Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors reveal the intrinsic skeleton of a linear transformation. Most vectors get both rotated and stretched when a matrix acts on them. But a few special vectors — the eigenvectors — only get stretched: their direction is preserved (or reversed), and the stretch factor is the eigenvalue.

Finding eigenvalues reduces to solving the characteristic equation $\det(A-\lambda I)=0$ , a polynomial in $\lambda$ . Each root is an eigenvalue; for each eigenvalue, the eigenvectors are the nonzero solutions to $(A-\lambda I)\mathbf{v}=\mathbf{0}$ .

Diagonalization — writing $A=PDP^{-1}$ — is the payoff: once you have eigenvectors and eigenvalues, complex operations like $A^{100}$ become trivial. This chapter also covers the full theory of linear transformations, which underpins every subsequent result.

Linear transformations and their matrices

A function $T:\mathbb{R}^n\to\mathbb{R}^m$ is linear if $T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$ and $T(c\mathbf{v})=cT(\mathbf{v})$ for all vectors and scalars. Together these imply $T(c\mathbf{u}+d\mathbf{v}) = cT(\mathbf{u})+dT(\mathbf{v})$ — linear maps preserve linear combinations.

Every linear transformation has a unique representing matrix: $A = [T(\mathbf{e}_1)\;|\;T(\mathbf{e}_2)\;|\;\cdots\;|\;T(\mathbf{e}_n)]$ , where the columns are the images of the standard basis vectors. Once you know what $T$ does to the basis, you know everything.

Geometric examples: rotations, reflections, projections, shears, and scalings are all linear. Non-linear examples: translations, squaring a vector, adding a constant.

Composition of transformations corresponds to matrix multiplication: $(T_A\circ T_B)(\mathbf{x}) = A(B\mathbf{x}) = (AB)\mathbf{x}$ . This is the geometric motivation for the matrix multiplication rule.

💡Explain it simply

A linear transformation reshapes space without bending, tearing, or moving the origin. Lines stay lines. Parallel lines stay parallel. The whole space gets rotated, stretched, or squished in a uniform way. Every such rule is exactly captured by a matrix.

Matrix of a 90° rotation

Find the matrix for $90°$ counterclockwise rotation in $\mathbb{R}^2$ .
$T(\mathbf{e}_1)=T(\langle 1,0\rangle) = \langle 0,1\rangle$ (the $x$ -axis vector rotates to the $y$ -axis).
$T(\mathbf{e}_2)=T(\langle 0,1\rangle) = \langle -1,0\rangle$ .
Matrix:
$R_{90} = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$
.
Apply to $\langle 3,2\rangle$ :
$\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}3\\2\end{pmatrix} = \begin{pmatrix}-2\\3\end{pmatrix}$
. Rotated $90°$ ✓.

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Kernel and image

The kernel of $T$ is $\ker(T) = \{\mathbf{x}: T(\mathbf{x})=\mathbf{0}\}$ — the set of inputs that $T$ collapses to zero. For $T_A$ , this is exactly $\text{Null}(A)$ .

$T$ is injective (one-to-one) iff $\ker(T)=\{\mathbf{0}\}$ — only zero maps to zero. Equivalently, different inputs always produce different outputs.

The image of $T$ is $\text{Im}(T) = \{T(\mathbf{x}): \mathbf{x}\in\mathbb{R}^n\} = \text{Col}(A)$ . $T$ is surjective (onto) iff $\text{Im}(T)=\mathbb{R}^m$ iff $\text{rank}(A)=m$ .

The rank-nullity theorem for transformations: $\dim(\ker(T))+\dim(\text{Im}(T))=n$ . Independent of any choice of basis or matrix representation.

💡Explain it simply

The kernel is the transformation's blind spot — inputs it can't distinguish from zero. The image is its range — all possible outputs. An injective map has a trivial kernel (no blind spots). A surjective map has an image that fills the entire output space.

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A nonzero vector $\mathbf{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$ if $A\mathbf{v}=\lambda\mathbf{v}$ . The matrix merely scales $\mathbf{v}$ — it does not change its direction (or it reverses it when $\lambda<0$ ).

To find eigenvalues: rewrite as $(A-\lambda I)\mathbf{v}=\mathbf{0}$ . For a nonzero solution to exist, the matrix $A-\lambda I$ must be singular: $\det(A-\lambda I)=0$ . This is the characteristic equation.

The characteristic polynomial of an $n\times n$ matrix has degree $n$ and exactly $n$ roots in $\mathbb{C}$ (counting multiplicity). Real matrices may have complex eigenvalues in conjugate pairs.

For each eigenvalue $\lambda_k$ , the eigenspace is $E_{\lambda_k}=\text{Null}(A-\lambda_k I)$ . Any nonzero vector in this eigenspace is an eigenvector. The eigenspace always has dimension at least $1$ .

Trace and determinant shortcuts: $\sum_i \lambda_i = \text{tr}(A)$ and $\prod_i \lambda_i = \det(A)$ . For a $2\times 2$ matrix, these two conditions fully determine the eigenvalues once you find the characteristic polynomial.

💡Explain it simply

Imagine pushing on a rubber sheet with a matrix. Most arrows move and spin. Eigenvectors are special arrows that only stretch or shrink — they never change direction. The eigenvalue is the stretch factor. If $\lambda=3$ , the arrow triples in length. If $\lambda=-1$ , it flips around.

Eigenvalues and eigenvectors of a 2×2 matrix

Find eigenvalues and eigenvectors of
$A=\begin{pmatrix}3&1\\0&2\end{pmatrix}$
.
Characteristic equation: $\det(A-\lambda I)=(3-\lambda)(2-\lambda)=0$ . Eigenvalues: $\lambda_1=3$ , $\lambda_2=2$ .
For $\lambda_1=3$ :
$(A-3I)\mathbf{v}=\mathbf{0}\Rightarrow\begin{pmatrix}0&1\\0&-1\end{pmatrix}\mathbf{v}=\mathbf{0}$
. Solution: $\mathbf{v}_1=\langle 1,0\rangle$ .
For $\lambda_2=2$ :
$(A-2I)\mathbf{v}=\mathbf{0}\Rightarrow\begin{pmatrix}1&1\\0&0\end{pmatrix}\mathbf{v}=\mathbf{0}$
. Solution: $\mathbf{v}_2=\langle 1,-1\rangle$ .
Check: $\text{tr}(A)=5=3+2$ ✓. $\det(A)=6=3\times 2$ ✓.

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Diagonalization

A matrix $A$ is diagonalizable if $A=PDP^{-1}$ , where the columns of $P$ are $n$ linearly independent eigenvectors and $D=\text{diag}(\lambda_1,\ldots,\lambda_n)$ contains the corresponding eigenvalues.

Criterion: $A$ is diagonalizable iff it has $n$ linearly independent eigenvectors. A sufficient condition: $n$ distinct eigenvalues. Repeated eigenvalues may or may not permit diagonalization.

Computing powers: $A^k = PD^kP^{-1}$ , and $D^k=\text{diag}(\lambda_1^k,\ldots,\lambda_n^k)$ is trivial. This turns computing $A^{100}$ into three matrix multiplications.

Geometric interpretation: $D=P^{-1}AP$ means that in the coordinate system of eigenvectors, the transformation $A$ acts as pure diagonal scaling. Diagonalisation finds the 'natural' coordinate system for $A$ .

Non-diagonalisable matrices (defective matrices) require Jordan normal form — beyond this course but worth knowing exists.

💡Explain it simply

Diagonalisation asks: is there a tilted coordinate system in which your transformation just stretches each axis independently? If you tilt your axes to align with the eigenvectors, the transformation becomes trivial. The tilt is $P$ ; the independent scalings are $D$ .

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Geometric transformations and their eigenvalues

Rotation by $\theta$ (counterclockwise): $R_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$ . $\det(R_\theta)=1$ (area preserved, orientation maintained). Eigenvalues $e^{\pm i\theta}$ are complex for $\theta\neq 0,\pi$ — no real eigenvectors, consistent with the fact that rotations spin everything.

Reflection across the $x$ -axis: $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ . Eigenvalues $\lambda_1=1$ (vectors on $x$ -axis are fixed) and $\lambda_2=-1$ (vectors on $y$ -axis are flipped). $\det=-1$ : orientation reversed.

Projection onto a line: eigenvalues $1$ (vectors on the line are fixed by projection) and $0$ (vectors orthogonal to the line collapse to zero). Idempotent: $P^2=P$ .

Shear $\begin{pmatrix}1&k\\0&1\end{pmatrix}$ : repeated eigenvalue $\lambda=1$ with only one independent eigenvector. Not diagonalisable — it is a canonical example of a defective matrix.

💡Explain it simply

Each transformation type has its own eigenvalue signature. Rotations have complex eigenvalues (no real fixed directions). Reflections have eigenvalues $\pm 1$ (one axis is fixed, the other flips). Projections have eigenvalues $0$ and $1$ (one subspace is kept, the complementary subspace is erased).

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⚠️

Common Mistakes to Avoid

Including the zero vector as an eigenvector. Eigenvectors must be nonzero by definition.
Writing $\det(A-\lambda)=0$ instead of $\det(A-\lambda I)=0$ . You cannot subtract a scalar from a matrix.
Assuming every matrix is diagonalisable. A matrix with $n$ distinct eigenvalues is, but repeated eigenvalues can prevent it.
Confusing eigenvalues (scalars $\lambda$ ) and eigenvectors (nonzero vectors $\mathbf{v}$ ). They are distinct objects that go together.
Forgetting to check the trace/determinant as a sanity check after computing eigenvalues.