This experiment deals with the study of eigenvalues and eigenvectors of a matrix and linear transformation which helps the users to understand their geometry in the context of linear transformations. Students will learn how to compute eigenvalues and eigenvectors, interpret their geometric meaning, and see how they describe the scaling or directional behaviour of a transformation.
Note: Matrices considered in this experiment have real entries. Recall (from Experiment 6) that an n×1 matrix is called a column vector. Its transpose is a 1×n matrix called a row vector. These are considered as elements of Rⁿ and are used interchangeably as and when needed.

1. Characteristic polynomial and characteristic equation of a matrix:

The characteristic polynomial of a matrix A of order n×n is defined as |A-λI|, where λ∈R, I is the n×n identity matrix and |A-λI| denotes the determinant of (A-λI). Clearly, degree of the characteristic polynomial is n. Characteristic polynomial of A is denoted by f(λ). The equation f(λ)=0, i.e. |A-λI|=0 is called the characteristic equation.

2. Eigenvalues and eigenvectors of a matrix:

Let A be a matrix of order n×n. Then λ∈R is called an eigenvalue of A, if there exists an n×1matrix X≠0 such that AX=λX and X is called an eigenvector corresponding to the eigenvalue λ.

3. Example:

(i) Let $A = \begin{pmatrix} 1 & 8 \\\ 1 & 3\end{pmatrix}.$ To find the eigenvalues of A, consider the characteristic equation |A-λI|=0, This gives (1-λ)(3-λ)-8=0, which implies that λ=5 and -1. Thus the eigenvalues of A are 1 and 3. To find eigenvectors of A for λ=5, consider (A-5.I)X=0. This gives $\begin{pmatrix} -4 & 8 \\\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\\ y \end{pmatrix} = \begin{pmatrix} 0 \\\ 0 \end{pmatrix}.$

Thus -4x+8y=0 and x-2y=0 which imply that x=2y. So, the eigenvectors corresponding to λ=5 are (2y, y), where 0≠y∈R. To find eigenvectors of A for λ=-1, consider (A-(-1).I)X=0. This gives $\begin{pmatrix} 2 & 8 \\\ 1 & 4 \end{pmatrix}\begin{pmatrix} x \\\ y \end{pmatrix} = \begin{pmatrix} 0 \\\ 0 \end{pmatrix}.$

Thus 2x+8y=0 and x+4y=0 which imply that x=-4y. So, the eigenvectors corresponding to λ=-1 are (-4y, y), where 0≠y∈R.

(ii) Let $A = \begin{pmatrix} 1 & 2 & 1 \\\ 0 & 2 & 0 \\\ 0 & 1 & 2 \end{pmatrix}.$ To find the eigenvalues of A, consider the characteristic equation |A-λI|=0. This gives (1-λ)(2-λ)(2-λ)=0, which implies that λ=1, 2 and 2. Thus the eigenvalues of A are 1, 2 and 2. To find eigenvectors of A for λ=1, consider (A-1.I)X=0. This gives $\begin{pmatrix} 0 & 2 & 1 \\\ 0 & 1 & 0 \\\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\\ y \\\ z \end{pmatrix} = \begin{pmatrix} 0 \\\ 0 \\\ 0 \end{pmatrix}.$

Thus 2y+z=0, 2y=0 and y+z=0 which imply that y=0 and z=0. So, the eigenvectors corresponding to λ=1 are (x, 0, 0), where 0≠x∈R. To find eigenvectors of A for λ=2, consider (A-2.I)X=0. This gives $\begin{pmatrix} -1 & 2 & 1 \\\ 0 & 0 & 0 \\\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\\ y \\\ z \end{pmatrix} = \begin{pmatrix} 0 \\\ 0 \\\ 0 \end{pmatrix}.$

Thus -x+2y+z=0, y=0 and y+z=0 which imply that x=z. So, the eigenvectors corresponding to λ=2 are (x, 0, x), where 0≠x∈R.

4. Eigenvalues and eigenvectors of a linear transformation:

Let V be a finite dimensional vector space over the field R and let T: V→V be a linear transformation. Then λ∈R is called an eigenvalue of T, if there exists x∈V such that x≠0 and T(x)=λx. Such an x is called an eigenvector corresponding to the eigenvalue λ.

5. Eigen space of a linear transformation:

Let V be a finite dimensional vector space over the field R. Let T: V→V be a linear transformation and λ∈R be an eigenvalue of T. Then E_λ≡{x∈V|T(x)=λx} is called the eigen space corresponding to λ. Clearly, E_λ consists of all the eigen vectors and the zero vector. Note that E_λ is a subspace of V.

6. Finding eigenvalues and eigenvectors of a linear transformation:

Let V be a finite dimensional vector space over the field R and let T: V→V be a linear transformation. Then to find the eigenvalues and eigenvectors of T, consider the matrix representation A associated with the linear transformation T w.r.t. a basis of V. Eigenvalues and eigenvectors of the matrix A are same as that of the linear transformation T.

7. Example:

Let T:R²→R² be a linear transformation such that T(x, y)=(y, x), where x, y∈R. Then to find the eigenvalues and eigenvectors of T, consider the matrix representation given by $\begin{pmatrix} 0 & 1 \\\ 1 & 0 \end{pmatrix}$ associated with the linear transformation T w.r.t. the basis of R². Thus the eigenvalues of T are obtained from the characteristic equation |A-λI|=0 which implies that λ=1 and -1 are the eigenvalues of T.
To find eigenvectors of T for λ=1, we solve the equation T(x, y)=(x, y) which implies that (y, x)=(x, y). Thus y=x. So, the eigenvectors corresponding to λ=1 are (x, x), where 0≠x∈R and the eigen space is E₁={(x, x), where x∈R}. To find eigenvectors of T for λ=-1, we solve the equation T(x, y)=-1(x, y) which implies that (y, x)=-1(x, y). Thus y=-x. So, the eigenvectors corresponding to λ=-1 are (x, -x), where 0≠x∈R and the eigen space is E_-1={(x, -x), where x∈R}.

8. Properties of eigenvalue and eigenvector:

Let A be matrix of order n×n and T: Rⁿ→Rⁿ be a linear transformation. Then
(i) Number of eigenvalues of A and T is less than or equal to n.
(ii) If 0 is an eigenvalue of A and T, then A and T are singular.
(iii) If x is an eigenvector of A and T corresponding to an eigenvalue λ, then αx is also an eigenvector corresponding to the eigenvalue λ, where α≠0.
(iv) If x and y are eigenvectors corresponding to an eigenvalue λ of A and T and -x≠y, then x+y is an eigenvector corresponding to the eigenvalue λ.

9. Diagonalizability:

Definition: A matrix A of order n×n is said to be diagonalizable if its eigenvectors form a basis of the vector space Rⁿ over R.
Some results: (i) Let A be diagonalizable matrix of order n×n such that its eigenvalues are λ₁, λ₂, λ₃, …, λ_n and B={x₁, x₂, x₃, …, x_n} be a basis of Rⁿ, where x₁, x₂, x₃, …, x_n are the corresponding eigenvectors. Let T: Rⁿ→Rⁿ be the linear transformation associated with the matrix A w.r.t. the standard basis of Rⁿ. Then the matrix of T w.r.t. the basis B is the diagonal matrix D having eigenvalues λ₁, λ₂, λ₃, …, λ_n as its diagonal entries: the matrices A and D are called similar. Thus A is similar to a diagonal matrix.
(ii) If a matrix A of order n×n has n distinct eigenvalues, then A is diagonalizable.
Note: If a matrix A of order n×n does not have n distinct eigenvalues, then A can still be diagonalizable. This is shown in Example 10 (ii) below.

10. Example:

$\text{(i) Let } A = \begin{pmatrix} 2 & 0 \\\ 1 & 1\end{pmatrix}$. Then its eigenvalues are λ=1 and 2. Thus the matrix A of order 2×2 has 2 distinct eigenvalues, Furthermore, the eigenvectors (0, 1) and (1, 1) of A form a basis of R². Hence A is a diagonalizable matrix and it is similar to $\text{D= } \begin{pmatrix} 1 & 0 \\\ 0 & 2\end{pmatrix}.$

$\text{(ii) Let A= }\begin{pmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & -1 \end{pmatrix}.$ Then its eigenvalues are λ=1, 1 and -1 which are not all distinct. Furthermore, the eigenvectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) of A form a basis of R³. Hence A is a diagonalizable matrix and it is similar to the matrix $\text{D= } \begin{pmatrix} 1 & 0 & 0 \\\ 0 & -1 & 0 \\\ 0 & 0 & 1 \end{pmatrix}.$