Matrices considered in this experiment have real entries. An n×1 matrix is called a column vector. Its transpose is a 1×n matrix called a row vector and it is also an element of Rⁿ. These 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 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 a column vector i.e. an n×1matrix X≠0 such that AX=λX and X is called an eigenvector corresponding to the eigenvalue λ.

3. Examples:

$$\begin{aligned} &\hspace{-9.2cm} \text{i.) Let } A = \begin{pmatrix} 1 & 8 \\\ 1 & 3\end{pmatrix} \end{aligned}$$

To find the eigenvalues of A, consider the characteristic equation |A-λI|=0, This gives (1-λ)(3-λ)-8=0, which implies 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 implies 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 implies that x=-4y. So, the eigenvector corresponding to λ=-1 are (-4y, y),where 0≠y∈R.

$$\begin{aligned} &\hspace{-9.2cm} \text{ii.) Let } A = \begin{pmatrix} 1 & 2 & 1 \\\ 0 & 2 & 0 \\\ 0 & 1 & 2 \end{pmatrix} \end{aligned}$$

To find the eigenvalues of A, consider the characteristic equation |A-λI|=0.

$$\begin{aligned} \text{ i.e. }\mid \begin{pmatrix} 1 & 2 & 1 \\\ 0 & 2 & 0 \\\ 0 & 1 & 2 \end{pmatrix} - \begin{pmatrix} λ & 0 & 0 \\\ 0 & λ & 0 \\\ 0 & 0 & λ \end{pmatrix} \mid= 0, \text{ i.e. }\mid \begin{matrix} 1-λ & 2 & 1 \\\ 0 & 2-λ & 0 \\\ 0 & 1 & 2-λ \end{matrix} \mid. \end{aligned}$$

This gives (1-λ)(2-λ)2=0, which implies 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

$$\begin{aligned} \begin{pmatrix} 0 & 2 & 1 \\\ 0 & 1 & 0 \\\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\\ y \\\ z \end{pmatrix} = \begin{pmatrix} 0 \\\ 0 \\\ 0 \end{pmatrix}. \end{aligned}$$

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

$$\begin{aligned} \begin{pmatrix} -1 & 2 & 1 \\\ 0 & 0 & 0 \\\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\\ y \\\ z \end{pmatrix} = \begin{pmatrix} 0 \\\ 0 \\\ 0 \end{pmatrix}. \end{aligned}$$

Thus x+2y+z=0, y=0 and y+z=0 which implies that x=-z. So, the eigenvector 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 eigenvalue and eigenvector of T, consider a matrix representation A associated to the linear transformation T w.r.t. the standard basis. Eigenvalues and eigenvectors of the matrix A are same as that of 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 to 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 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 eigenvector corresponding to λ=-1 are (x, -x), where 0≠x∈R and its 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.
(i.) Number of eigenvalues of a matrix A is less than or equal to n.
(ii.) If 0 is an eigenvalue of a matrix A or a linear transformation T, then the matrix A or the linear transformation T is singular.
(iii.) If x is an eigenvector of a matrix A or a linear transformation T corresponding to an eigenvalue λ, then αx, where α≠0 is also an eigenvector corresponding to the eigenvalue λ
(iv.) If x and y are eigenvectors corresponding to an eigenvalue λ of a matrix A or a linear transformation 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.
(i) Let A be diagonalizable such that λ₁, λ₂, λ₃, …, λ_n are its eigenvalues and B={x₁, x₂, x₃, …, x_n} is 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 matrix of T w.r.t. B is the diagonal matrix D having eigenvalues λ₁, λ₂, λ₃, …, λ_n as its diagonal entries.
In such a case A and D are called similar.
(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 the example 9(ii) below.

10. Examples:

$$\begin{aligned} &\hspace{-9.2cm} \text{i.) Let } A = \begin{pmatrix} 2 & 0 \\\ 1 & 1\end{pmatrix} \end{aligned}$$

Clearly, 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 as it is similar to

$$\begin{aligned} &\hspace{-3cm} \text{D= } \begin{pmatrix} 1 & 0 \\\ 0 & 2\end{pmatrix}. \\ &\hspace{-9.2cm} \text{ii.) Let A= }\begin{pmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & -1 \end{pmatrix}. \end{aligned}$$ Clearly, 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 as it is similar to the matrix $$\begin{aligned} &\hspace{-3cm} \text{D= } \begin{pmatrix} 1 & 0 & 0 \\\ 0 & -1 & 0 \\\ 0 & 0 & 1 \end{pmatrix}. \end{aligned}$$

Remark:

Thus it may be noted that the eigenvalues of a matrix may not be distinct but the matrix is diagonalizable.