Eigenvalues and Eigenvectors Questions with Solutions

Examples and questions on the eigenvalues and eigenvectors of square matrices along with their solutions are presented. The properties of the eigenvalues and their corresponding eigenvectors are also discussed and used in solving questions.

Definition of Eigenvalues and Eigenvectors

Let A be an n n square matrix. If there exist a non trivial (not all zeroes) column vector X solution to the matrix equation A X = λ X ; where λ is a scalar, then X is called the eigenvector of matrix A and the corresponding value of λ is called the eigenvalue of matrix A.
Let us rewrite the matrix equation in standard form:
A X - λ X = 0
Let
I be the n n identity matrix and substitute X by I X in the above equation
A X - λ I X = 0
Rewrite as
(A - λ I) X = 0
The above matrix equation has non trivial solutions if and only if the
determinant of the matrix (A - λ I) is equal to zero. Det (A - λ I) = 0 is called the characteristic equation of A.
If A is an n by n matrix, when
(A - λ I) is expanded, it is a polynomial of degree n and therefore (A - λ I) is called the characteristic polynomial of A.

Examples with Solutions on Eigenvalues and Eigenvectors

Example 1
Find all eigenvalues and eigenvectors of matrix \[ A = \begin{bmatrix} -2 & 1 \\ 12 & -3 \end{bmatrix} \]
Solution
We first calculate the eigenvalues and then the eigenvectors.
Find Eigenvalues
We substitute \( A, \lambda \; \text{and} \; I \) in the matrix \( A - \lambda I \) as follows
\( A - \lambda I = \begin{bmatrix} -2 & 1 \\ 12 & -3 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 - \lambda & 1 \\ 12 & -3 - \lambda \end{bmatrix}\)
Solve the equation \( Det( A - \lambda I ) = 0 \)
Calculate the determinant and substitute in the above equation
\( (-2 - \lambda)(-3 - \lambda) - (1)(12) = 0 \)
Expand and rewrite as
\( \lambda^2 +5\lambda - 6 = 0 \)
Solve the above quadratic equation to find two eigenvalues
\( \lambda = 1 \) and \( \lambda = -6 \)
Find Eigenvectors
Eigenvectors for \( \lambda = 1 \)
Substitute \( \lambda \) by 1 in the matrix equation \( (A - \lambda I) X = 0 \)
\( \left( \begin{bmatrix} -2 & 1 \\ 12 & -3 \end{bmatrix} - 1 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) X = 0 \)
Simplify the above
\( \begin{bmatrix} -3 & 1 \\ 12 & -4 \end{bmatrix} X = 0 \)
Let \( X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \) and rewrite the above matrix equation as
\( \begin{bmatrix} -3 & 1 \\ 12 & -4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0 \)
Multiply the top equation by 4 and add it to the second equation and rewrite the system of equations as follows
\( \begin{bmatrix} -3 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0 \)
A solution for \( x_2 \) could be written as \( x_2 = t\) where t takes all real numbers.
Use the top equation \( - 3 x_1 + x_2 = 0 \)
to find \( x_1\) as follows
\( x_1 = \dfrac{x_2}{3} \)
substitute \( x_2 \) by t to obtain
\( x_1 = \dfrac{1}{3} t \)
Hence the eigenvector X corresponding the eigenvalue \( \lambda = 1 \) may be written as
\( X = t \begin{bmatrix} \dfrac{1}{3} \\ 1 \end{bmatrix} , t \in \mathbb{R} \)
Eigenvectors for \( \lambda = -6 \)
Substitute \( \lambda \) by 6 in the matrix equation \( (A - \lambda I) X = 0 \)
\( \left( \begin{bmatrix} -2 & 1 \\ 12 & -3 \end{bmatrix} -( - 6) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) X = 0 \)
which may be simplified to
\( \left( \begin{bmatrix} 4 & 1 \\ 12 & 3 \end{bmatrix} \right) X = 0 \)
Subtract 3 times the top row from the second row to obtain
\( \left( \begin{bmatrix} 4 & 1 \\ 0 & 0 \end{bmatrix} \right) X = 0 \)
A solution for \( x_2 \) could be written as \( x_2 = t\) where t takes all real numbers.
Use the top equation \( 4 x_1 + x_2 = 0 \)
to find \( x_1\) as follows
\( x_1 = - \dfrac{x_2}{4} \)
substitute \( x_2 \) by t to obtain
\( x_1 = - \dfrac{1}{4} t \)
Hence the eigenvector X corresponding the eigenvalue \( \lambda = - 6 \) may be written as
\( X = t \begin{bmatrix} - \dfrac{1}{4} \\ 1 \end{bmatrix} , t \in \mathbb{R} \)


Example 2
Find all eigenvalues and eigenvectors of matrix \[ A = \begin{bmatrix} 1 & 0 & -1 \\ 1 & 0 & 0 \\ -2 & 2 & 1 \end{bmatrix} \] Solution
Find Eigenvalues
We first find the matrix \( A - \lambda I \).
\( A - \lambda I = \begin{bmatrix} 1 & 0 & -1 \\ 1 & 0 & 0 \\ -2 & 2 & 1 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0& 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 - \lambda & 0 & -1 \\ 1 & - \lambda & 0 \\ -2 & 2 & 1 - \lambda \end{bmatrix}\)

Write the characteristic equation.
\( Det(A - \lambda I) = (1-\lambda)(-\lambda(1-\lambda)) - 1(2 - 2\lambda) = 0 \)
factor and rewrite the equation as
\( (1 - \lambda)(\lambda - 2)(\lambda+1) = 0 \)
which gives 3 solutions
\( \lambda = - 1 , \lambda = 1 , \lambda = 2 \)
Find Eigenvectors
Eigenvectors for \( \lambda = - 1 \)
Substitute \( \lambda \) by - 1 in the matrix equation \( (A - \lambda I) X = 0 \) with \( X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \)
\( \begin{bmatrix} 2 & 0 & -1 \\ 1 & 1 & 0 \\ -2 & 2 & 2 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \)
Row reduce to echelon form gives
\( \begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \)
The solutions to the above system and are given by
\( x_3 = t , x_2 = -t/2 , x_1 = t/2, t \in \mathbb{R} \)
Hence the eigenvector corresponding to the eigenvalue \( \lambda = -2 \) is given by
\( X = t \begin{bmatrix} 1/2 \\ -1/2 \\ 1 \end{bmatrix} \)

Eigenvectors for \( \lambda = 1 \)
Substitute \( \lambda \) by \( 1 \) in the matrix equation \( (A - \lambda I) X = 0 \).
\( \begin{bmatrix} 0 & 0 & -1 \\ 1 & - 1 & 0 \\ -2 & 2 & 0 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \)
Row reduce to echelon form gives
\( \begin{bmatrix} 1 & - 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \)
The solutions to the above system and are given by
\( x_3 = 0 , x_2 = t , x_1 = t , t \in \mathbb{R} \)
Hence the eigenvector corresponding to the eigenvalue \( \lambda = 1 \) is given by
\( X = t \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \)

Eigenvectors for \( \lambda = 2 \)
Substitute \( \lambda \) by \( 1 \) in the matrix equation \( (A - \lambda I) X = 0 \).
\( \begin{bmatrix} -1 & 0 & -1 \\ 0& - 2 & 0 \\ -2 & 2 & -1 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \)
Row reduce to echelon form gives
\( \begin{bmatrix} 1 & 0 & -1 \\ 0 & -2 & -1 \\ 0 & 0 & 0 \end{bmatrix} \ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \)
The solutions to the above system and are given by
\( x_3 = t , x_2 = -t/2 , x_1 = - t , t \in \mathbb{R} \)
Hence the eigenvector corresponding to the eigenvalue \( \lambda = 2 \) is given by
\( X = t \begin{bmatrix} -1 \\ -1/2 \\ 1 \end{bmatrix} \)

Eigenvalues of Triangular Matrices

let us find the eigenvalues of matrix \( A = \begin{bmatrix} b & c & d \\ 0 & e & f \\ 0 & 0 & g \end{bmatrix} \)
The characteristic equation is given by
\( Det (A - \lambda I) = Det \begin{bmatrix} b - \lambda & c & d \\ 0 & e -\lambda & f \\ 0 & 0 & g - \lambda \end{bmatrix} = (b - \lambda)(e - \lambda)(g - \lambda) = 0 \)
The eigenvalues values for a triangular matrix are equal to the entries in the given triangular matrix. In this example the eigenvalues are: a , e and g.

Eigenvalues of the Power of a Matrix

If \( \lambda \) is an eigenvalue of matrix A, then we can write
\( AX = \lambda X \), where X is the eigenvector corresponding to the eigenvalue \( \lambda \).
Left multiply both sides of the above equation by matrix A.
\( A (A X) = A (\lambda X) \)
Simplify to
\( A^2 X = A (\lambda X) = \lambda (A X)\)
Substitute \( A X \) on the right side by \( \lambda X \) to obtain
\( A^2 X = \lambda^2 X\)
We can continue multiplying by A and simplifying to obtain
\( A^n X = \lambda^n X\)
If \( \lambda \) is an eigenvalue of matrix A and X the corresponding eigenvector, then \( \lambda^n \) is an eigenvalue of matrix \( A ^n\) and X its corresponding eigenvector.

Properties of Eigenvalues and Eigenvectors

  1. Matrix A is singular if and only if \( \lambda = 0 \) is an eigenvalue value of matrix A.
    or
    If matrix A is invertible, then none of its eigenvalues is equal to zero.
  2. If \( \lambda \) is an eigenvalue of matrix A and X the corresponding eigenvector, then the eigenvalue of matrix \( A ^n\) is equal to \( \lambda^n \) and the corresponding eigenvector is X.
  3. The product of all the eigenvalues of a matrix is equal to its determinant.
  4. The sum of all the eigenvalues of a matrix is equal to its trace (the sum of all entries in the main diagonal). You may check the examples above.
  5. The eigenvalues of matrix A and its transpose are the same.
  6. If A is a square invertible matrix with \( \lambda \) its eigenvalue and X its corresponding eigenvector, then \( 1/\lambda \) is an eigenvalue of \( A^{-1} \) and X is a corresponding eigenvector.
  7. If \( \lambda \) is an eigenvalue of matrix A and X a corresponding eigenvalue, then \( \lambda - t \) , where t is a scalar, is an eigenvalue of \( A - t I \) and X is a corresponding eigenvector.

Questions

  • Part 1
    1) Find all eigenvalues and their corresponding eigenvectors for the matrices:
    a) \( A = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} \) , b) \( B = \begin{bmatrix} -1 & 1 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} \)
    Part 2
    1) Find all values of parameters p and q for which the matrix \( A = \begin{bmatrix} 2 & p\\ 2 & q \end{bmatrix} \) has eigenvalues equal to - 1 and -3.
    2) Find all values of parameters p which the matrix \( A = \begin{bmatrix} 1 & -33 & -1\\ 0 & p - 1 & 3 \\ 0 & 0 & p + 1 \end{bmatrix} \) has eigenvalues equal to 1 and 2 and 3.
  • Part 3
    Matrix \( A = \begin{bmatrix} 1 & - 1\\ 2 & p \end{bmatrix} \) , where p is a parameter, has an eigenvector \( X = \begin{bmatrix} 1\\ -1 \end{bmatrix} \). Find the all eigenvalues and eigenvectors of matrix A.
  • Part 4
    Matrix \( A = \begin{bmatrix} a & b\\ 1 & -1 \end{bmatrix} \) has eigenvalues 3 and 4. Find the eigenvectors of matrix A.
  • Part 5
    A 3 by 3 matrix A has eigenvalues 1 ,2 and 3 and their corresponding eigenvectors \( \begin{bmatrix} 1\\ 0 \\ 1 \end{bmatrix} \), \( \begin{bmatrix} 0\\ -1 \\ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1/2\\ 1 \\ 0 \end{bmatrix} \). Find the product \( A \begin{bmatrix} 3\\ -1 \\ 2 \end{bmatrix} \).
  • Part 6
    Matrix \( A \) has eigenvalues 2 and 3 and their corresponding eigenvectors \( \begin{bmatrix} -1\\ 1 \end{bmatrix} \) and \( \begin{bmatrix} -1/2\\ 1 \end{bmatrix} \).
    Find the eigenvalues and the corresponding eigenvectors of \( A^{-3} \).

Solutions to the Above Questions

  • Part 1
    Matrix A
    \( A - \lambda I = \begin{bmatrix} 1 - \lambda & 0\\ -1 & 2-\lambda \end{bmatrix} \)
    Characteristic equation
    \( (1-\lambda)(2-\lambda) = 0 \)
    Eigenvalues are solutions to the above equation; there are two solutions.
    \( \lambda = 1 \) and \( \lambda = 2 \)
  • Eigenvectors for \( \lambda = 1 \)
    \( A - \lambda I = \begin{bmatrix} 0 & 0\\ -1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0\)
    Eigenvector is the solution to the above system which can be written as
    \( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = t \begin{bmatrix} 1 \\ 1 \end{bmatrix} , t \in \mathbb{R} \)
    Eigenvectors for \( \lambda = 2 \)
    \( A - \lambda I = \begin{bmatrix} -1 & 0\\ -1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0\)
    Eigenvector is the solution to the above system which can be written as
    \( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = t \begin{bmatrix} 0 \\ 1 \end{bmatrix} , t \in \mathbb{R} \)

    Matrix B
    \( B - \lambda I = \begin{bmatrix} -1 - \lambda & 1 & 1\\ 2 & 1 -\lambda & 1 \\ 1 & 0 & -\lambda \end{bmatrix} \)
    Using the last row to find determinant, the characteristic equation
    \( - \lambda^3 + 4\lambda = 0 \)
    Eigenvalues are solutions to the above equation and are given by:
    \( \lambda = 0 \) , \( \lambda = 2 \) and \( \lambda = - 2 \)
    Eigenvectors for \( \lambda = 0 \)
    \( B - \lambda I = \begin{bmatrix} -1 & 1 & 1\\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \)
    Rewrite in row echelon form
    \( \begin{bmatrix} -1 & 1 & 1\\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0\)
    Eigenvector is the solution to the above system which can be written as
    \( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = t \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} , t \in \mathbb{R} \)
    Eigenvectors for \( \lambda = 2 \)
    \( B - \lambda I = \begin{bmatrix} -3 & 1 & 1\\ 2 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0\)
    Rewrite in row echelon form
    \( \begin{bmatrix} 1 & 0 & - 2\\ 0 & - 1 & 5 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0\)
    Eigenvector is the solution to the above system which can be written as
    \( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = t \begin{bmatrix} 2 \\ 5 \\ 1 \end{bmatrix} , t \in \mathbb{R} \)
    Eigenvectors for \( \lambda = - 2 \)
    \( B - \lambda I = \begin{bmatrix} 1 & 1 & 1\\ 2 & 3 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0\)
    Rewrite in row echelon form
    \( \begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0\)
    Eigenvector is the solution to the above system which can be written as
    \( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = t \begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix} , t \in \mathbb{R} \)

  • Part 2
    \( A - \lambda I = \begin{bmatrix} 2 - \lambda & p\\ 2 & q -\lambda \end{bmatrix} \)
    The characteristic equation is given by
    \( (2 - \lambda)(q - \lambda ) - 2p = 0 \)
    The eigenvalues are given as - 1 and -3 and are solutions to the characteristic equation. Substitute \( \lambda \) by - 1 and -3 to obtain a system of equations in p and q.
    \( 3(q + 1 ) - 2p = 0 \) and \( 5(q + 3 ) - 2p = 0 \)
    Solve to obtain p = -15/2 and q = -6.

  • Part 3
    Let \( \lambda \) be the eigenvalue corresponding to the given eigenvector. Hence
    \( \begin{bmatrix} 1 & - 1\\ 2 & p \end{bmatrix} \begin{bmatrix} 1\\ -1 \end{bmatrix} = \lambda \begin{bmatrix} 1\\ -1 \end{bmatrix} \)
    Use the above matrix equation to write a system of equation in p and \( \lambda \) as follows:
    \( 1 + 1 = \lambda \) and \( 2 - p = - \lambda \)
    Solve to obtain
    p = 4 and \( \lambda = 2 \)
    We can write matrix A as
    \( A = \begin{bmatrix} 1 & - 1\\ 2 & 4 \end{bmatrix} \)
    The product of the eigenvalues is equal to the determinant of A (property 3 above). Hence
    \( Det (A) = 4 + 2 = 2 \lambda \)
    gives the second eigenvalue as
    \( \lambda = 3 \)
    The eigenvector corresponding to \( \lambda = 3 \) is given by
    \( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = t \begin{bmatrix} -1/2 \\ 1 \end{bmatrix} , t \in \mathbb{R} \)

  • Part 4
    The product of the eigenvalues is equal to the determinant of A. Hence
    \( Det (A) = - a - b = 3 (4) = 12 \)
    The sum of the eigenvalues is equal to the trace. Hence
    a - 1 = 3 + 4 = 7
    Solve the two equations in a and b simultaneously to find
    a = 8 and b = -20
    Hence matrix A is given by
    \( A = \begin{bmatrix} 8 & -20\\ 1 & -1 \end{bmatrix} \)
    and its eigenvectors of are given by: \( \begin{bmatrix} 4\\ 1 \end{bmatrix} \) and \( \begin{bmatrix} 5\\ 1 \end{bmatrix} \).

  • Part 5
    Combine the three eigenvalues and eigenvectors to write
    \( A \begin{bmatrix} 1 & 0 & 1/2\\ 0 & -1 & 1\\ 1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 \begin{bmatrix} 1\\ 0 \\ 1 \end{bmatrix} & 2 \begin{bmatrix} 0\\ -1 \\ 1 \end{bmatrix} & 3\begin{bmatrix} 1/2\\ 1 \\ 0 \end{bmatrix} \end{bmatrix} \)
    Hence
    \( A = \begin{bmatrix} 1 & 0 & 3/2\\ 0 & -2 & 3 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1/2\\ 0 & -1 & 1\\ 1 & 1 & 0 \end{bmatrix}^{-1}\)
    which then gives
    \( A \begin{bmatrix} 3\\ -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 3/2\\ 0 & -2 & 3 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1/2\\ 0 & -1 & 1\\ 1 & 1 & 0 \end{bmatrix}^{-1} \begin{bmatrix} 3\\ -1 \\ 2 \end{bmatrix} = \begin{bmatrix}-1\\ -6\\ -1\end{bmatrix}\)

  • Part 6
    According to property 6 above, the eigenvalues of \( A^{-1} \) are 1/2 and 1/3 and the corresponding eigenvectors are \( \begin{bmatrix} -1\\ 1 \end{bmatrix} \) and \( \begin{bmatrix} -1/2\\ 1 \end{bmatrix} \).
    By definition \( A^{-3} \) = \( (A^{-1})^3 \). Hence according to property 2 above, the eigenvalues of \( A^{-3} \) are \( (1/2)^3 = 1/8 \) and \( (1/3)^3 = 1/27 \) and the corresponding eigenvalues are \( \begin{bmatrix} -1\\ 1 \end{bmatrix} \) and \( \begin{bmatrix} -1/2\\ 1 \end{bmatrix} \).

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