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The Eigenvalue Equation

Start from $A\mathbf{v} = \lambda\mathbf{v}$ and rearrange: $A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}$ . Writing $\lambda\mathbf{v} = \lambda I\mathbf{v}$ , this becomes $(A - \lambda I)\mathbf{v} = \mathbf{0}$ . So finding an eigenvector for $\lambda$ is the same as finding a nonzero vector in the null space of the matrix $A - \lambda I$ .

The null space of $A - \lambda I$ contains a nonzero vector if and only if $A - \lambda I$ is singular — that is, not invertible. A square matrix is singular precisely when its determinant is zero. So the condition for $\lambda$ to be an eigenvalue is: $\det(A - \lambda I) = 0$ .

This is the key insight: every eigenvalue problem reduces to a determinant equation. Instead of searching blindly for vectors satisfying $A\mathbf{v} = \lambda\mathbf{v}$ , we first find all valid $\lambda$ by solving $\det(A - \lambda I) = 0$ , then find the corresponding vectors.

Formal View

Theorem 7.3 — Eigenvalue Condition

A scalar

\lambda

is an eigenvalue of

A

if and only if

\det(A - \lambda I) = 0

Equivalently: $\lambda$ is an eigenvalue iff $A - \lambda I$ is singular iff $\text{Null}(A - \lambda I) \neq \{\mathbf{0}\}$ .

Remark 7.4

This is a deep connection between three equivalent conditions: (1)

(A - \lambda I)\mathbf{v} = \mathbf{0}

has a nonzero solution, (2)

A - \lambda I

is not invertible, and (3)

\det(A - \lambda I) = 0

. These are precisely the conditions from the Invertible Matrix Theorem (Chapter 4) applied to the matrix

A - \lambda I

Why This Matters

Reducing eigenvalue finding to a determinant condition gives a systematic algorithm that works for any square matrix.

Stability analysis of linear dynamical systems depends on whether eigenvalues have positive or negative real parts
Numerical methods like the QR algorithm find all eigenvalues of large matrices iteratively
The condition $\det(A - \lambda I) = 0$ underlies the characteristic equation used in control theory
Differential equations: $e^{At}\mathbf{x}_0$ evolves along eigenvector directions, with rates given by eigenvalues

Learning Resources

Finding eigenvalues and eigenvectors

Khan Academy

Walks through the determinant condition and the full process for finding eigenvalues.

16 min

Eigenvalues — the full picture

Trefor Bazett

Visual and algebraic explanation of why the determinant condition works.

12 min

Quiz

Question 1

$\lambda$ is an eigenvalue of $A$ if and only if:

Question 2

If $A - \lambda I$ is invertible, then $\lambda$ is an eigenvalue of $A$ .

Question 3

For the diagonal matrix $A = \begin{pmatrix}5 & 0 \\ 0 & 3\end{pmatrix}$ , which values are eigenvalues?

Common Mistakes

Writing the eigenvalue equation as $A\lambda = \mathbf{v}$ instead of $A\mathbf{v} = \lambda\mathbf{v}$ — $\lambda$ is a scalar, not a matrix input.
Confusing $A - \lambda I$ with $A - \lambda$ (you must subtract a matrix $\lambda I$ , not a scalar from $A$ ).
Forgetting the $I$ : computing $\det(A - \lambda)$ instead of $\det(A - \lambda I)$ .