Lecture 7 — Introduction to Eigenvalues & Eigenvectors

1. Definition

For a square matrix \(A\in\mathbb{R}^{n\times n}\), a non-zero vector \(v\) is an eigenvector and \(\lambda\) is the corresponding eigenvalue if:

\[A v = \lambda v.\]

2. How to compute

Eigenvalues are roots of the characteristic polynomial:

\[\det(A - \lambda I) = 0.\]

Eigenvectors are non-zero solutions to \((A-\lambda I)v=0\) for each eigenvalue \(\lambda\).

3. Worked Examples

Example: Consider the matrix

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

Step 1: Characteristic Equation

The eigenvalues are obtained by solving \[ \det(A - \lambda I) = 0 \]

\[ \det \begin{bmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{bmatrix} = (2-\lambda)(2-\lambda) - (1)(1) \]

\[ = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 \]

Thus, the characteristic equation is \[ \lambda^2 - 4\lambda + 3 = 0 \]

Step 2: Solve for Eigenvalues

\[ \lambda^2 - 4\lambda + 3 = 0 \]

Factoring: \[ (\lambda - 1)(\lambda - 3) = 0 \]

Hence, the eigenvalues are \[ \lambda_1 = 1, \quad \lambda_2 = 3 \]

Step 3: Find Eigenvectors

(a) For \(\lambda_1 = 1\):

Solve \[ (A - I)x = 0 \]

\[ \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

This gives equation \(x_1 + x_2 = 0\), so \(x_1 = -x_2\).

Thus, the eigenvector is \[ v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \]

(b) For \(\lambda_2 = 3\):

Solve \[ (A - 3I)x = 0 \]

\[ \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

This gives equation \(-x_1 + x_2 = 0\), so \(x_1 = x_2\).

Thus, the eigenvector is \[ v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

Final Answer:

The eigenvalues are \(\lambda_1 = 1, \lambda_2 = 3\).

Corresponding eigenvectors are \[ v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \quad v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

Example 2:

Consider the matrix

\[ B = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \]

Step 1: Characteristic Equation

\[ \det(B - \lambda I) = \det \begin{bmatrix} 4-\lambda & 2 \\ 1 & 3-\lambda \end{bmatrix} \]

\[ = (4-\lambda)(3-\lambda) - (2)(1) \]

\[ = (12 - 7\lambda + \lambda^2) - 2 = \lambda^2 - 7\lambda + 10 \]

So the characteristic equation is \[ \lambda^2 - 7\lambda + 10 = 0 \]

Step 2: Solve for Eigenvalues

\[ \lambda^2 - 7\lambda + 10 = 0 \]

Factoring: \[ (\lambda - 5)(\lambda - 2) = 0 \]

Thus, the eigenvalues are \[ \lambda_1 = 5, \quad \lambda_2 = 2 \]

Step 3: Find Eigenvectors

(a) For \(\lambda_1 = 5\):

Solve \[ (B - 5I)x = 0 \]

\[ \begin{bmatrix} -1 & 2 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

Equation: \(-x_1 + 2x_2 = 0 \quad \Rightarrow \quad x_1 = 2x_2\).

So eigenvector is \[ v_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \]

(b) For \(\lambda_2 = 2\):

Solve \[ (B - 2I)x = 0 \]

\[ \begin{bmatrix} 2 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

Equation: \(2x_1 + 2x_2 = 0 \quad \Rightarrow \quad x_1 = -x_2\).

So eigenvector is \[ v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \]

Final Answer:

The eigenvalues are \(\lambda_1 = 5, \lambda_2 = 2\).

Corresponding eigenvectors are \[ v_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \quad v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \]

3×3 Worked Examples

Example 1:

Consider the symmetric matrix

\[ A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & 9 \end{bmatrix} \]

Step 1: Characteristic Equation

We compute \[ \det(A - \lambda I) = \det \begin{bmatrix} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & 4 \\ 0 & 4 & 9-\lambda \end{bmatrix} \]

\[ = (2-\lambda) \cdot \det \begin{bmatrix} 3-\lambda & 4 \\ 4 & 9-\lambda \end{bmatrix} \]

\[ = (2-\lambda) \big[(3-\lambda)(9-\lambda) - 16\big] \]

\[ = (2-\lambda)(\lambda^2 - 12\lambda + 11) \]

Step 2: Solve for Eigenvalues

Characteristic polynomial: \[ (2-\lambda)(\lambda^2 - 12\lambda + 11) = 0 \]

Eigenvalues are \[ \lambda_1 = 2, \quad \lambda_{2,3} = \frac{12 \pm \sqrt{144 - 44}}{2} = \frac{12 \pm 10}{2} \]

So \[ \lambda_2 = 11, \quad \lambda_3 = 1 \]

Step 3: Find Eigenvectors

(a) For \(\lambda_1 = 2\):

\[ (A - 2I)x = 0 \quad \Rightarrow \quad \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 4 & 7 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \]

Equation: \(x_2 + 4x_3 = 0\), so \(x_2 = -4x_3\). Let \(x_3=1\), then \(x_2=-4\).

Eigenvector: \[ v_1 = \begin{bmatrix} 1 \\ -4 \\ 1 \end{bmatrix} \]

(b) For \(\lambda_2 = 11\):

\[ (A - 11I)x = 0 \quad \Rightarrow \quad \begin{bmatrix} -9 & 0 & 0 \\ 0 & -8 & 4 \\ 0 & 4 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \]

From equations: \(-8x_2 + 4x_3 = 0 \Rightarrow 2x_2 = x_3\). Let \(x_2 = 1\), then \(x_3 = 2\).

Eigenvector: \[ v_2 = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \]

(c) For \(\lambda_3 = 1\):

\[ (A - I)x = 0 \quad \Rightarrow \quad \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 4 \\ 0 & 4 & 8 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \]

Equation: \(2x_2 + 4x_3 = 0 \Rightarrow x_2 = -2x_3\). Let \(x_3=1\), then \(x_2=-2\).

Eigenvector: \[ v_3 = \begin{bmatrix} 0 \\ -2 \\ 1 \end{bmatrix} \]

Final Answer:

Eigenvalues: \(\lambda_1=2, \lambda_2=11, \lambda_3=1\)
Eigenvectors: \[ v_1=\begin{bmatrix}1\\-4\\1\end{bmatrix}, \quad v_2=\begin{bmatrix}0\\1\\2\end{bmatrix}, \quad v_3=\begin{bmatrix}0\\-2\\1\end{bmatrix} \]

Example 2:

Consider the non-symmetric matrix

\[ B = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & 2 \end{bmatrix} \]

Step 1: Characteristic Equation

Since \(B\) is upper triangular, the eigenvalues are just the diagonal entries: \[ \lambda_1 = 1, \quad \lambda_2 = 3, \quad \lambda_3 = 2 \]

Step 2: Find Eigenvectors

(a) For \(\lambda_1 = 1\):

\[ (B - I)x = \begin{bmatrix} 0 & 2 & 0 \\ 0 & 2 & 4 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \]

From last row: \(x_3=0\). From first row: \(2x_2=0 \Rightarrow x_2=0\). So eigenvector: \[ v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \]

(b) For \(\lambda_2 = 3\):

\[ (B - 3I)x = \begin{bmatrix} -2 & 2 & 0 \\ 0 & 0 & 4 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \]

From last row: \(-x_3=0 \Rightarrow x_3=0\). From first row: \(-2x_1 + 2x_2 = 0 \Rightarrow x_1=x_2\). So eigenvector: \[ v_2 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \]

(c) For \(\lambda_3 = 2\):

\[ (B - 2I)x = \begin{bmatrix} -1 & 2 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \]

From first row: \(-x_1 + 2x_2 = 0 \Rightarrow x_1=2x_2\). From second row: \(x_2 + 4x_3=0 \Rightarrow x_2=-4x_3\). Let \(x_3=1\), then \(x_2=-4, x_1=-8\). So eigenvector: \[ v_3 = \begin{bmatrix} -8 \\ -4 \\ 1 \end{bmatrix} \]

Final Answer:

Eigenvalues: \(\lambda_1=1, \lambda_2=3, \lambda_3=2\)
Eigenvectors: \[ v_1=\begin{bmatrix}1\\0\\0\end{bmatrix}, \quad v_2=\begin{bmatrix}1\\1\\0\end{bmatrix}, \quad v_3=\begin{bmatrix}-8\\-4\\1\end{bmatrix} \]

Example 4

Consider the matrix

\[ A = \begin{bmatrix} 1 & -1 & 1 \\ -3 & 1 & 3 \\ -3 & -1 & 5 \end{bmatrix} \]

Step 1: Characteristic Polynomial

We compute the characteristic polynomial \(\det(A - \lambda I)\):

\[ \det\begin{bmatrix} 1-\lambda & -1 & 1 \\ -3 & 1-\lambda & 3 \\ -3 & -1 & 5-\lambda \end{bmatrix} \]

Expanding (you may expand along the first row) gives a cubic that factors nicely: \[ \det(A-\lambda I) = -(\lambda-1)(\lambda-2)(\lambda-4) \] so the characteristic polynomial is \[ (\lambda-1)(\lambda-2)(\lambda-4)=0. \]

Step 2: Eigenvalues

Therefore the eigenvalues are \[ \lambda_1 = 1,\qquad \lambda_2 = 2,\qquad \lambda_3 = 4. \]

Step 3: Eigenvectors

(a) For \(\lambda_1 = 1\):

Solve \((A - I)x = 0\): \[ \begin{bmatrix} 0 & -1 & 1 \\ -3 & 0 & 3 \\ -3 & -1 & 4 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}. \] From the equations we get \(x_1 = x_2 = x_3\). Choosing \(x_1=1\) gives \[ v_1 = \begin{bmatrix}1\\1\\1\end{bmatrix}. \]

(b) For \(\lambda_2 = 2\):

Solve \((A - 2I)x = 0\): \[ \begin{bmatrix} -1 & -1 & 1 \\ -3 & -1 & 3 \\ -3 & -1 & 3 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} = 0. \] This system yields \(x_1 = x_3\) and \(x_2 = 0\). Choosing \(x_1=1\) gives \[ v_2 = \begin{bmatrix}1\\0\\1\end{bmatrix}. \]

(c) For \(\lambda_3 = 4\):

Solve \((A - 4I)x = 0\): \[ \begin{bmatrix} -3 & -1 & 1 \\ -3 & -3 & 3 \\ -3 & -1 & 1 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} = 0. \] This gives \(x_1 = 0\) and \(x_2 = x_3\). Choosing \(x_3 = 1\) yields \[ v_3 = \begin{bmatrix}0\\1\\1\end{bmatrix}. \]

Final Answer:

Eigenvalues: \[ \lambda_1 = 1,\quad \lambda_2 = 2,\quad \lambda_3 = 4. \] Corresponding eigenvectors (one convenient choice each): \[ v_1=\begin{bmatrix}1\\1\\1\end{bmatrix},\quad v_2=\begin{bmatrix}1\\0\\1\end{bmatrix},\quad v_3=\begin{bmatrix}0\\1\\1\end{bmatrix}. \]

Properties of Eigenvalues and Eigenvectors

• Definition: For a square matrix \(A\), a scalar \(\lambda\) and nonzero vector \(x\) satisfy \[ A x = \lambda x \] where \(\lambda\) is called an eigenvalue and \(x\) is the corresponding eigenvector.
• Number of Eigenvalues: An \(n \times n\) matrix has exactly \(n\) eigenvalues (counting multiplicities).
• Characteristic Polynomial: Eigenvalues are roots of \[ \det(A - \lambda I) = 0. \]
• Trace Property: The sum of eigenvalues equals the trace of \(A\) (sum of diagonal elements): \[ \text{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n. \]
• Determinant Property: The product of eigenvalues equals the determinant of \(A\): \[ \det(A) = \lambda_1 \cdot \lambda_2 \cdots \lambda_n. \]
• Similarity Invariance: If \(B = P^{-1}AP\), then \(A\) and \(B\) have the same eigenvalues.
• Independence of Eigenvectors: Eigenvectors corresponding to distinct eigenvalues are linearly independent.
• Diagonalization: If \(A\) has \(n\) linearly independent eigenvectors, then \[ A = P D P^{-1} \] where \(D\) is a diagonal matrix of eigenvalues and \(P\) is the matrix of eigenvectors.
• Symmetric Matrices: For real symmetric matrices:
  • All eigenvalues are real.
  • Eigenvectors corresponding to different eigenvalues are orthogonal.
• Scaling Property: If \(x\) is an eigenvector, any nonzero scalar multiple \(\alpha x\) is also an eigenvector (direction matters, not magnitude).
• Zero Eigenvalue: If \(\lambda = 0\) is an eigenvalue, then \(A\) is singular (non-invertible).
• Complex Eigenvalues: Real matrices may have complex eigenvalues (conjugate pairs).

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