eigenvalues encode information about oscillation and rotation.
What is the characteristic polynomial actually used for in real applications?: Eigenvalues determine stability in differential equations (medicine, engineering), natural frequencies in vibrating systems (buildings, bridges), ranking in search algorithms (Google PageRank), compression in data analysis (principal component analysis), and long-term behavior in Markov chains (finance, genetics). The characteristic polynomial is the gateway to all these applications.

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Reviewed 2026-09-18