Eigenvalues of a Matrix

In linear algebra, eigenvalues are special numbers associated with a square matrix. They reveal important characteristics like stability, direction, and scaling of vector transformations. Widely used in mathematics, physics, and data science, eigenvalues simplify complex matrix operations. Understanding eigen values is key for solving systems of equations, quantum mechanics, and even in machine learning techniques like PCA (Principal Component Analysis).

1.0Definition of Eigenvalues

An eigenvalue of a square matrix A is a scalar λ such that:

$Ax=\lambda x$

Here, A is an n × n matrix, x is a non-zero vector (called the eigenvector), and λ is the eigenvalue. This equation means that multiplying a matrix A by a vector x results in a scaled version of x, not a rotated or otherwise transformed one.

2.0Eigenvalue Formula

To find the eigenvalues of a matrix, use the characteristic equation:

$det(A-\lambda l)=0$

Where:

• A is a square matrix,
• I is the identity matrix of the same size,
• λ represents eigenvalues,
• det stands for determinant.

This results in a polynomial equation in λ. The roots of this polynomial are the eigenvalues of the matrix.

3.0Eigenvalue of a 2×2 Matrix

For a 2×2 matrix: $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

The characteristic equation becomes:

$\det \left(\left[\begin{array}{c}a-\lambda b\\ cd-\lambda \end{array}\right]\right)=0$

$\Rightarrow (a-\lambda )(d-\lambda )-bc=0$

$\Rightarrow {\lambda }^2-(a+d)\lambda +(ad-bc)=0$

The roots of this quadratic equation give the eigenvalues of the 2×2 matrix.

4.0Properties of Eigen Values

Here are some key properties of eigen values for square matrices:

1. Only Square Matrices Have Eigenvalues: Non-square matrices do not have eigenvalues.
2. Sum of Eigenvalues = Trace of Matrix: $\sum {\lambda }_1=Tr(A)$
3. Product of Eigenvalues = Determinant of Matrix: $\prod {\lambda }_i=\det (A)$
4. Eigenvalues of a Diagonal Matrix are the Diagonal Elements.
5. If A is symmetric, all eigenvalues are real.
6. Eigenvalues of A⁻¹ are reciprocal of A’s eigenvalues.
7. If A is triangular, eigenvalues are its diagonal entries.

5.0Example: Finding Eigenvalues of a 2×2 Matrix

Example 1: Let  $A=\left[\begin{array}{cc}4& 2\\ 1& 3\end{array}\right]$

Solution: 

Step 1: Set up the characteristic equation:

$\det (A-\lambda I)=0$

$=\det \left(\left[\begin{array}{cc}4-\lambda & 2\\ 1& 3-\lambda \end{array}\right]\right)$

$=(4-\lambda )(3-\lambda )-(2)(1)$

$={\lambda }^2-7\lambda +10=0$

Step 2: Solve the quadratic equation:

$\lambda =\frac{7\pm \sqrt{49-40}}{2}=\frac{7\pm 3}{2}$

$\Rightarrow {\lambda }_1=5,{\lambda }_2=2$ 

Hence, the eigenvalues are 5 and 2.

Example 2: Let $A=\left[\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right].$ Find the eigenvalues of matrix A.

Solution:

This is a symmetric tridiagonal matrix. Use the characteristic equation:

$\det (A-\lambda I)=0$

$\Rightarrow \left|\begin{array}{ccc}2-\lambda & -1& 0\\ -1& 2-\lambda & -1\\ 0& -1& 2-\lambda \end{array}\right|=0$ 

Let’s evaluate the determinant using cofactor expansion:

Let $D(\lambda )=\det (A-\lambda I)$

Use shortcut tridiagonal determinant formula or compute directly:

$$\begin{gathered} D(\lambda )=(2-\lambda )[(2-\lambda {)}^2-1]-(-1{)}^2(2-\lambda ) \\ D(\lambda )=(2-\lambda )[(2-\lambda {)}^2-1]-(2-\lambda ) \\ D(\lambda )=(2-\lambda )[(2-\lambda {)}^2-1]-(2-\lambda ) \end{gathered}$$

Factor out : $(2-\lambda )$

$$\begin{gathered} =(2-\lambda )[(2-\lambda {)}^2-1-1] \\ =(2-\lambda )[(2-\lambda {)}^2-2] \end{gathered}$$

Let  $x=2-\lambda$

$$\begin{gathered} \Rightarrow D=x(x^2-2)=x^3-2x=0 \\ \Rightarrow x(x^2-2)=0 \\ \Rightarrow x=0,x=\pm \sqrt{2} \end{gathered}$$

So,

$$\begin{gathered} 2-\lambda =0 \\ \Rightarrow \lambda =22-\lambda =\sqrt{2} \\ \Rightarrow \lambda =2-\sqrt{22}-\lambda =-\sqrt{2} \\ \Rightarrow \lambda =2+\sqrt{2} \end{gathered}$$

Eigenvalues:

$2,2+\sqrt{2},2-\sqrt{2}$

Example 3: Find the eigenvalues of  $A=\left[\begin{array}{ccc}7& 0& 0\\ 0& -3& 0\\ 0& 0& 5\end{array}\right].$

Solution:

For a diagonal matrix, the eigenvalues are the diagonal elements.

So, the eigenvalues are:

7, −3, 5  

Example 4: Let $A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right].$ Find the eigenvalues of A.

Solution:

$\det (A-\lambda I)=\left|\begin{array}{cc}-\lambda & -1\\ 1& -\lambda \end{array}\right|={\lambda }^2+1=0$

$\lambda =\pm i$

Eigenvalues: i, -i (Complex)

Example 5: Let $A=\left[\begin{array}{cc}a& b\\ b& a\end{array}\right].$ Show that the eigenvalues of A are a + b and a - b. 

Solution:

$A=\left[\begin{array}{cc}a& b\\ b& a\end{array}\right]\Rightarrow A-\lambda I=\left[\begin{array}{cc}a-\lambda & b\\ b& a-\lambda \end{array}\right]$

$\det (A-\lambda I)=(a-\lambda {)}^2-b^2=0$

$$\begin{gathered} \Rightarrow (a-\lambda {)}^2=b^2 \\ \Rightarrow a-\lambda =\pm b \\ \Rightarrow \lambda =a\pm b \end{gathered}$$

Eigenvalues: $a+b,a-b$

Example 6: Let $A=\left[\begin{array}{cc}6& 2\\ 2& 3\end{array}\right].$ Find the eigenvalues and verify:

• Sum of eigenvalues = Trace(A)
• Product of eigenvalues = det(A)

Solution:

Characteristic equation:

$\det (A-\lambda I)=\left|\begin{array}{cc}6-\lambda & 2\\ 2& 3-\lambda \end{array}\right|=0$

$\Rightarrow (6-\lambda )(3-\lambda )-4=0$

$\Rightarrow {\lambda }^2-9\lambda +14=0$

$\Rightarrow \lambda =\frac{9\pm \sqrt{81-56}}{2}=\frac{9\pm 5}{2}$

$\Rightarrow {\lambda }_1=7,{\lambda }_2=2$

Verification:

• Sum = 7 + 2 = 9 = Trace(A) 
• Product = 7⋅2 = 14 = det(A) 

Verified

6.0Applications of Eigenvalues

• Physics: Quantum mechanics, vibration analysis
• Mathematics: Stability of systems, differential equations
• Engineering: Stress-strain analysis
• Data Science: Principal Component Analysis (PCA)

7.0Practice Question on Eigenvalues

1. Find the eigen values of the matrix: $A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right].$ (Hint: Use the characteristic equation: $det(A-\lambda I)=0$)