Eigenvalues and Eigenvectors Problems

Author

Siju Swamy

Published

07 October 2024

Instructions: Solve the problems given in the course resource page. Some of the solutions are wrong. Identify the mistakes in the solution and correct it in your work.

Eigenvalues and Eigenvectors of 2X2 Matrices

Problem 1

Find the eigenvalues and eigenvectors of the matrix: \[ A = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} \]

Solution 1

  1. Characteristic Polynomial: \[ \text{det}(A - \lambda I) = \begin{vmatrix} 1 - \lambda & 2 \\ 2 & 3 - \lambda \end{vmatrix} = (1 - \lambda)(3 - \lambda) - 4 = \lambda^2 - 4\lambda - 1 = 0 \] Eigenvalues: \(\lambda_1 = 2 + \sqrt{5}, \lambda_2 = 2 - \sqrt{5}\)

  2. Eigenvectors:

    • For \(\lambda_1 = 2 + \sqrt{5}\): \[ (A - (2 + \sqrt{5})I)v = 0 \implies \begin{pmatrix} -1 - \sqrt{5} & 2 \\ 2 & 1 - \sqrt{5} \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 2 \\ 1 + \sqrt{5} \end{pmatrix} \]
    • For \(\lambda_2 = 2 - \sqrt{5}\): \[ (A - (2 - \sqrt{5})I)v = 0 \implies \begin{pmatrix} -1 + \sqrt{5} & 2 \\ 2 & 1 + \sqrt{5} \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_2 = \begin{pmatrix} 2 \\ 1 - \sqrt{5} \end{pmatrix} \]

Problem 2

Find the eigenvalues and eigenvectors of the matrix: \[ B = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} \]

Solution 2

  1. Characteristic Polynomial: \[ \text{det}(B - \lambda I) = \begin{vmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{vmatrix} = (4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 10 = 0 \] Eigenvalues: \(\lambda_1 = 5, \lambda_2 = 2\)

  2. Eigenvectors:

    • For \(\lambda_1 = 5\): \[ (B - 5I)v = 0 \implies \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \]
    • For \(\lambda_2 = 2\): \[ (B - 2I)v = 0 \implies \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_2 = \begin{pmatrix} -1 \\ 2 \end{pmatrix} \]

Problem 3

Find the eigenvalues and eigenvectors of the matrix: \[ C = \begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix} \]

Solution 3

  1. Characteristic Polynomial: \[ \text{det}(C - \lambda I) = \begin{vmatrix} 3 - \lambda & 5 \\ 2 & 4 - \lambda \end{vmatrix} = (3 - \lambda)(4 - \lambda) - 10 = \lambda^2 - 7\lambda + 2 = 0 \] Eigenvalues: \(\lambda_1 = 6, \lambda_2 = 1\)

  2. Eigenvectors:

    • For \(\lambda_1 = 6\): \[ (C - 6I)v = 0 \implies \begin{pmatrix} -3 & 5 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 5 \\ 3 \end{pmatrix} \]
    • For \(\lambda_2 = 1\): \[ (C - 1I)v = 0 \implies \begin{pmatrix} 2 & 5 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_2 = \begin{pmatrix} -5 \\ 2 \end{pmatrix} \]

Problem 4

Find the eigenvalues and eigenvectors of the matrix: \[ D = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \]

Solution 4

  1. Characteristic Polynomial: \[ \text{det}(D - \lambda I) = \begin{vmatrix} -\lambda & 1 \\ 0 & -\lambda \end{vmatrix} = \lambda^2 = 0 \] Eigenvalues: \(\lambda_1 = 0\) (with algebraic multiplicity 2)

  2. Eigenvectors:

    • For \(\lambda_1 = 0\): \[ (D - 0I)v = 0 \implies \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \]

Problem 5

Find the eigenvalues and eigenvectors of the matrix: \[ E = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} \]

Solution 5

  1. Characteristic Polynomial: \[ \text{det}(E - \lambda I) = \begin{vmatrix} 2 - \lambda & 0 \\ 0 & 3 - \lambda \end{vmatrix} = (2 - \lambda)(3 - \lambda) = 0 \] Eigenvalues: \(\lambda_1 = 2, \lambda_2 = 3\)

  2. Eigenvectors:

    • For \(\lambda_1 = 2\): \[ (E - 2I)v = 0 \implies \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \]
    • For \(\lambda_2 = 3\): \[ (E - 3I)v = 0 \implies \begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \]

Problem 6

Find the eigenvalues and eigenvectors of the matrix: \[ F = \begin{pmatrix} 4 & 1 \\ 2 & 5 \end{pmatrix} \]

Solution 6

  1. Characteristic Polynomial: \[ \text{det}(F - \lambda I) = \begin{vmatrix} 4 - \lambda & 1 \\ 2 & 5 - \lambda \end{vmatrix} = (4 - \lambda)(5 - \lambda) - 2 = \lambda^2 - 9\lambda + 18 = 0 \] Eigenvalues: \(\lambda_1 = 6, \lambda_2 = 3\)

  2. Eigenvectors:

    • For \(\lambda_1 = 6\): \[ (F - 6I)v = 0 \implies \begin{pmatrix} -2 & 1 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \]
    • For \(\lambda_2 = 3\): \[ (F - 3I)v = 0 \implies \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix} \]

Problem 7

Find the eigenvalues and eigenvectors of the matrix: \[ G = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \]

Solution 7

  1. Characteristic Polynomial: \[ \text{det}(G - \lambda I) = \begin{vmatrix} 1 - \lambda & 2 \\ 2 & 1 - \lambda \end{vmatrix} = (1 - \lambda)^2 - 4 = \lambda^2 - 2\lambda - 3 = 0 \] Eigenvalues: \(\lambda_1 = 3, \lambda_2 = -1\)

  2. Eigenvectors:

    • For \(\lambda_1 = 3\): \[ (G - 3I)v = 0 \implies \begin{pmatrix} -2 & 2 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \]
    • For \(\lambda_2 = -1\): \[ (G + I)v = 0 \implies \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \]

Problem 8

Find the eigenvalues and eigenvectors of the matrix: \[ H = \begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix} \]

Solution 8

  1. Characteristic Polynomial: \[ \text{det}(H - \lambda I) = \begin{vmatrix} 2 - \lambda & 3 \\ 2 & 1 - \lambda \end{vmatrix} = (2 - \lambda)(1 - \lambda) - 6 = \lambda^2 - 3\lambda - 4 = 0 \] Eigenvalues: \(\lambda_1 = 4, \lambda_2 = -1\)

  2. Eigenvectors:

    • For \(\lambda_1 = 4\): \[ (H - 4I)v = 0 \implies \begin{pmatrix} -2 & 3 \\ 2 & -3 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \]
    • For \(\lambda_2 = -1\): \[ (H + I)v = 0 \implies \begin{pmatrix} 3 & 3 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix} \]

Problem 9

Find the eigenvalues and eigenvectors of the matrix: \[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Solution 9

  1. Characteristic Polynomial: \[ \text{det}(I - \lambda I) = \begin{vmatrix} 1 - \lambda & 0 \\ 0 & 1 - \lambda \end{vmatrix} = (1 - \lambda)^2 = 0 \] Eigenvalues: \(\lambda_1 = 1\) (with algebraic multiplicity 2)

  2. Eigenvectors:

    • For \(\lambda_1 = 1\): \[ (I - I)v = 0 \implies \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \]

Eigenvalues and Eigenvectors of 3x3 Matrices

Problem 1

Find the eigenvalues and eigenvectors of the matrix: \[ A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix} \]

Solution 1

  1. Characteristic Polynomial: \[ \text{det}(A - \lambda I) = \begin{vmatrix} 2 - \lambda & 1 & 0 \\ 1 & 2 - \lambda & 1 \\ 0 & 1 & 2 - \lambda \end{vmatrix} = (2 - \lambda)^3 - 2 = 0 \] Eigenvalues: \(\lambda_1 = 3, \lambda_2 = 1\) (with algebraic multiplicity 1)

  2. Eigenvectors:

    • For \(\lambda_1 = 3\): \[ (A - 3I)v = 0 \implies \begin{pmatrix} -1 & 1 & 0 \\ 1 & -1 & 1 \\ 0 & 1 & -1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]
    • For \(\lambda_2 = 1\): \[ (A - I)v = 0 \implies \begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies v_2 = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \]

Problem 2

Find the eigenvalues and eigenvectors of the matrix: \[ B = \begin{pmatrix} 4 & 1 & 2 \\ 1 & 4 & 3 \\ 2 & 3 & 4 \end{pmatrix} \]

Solution 2

  1. Characteristic Polynomial: \[ \text{det}(B - \lambda I) = \begin{vmatrix} 4 - \lambda & 1 & 2 \\ 1 & 4 - \lambda & 3 \\ 2 & 3 & 4 - \lambda \end{vmatrix} = (4 - \lambda)^3 - 6(4 - \lambda) = 0 \] Eigenvalues: \(\lambda_1 = 6, \lambda_2 = 3\) (with algebraic multiplicity 1)

  2. Eigenvectors:

    • For \(\lambda_1 = 6\): \[ (B - 6I)v = 0 \implies \begin{pmatrix} -2 & 1 & 2 \\ 1 & -2 & 3 \\ 2 & 3 & -2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \]
    • For \(\lambda_2 = 3\): \[ (B - 3I)v = 0 \implies \begin{pmatrix} 1 & 1 & 2 \\ 1 & 1 & 3 \\ 2 & 3 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies v_2 = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} \]

Problem 3

Find the eigenvalues and eigenvectors of the matrix: \[ C = \begin{pmatrix} 5 & 4 & 2 \\ 4 & 6 & 3 \\ 2 & 3 & 4 \end{pmatrix} \]

Solution 3

  1. Characteristic Polynomial: \[ \text{det}(C - \lambda I) = \begin{vmatrix} 5 - \lambda & 4 & 2 \\ 4 & 6 - \lambda & 3 \\ 2 & 3 & 4 - \lambda \end{vmatrix} = (5 - \lambda)((6 - \lambda)(4 - \lambda) - 9) - 4(4(4 - \lambda) - 6) + 2(12 - 3(6 - \lambda)) \] Solving the determinant gives the eigenvalues: \(\lambda_1 = 12, \lambda_2 = 3, \lambda_3 = 0\)

  2. Eigenvectors:

    • For \(\lambda_1 = 12\): \[ (C - 12I)v = 0 \implies \begin{pmatrix} -7 & 4 & 2 \\ 4 & -6 & 3 \\ 2 & 3 & -8 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \]
    • For \(\lambda_2 = 3\): \[ (C - 3I)v = 0 \implies \begin{pmatrix} 2 & 4 & 2 \\ 4 & 3 & 3 \\ 2 & 3 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies v_2 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \]
    • For \(\lambda_3 = 0\): \[ (C - 0I)v = 0 \implies \begin{pmatrix} 5 & 4 & 2 \\ 4 & 6 & 3 \\ 2 & 3 & 4 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies v_3 = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} \]

Problem 4

Find the eigenvalues and eigenvectors of the matrix: \[ D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix} \]

Solution 4

  1. Characteristic Polynomial: \[ \text{det}(D - \lambda I) = \begin{vmatrix} 2 - \lambda & 0 & 0 \\ 0 & 2 - \lambda & 1 \\ 0 & 0 & 2 - \lambda \end{vmatrix} = (2 - \lambda)^3 = 0 \] Eigenvalues: \(\lambda_1 = 2\) (with algebraic multiplicity 3)

  2. Eigenvectors:

    • For \(\lambda_1 = 2\): \[ (D - 2I)v = 0 \implies \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, v_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \]

Problem 5

Find the eigenvalues and eigenvectors of the matrix: \[ E = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{pmatrix} \]

Solution 5

  1. Characteristic Polynomial: \[ \text{det}(E - \lambda I) = \begin{vmatrix} 3 - \lambda & 0 & 0 \\ 0 & 3 - \lambda & 1 \\ 0 & 0 & 3 - \lambda \end{vmatrix} = (3 - \lambda)^3 = 0 \] Eigenvalues: \(\lambda_1 = 3\) (with algebraic multiplicity 3)

  2. Eigenvectors:

    • For \(\lambda_1 = 3\): \[ (E - 3I)v = 0 \implies \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, v_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \]

Diagonalization of a matrix

Let \(A\) be square matrix, with all eigen values satisfies the relation \(A.M=G.M\), then the diagonal form of \(A\) is \[D=B^{-1}DB\]

where \(D\) is the diagonal matrix of eigen values of \(A\), \(B\) is the modal matrix, which contains eigen values of as columns.

When \(A\) is diagonalizable

The matrx \(A\) is diagonalizable only if Algebraic Multiplicity (A.M) and Geometric Multiplicity (G.M) of each eigen value \(\lambda\) of \(A\) are same.

Algebraic Multiplicity: Number of times the eigen value \(\lambda\) repeats in the solution of the characteristic polynomial of \(A\).

Geometric Multiplicity: Number of vectors in the eigen space of \(\lambda\).

Problems

Diagonalize the following matrices if possible.

  1. \(A=\begin{bmatrix} 3&-2&0\\ -2&3&0\\ 0&0&5\end{bmatrix}\).

  2. \(A=\begin{bmatrix}2&1&-1\\ 1&1&-2\\ -1&-2&1\end{bmatrix}\).

  3. \(A=\begin{bmatrix}-2& 2&-3\\ 2&1&-6\\ -1&-2&0\end{bmatrix}\).

  4. \(A=\begin{bmatrix} 2&1&0\\ 0&1&-1\\ 0&2&4\end{bmatrix}\).

  5. \(A=\begin{bmatrix}-3&-7&-5\\ 2&4&3\\ 1&2&2\end{bmatrix}\).

  6. If \(2\) is an eigen value of \(\begin{bmatrix}3 &-1&1\\ -1&5&-1\\ 1&-1&3\end{bmatrix}\), without using its characteristic polynomial, find the other eigen values. Also find the eigen values of \(A^3,A^T,A^{-1}, 5A, A-3I\) and \(\text{adj}(A)\).

  7. Examine whether \(A=\begin{bmatrix}7&-1\\4 &3\end{bmatrix}\) is diagonalizable or not.

  8. Diagonalize \(A=\begin{bmatrix}2&0&1\\ 0&2&0\\ 1&0&2\end{bmatrix}\).

  9. Examine whether the matrix \(A=\begin{bmatrix}1&-3&3\\ 0&-5&6\\ 0&-3&4\end{bmatrix}\) is diagonalizable. If yes find the diagonal form.

  10. Examine whether \(A=\begin{bmatrix}1&2&2\\ 0&2&1\\ -1&2&2\end{bmatrix}\) is diagonalizable or not. If so diagonalize it.