Eigenvalues and eigenvectors

Lecture 26

Minjae Park

Auburn University
MATH 2660 - Spring 2026

March 20, 2026

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Eigenvalues and eigenvectors

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Special linear transformations

• In general, describing the effect of an arbitrary linear transformation is not an easy task.
• However, we have seen some special cases where the geometric description is clear.
• For example, consider how a diagonal matrix \(D=\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}\) changes a shape in \(\mathbb{R}^2\).

\(2\times 2\) Diagonal matrices

• In this case, the effect of \(D\) is easy to understand because:
  • \(D\vec{e}_1 = a\vec{e}_1\)
  • \(D\vec{e}_2 = b\vec{e}_2\)
• Note that \(\{\vec{e}_1, \vec{e}_2\}\) is an orthonormal basis, so each direction is independent.
• Therefore:
  • Stretch by a factor \(a\) in the \(\vec{e}_1\) direction
  • Stretch by a factor \(b\) in the \(\vec{e}_2\) direction

\(n\times n\) diagonal matrices

• Let \(D=\operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_n)\).
• Then the geometric effect of \(D:\mathbb{R}^n \to \mathbb{R}^n\) is:
  • Stretching by a factor \(\lambda_i\) in the \(\vec{e}_i\) direction for each \(i=1,2,\dots,n\)
• Since \(\{\vec{e}_1,\dots,\vec{e}_n\}\) is an orthonormal basis, each stretching is independent.
• In this case, we have \(D\vec{e}_i = \lambda_i \vec{e}_i\)

Linear transformations

• Recall that \(A=[\vec{u}_1 \ \cdots \ \vec{u}_n]\) represents a linear transformation that sends \(\vec{e}_i\) to \(\vec{u}_i\).
• We can think of \(\{\vec{u}_1,\dots,\vec{u}_n\}\) as defining a new coordinate system.
• In this coordinate system, a “diagonal-like” behavior would mean:
  • \(A\vec{u}_i = \lambda_i \vec{u}_i\) for each \(i=1,2,\dots,n\)
• That is, each \(\vec{u}_i\) is simply stretched (not rotated or mixed with others).

Eigenvalues and eigenvectors

Let \(A\) be an \(n\times n\) matrix. If a nonzero vector \(\vec{u}\in\mathbb{R}^n\) satisfies \[ A\vec{u} = \lambda \vec{u} \] for some \(\lambda\in\mathbb{R}\), then:

• \(\vec{u}\) is called an eigenvector of \(A\)
• \(\lambda\) is called an eigenvalue of \(A\)

Eigenbasis

• If \(\{\vec{u}_1,\dots,\vec{u}_n\}\) is a basis consisting of eigenvectors, it is called an eigenbasis.
• In that case, \(A\) behaves like a diagonal matrix in this new coordinate system.
• Question: Can we always find an eigenbasis for any matrix \(A\)?
• Answer: Not always. We will see examples later.

Visualization

• The standard basis is not always the best frame for understanding a linear transformation.
• This visualization illustrates the difference between the standard basis and an eigenbasis.

Finding eigenvalues for 2x2

• Let \(A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\).
• Suppose \(\vec{u}=\langle x,y \rangle\neq \vec{0}\) is an eigenvector.
• Then \[ A\vec{u} = \lambda \vec{u} = (\lambda I_2) \vec{u} \]
• This is equivalent to: \[ (A-\lambda I_2)\vec{u} = \vec{0} \]
• When does this have a nonzero solution?

Characteristic equation

• A homogeneous system has a nonzero solution if and only if the matrix is singular, i.e. its determinant is zero.
• Therefore: \[ \det(A-\lambda I_2)=0 \]
• Compute: \[ \det\begin{pmatrix}a-\lambda & b \\ c & d-\lambda\end{pmatrix} = (a-\lambda)(d-\lambda)-bc = \lambda^2 - (a+d)\lambda - bc \]
• This gives a quadratic equation in \(\lambda\).
• It is called the characteristic equation.

Finding eigenvectors for 2x2

• Once an eigenvalue \(\lambda\) is found, solve: \[ (A-\lambda I_2)\vec{u} = \vec{0} \]
• Since \(\det(A-\lambda I_2)=0\), there are infinitely many solutions.
• Typically, the solution space is one-dimensional.
• Any nonzero vector in this space is an eigenvector.

Non-uniqueness of eigenvectors

• If \(\vec{u}\) is an eigenvector, then any nonzero multiple \(t\vec{u}\) is also an eigenvector.
• Indeed: \[ A(t\vec{u})=t(A\vec{u})=t\lambda\vec{u}=\lambda(t\vec{u}) \]
• Therefore, eigenvectors are not unique—only their direction matters.

Example

• Let \[ A=\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \]

• Step 1: Find eigenvalues
  \[ \det(A-\lambda I_2)=\begin{vmatrix}2-\lambda & 1 \\ 1 & 2-\lambda\end{vmatrix} =(2-\lambda)^2-1 =\lambda^2-4\lambda+3 \] \[ =(\lambda-1)(\lambda-3)=0 \] So \(\lambda_1=1\), \(\lambda_2=3\)

• Step 2: Find eigenvectors

  For \(\lambda=1\): \[ (A-I)=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \] Solve \(x+y=0 \Rightarrow \vec{u}_1=\langle 1,-1 \rangle\)

  For \(\lambda=3\): \[ (A-3I)=\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \] Solve \(-x+y=0 \Rightarrow \vec{u}_2=\langle 1,1 \rangle\)

• Step 3: Geometric interpretation

  • Along direction \(\langle 1,1 \rangle\): stretch by factor 3
    
  • Along direction \(\langle 1,-1 \rangle\): stretch by factor 1 (no change)

• So \(A\) stretches space along two special directions (its eigenvectors).

Example: No real eigenvalues

• Let \[ A=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \]

• Step 1: Find eigenvalues
  \[ \det(A-\lambda I)=\begin{vmatrix}-\lambda & -1 \\ 1 & -\lambda\end{vmatrix} =\lambda^2+1=0 \]

• Solutions: \[ \lambda=\pm i \]

• These are not real numbers, so there are no real eigenvalues.

• Conclusion:

  • There are no real eigenvectors in \(\mathbb{R}^2\)
  • This transformation represents a rotation by \(90^\circ\)
  • No direction is preserved (everything is rotated)

Example: Repeated eigenvalue with eigenbasis

• Let \[ A=\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}=2I \]

• Step 1: Find eigenvalues
  \[ \det(A-\lambda I)=(2-\lambda)^2=0 \]

• So the only eigenvalue is \(\lambda=2\) (with multiplicity 2)

• Step 2: Find eigenvectors
  \[ (A-2I)=0 \]

  • Every nonzero vector satisfies \(A\vec{u}=2\vec{u}\)

• Conclusion:

  • Every direction is an eigenvector
  • We can choose infinitely many eigenbases (e.g., \(\{\vec{e}_1,\vec{e}_2\}\))
  • Geometrically: uniform scaling by factor 2 in all directions

Example: Repeated eigenvalue with no eigenbasis

• Let \[ A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \]

• Step 1: Find eigenvalues
  \[ \det(A-\lambda I)=\begin{vmatrix}1-\lambda & 1 \\ 0 & 1-\lambda\end{vmatrix} =(1-\lambda)^2=0 \]

• So \(\lambda=1\) (with multiplicity 2)

• Step 2: Find eigenvectors
  \[ (A-I)=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \] Solve: \[ y=0 \] So eigenvectors are: \[ \vec{u}=\langle x,0 \rangle \]

• Conclusion:

  • Only one independent eigenvector
  • Cannot form a basis of \(\mathbb{R}^2\)
  • No eigenbasis exists

• Geometric interpretation:

  • Horizontal direction is preserved
  • Other vectors are “sheared” (shifted sideways)
  • This is a shear transformation, not pure stretching

Summary

• Eigenvalues and eigenvectors provide important geometric insight into a linear transformation
• If a matrix \(A\) admits an eigenbasis (a basis consisting of eigenvectors), its action can be understood as independent scaling along those directions
• Eigenvalues are obtained from the characteristic equation, and eigenvectors are found by solving \((A-\lambda I)\vec{u}=\vec{0}\)
• Not every matrix admits an eigenbasis, as illustrated by previous examples
• Next class: We will study conditions under which a matrix admits an eigenbasis