Question

Determine the Eigen Values and Eigen Vectors of the matrix $A = \begin{bmatrix} 3 & 10 & 5 \\ -2 & -3 & -4 \\ 3 & 5 & 7 \end{bmatrix}$

Answer 1

Eigenvalues: λ=2 (multiplicity 2), λ=3. Eigenvectors: For λ=3, x_1 = \begin{bmatrix} 1 \ 1 \ -2 \end{bmatrix}. For λ=2, x_2 = \begin{bmatrix} 5 \ 2 \ -5 \end{bmatrix}.

Answer 2

To determine the eigenvalues and eigenvectors of the matrix $A = \begin{bmatrix} 3 & 10 & 5 \\ -2 & -3 & -4 \\ 3 & 5 & 7 \end{bmatrix}$, we follow these steps:

1. Determine Eigenvalues: The eigenvalues $\lambda$ are found by solving the characteristic equation $det(A - \lambda I) = 0$. $A - \lambda I = \begin{bmatrix} 3-\lambda & 10 & 5 \\ -2 & -3-\lambda & -4 \\ 3 & 5 & 7-\lambda \end{bmatrix}$

Calculate the determinant: $det(A - \lambda I) = (3-\lambda)[(-3-\lambda)(7-\lambda) - (5)(-4)] - 10[(-2)(7-\lambda) - (3)(-4)] + 5[(-2)(5) - (3)(-3-\lambda)]$ $= (3-\lambda)[\lambda^2 - 4\lambda - 21 + 20] - 10[-14 + 2\lambda + 12] + 5[-10 + 9 + 3\lambda]$ $= (3-\lambda)[\lambda^2 - 4\lambda - 1] - 10[2\lambda - 2] + 5[3\lambda - 1]$ $= (3\lambda^2 - 12\lambda - 3 - \lambda^3 + 4\lambda^2 + \lambda) - (20\lambda - 20) + (15\lambda - 5)$ $= -\lambda^3 + (3+4)\lambda^2 + (-12+1-20+15)\lambda + (-3+20-5)$ $= -\lambda^3 + 7\lambda^2 - 16\lambda + 12$

Set the characteristic polynomial to zero: $-\lambda^3 + 7\lambda^2 - 16\lambda + 12 = 0$ $\lambda^3 - 7\lambda^2 + 16\lambda - 12 = 0$

By inspection, test integer factors of 12: For $\lambda = 2$: $2^3 - 7(2^2) + 16(2) - 12 = 8 - 28 + 32 - 12 = 40 - 40 = 0$. So $\lambda = 2$ is an eigenvalue. Divide the polynomial by $(\lambda - 2)$ using synthetic division:

2 | 1  -7   16  -12
  |    2  -10   12
  ------------------
    1  -5    6    0

The depressed polynomial is $\lambda^2 - 5\lambda + 6 = 0$. Factoring this quadratic equation: $(\lambda - 2)(\lambda - 3) = 0$. So, the roots are $\lambda = 2$ and $\lambda = 3$.

Thus, the eigenvalues of the matrix A are $\lambda_1 = 2$ (with algebraic multiplicity 2) and $\lambda_2 = 3$.

2. Determine Eigenvectors: For each eigenvalue $\lambda$, we solve the equation $(A - \lambda I)x = 0$ for the eigenvector $x$.

Case 1: For $\lambda = 3$ Substitute $\lambda = 3$ into $(A - \lambda I)x = 0$: $\begin{bmatrix} 3-3 & 10 & 5 \\ -2 & -3-3 & -4 \\ 3 & 5 & 7-3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ $\begin{bmatrix} 0 & 10 & 5 \\ -2 & -6 & -4 \\ 3 & 5 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

From the first row: $10x_2 + 5x_3 = 0 \implies 2x_2 + x_3 = 0 \implies x_3 = -2x_2$. From the second row: $-2x_1 - 6x_2 - 4x_3 = 0 \implies x_1 + 3x_2 + 2x_3 = 0$. Substitute $x_3 = -2x_2$ into the second equation: $x_1 + 3x_2 + 2(-2x_2) = 0$ $x_1 + 3x_2 - 4x_2 = 0$ $x_1 - x_2 = 0 \implies x_1 = x_2$.

Let $x_2 = k$ (where $k$ is a non-zero scalar). Then $x_1 = k$ and $x_3 = -2k$. The eigenvector for $\lambda = 3$ is $x = k \begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix}$. Choosing $k=1$, a valid eigenvector is $\begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix}$.

Case 2: For $\lambda = 2$ Substitute $\lambda = 2$ into $(A - \lambda I)x = 0$: $\begin{bmatrix} 3-2 & 10 & 5 \\ -2 & -3-2 & -4 \\ 3 & 5 & 7-2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ $\begin{bmatrix} 1 & 10 & 5 \\ -2 & -5 & -4 \\ 3 & 5 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Perform row operations to simplify the matrix: $R_2 \to R_2 + 2R_1$: $\begin{bmatrix} 1 & 10 & 5 \\ 0 & 15 & 6 \\ 3 & 5 & 5 \end{bmatrix}$ $R_3 \to R_3 - 3R_1$: $\begin{bmatrix} 1 & 10 & 5 \\ 0 & 15 & 6 \\ 0 & -25 & -10 \end{bmatrix}$

From the second row: $15x_2 + 6x_3 = 0 \implies 5x_2 + 2x_3 = 0 \implies x_3 = -\frac{5}{2}x_2$. From the first row: $x_1 + 10x_2 + 5x_3 = 0$. Substitute $x_3 = -\frac{5}{2}x_2$ into the first equation: $x_1 + 10x_2 + 5(-\frac{5}{2}x_2) = 0$ $x_1 + 10x_2 - \frac{25}{2}x_2 = 0$ $x_1 + (\frac{20-25}{2})x_2 = 0$ $x_1 - \frac{5}{2}x_2 = 0 \implies x_1 = \frac{5}{2}x_2$.

Let $x_2 = 2k'$ (where $k'$ is a non-zero scalar, chosen to avoid fractions). Then $x_1 = \frac{5}{2}(2k') = 5k'$ and $x_3 = -\frac{5}{2}(2k') = -5k'$. The eigenvector for $\lambda = 2$ is $x = k' \begin{bmatrix} 5 \\ 2 \\ -5 \end{bmatrix}$. Choosing $k'=1$, a valid eigenvector is $\begin{bmatrix} 5 \\ 2 \\ -5 \end{bmatrix}$.

Note: The algebraic multiplicity of $\lambda = 2$ is 2, but its geometric multiplicity (number of linearly independent eigenvectors) is 1. This means the matrix is not diagonalizable.