Question

If Ax =tx
Where t = 2 2 -2
1 3 1
1 2 1
Determine the eigen values of the matrix A, and an eigen vector corresponding to each eigen value. If t = 4

Answer 1

Answer on Question #56218 – Math – Linear Algebra

Question

If Ax = tx

Where t = 2 2 - 2

1 3 1

1 2 1

Determine the eigen values of the matrix A, and an eigen vector corresponding to each eigen value.

If t = 4

Solution

Given a $3 \times 3$ matrix $A$, its eigenvector is a nonzero vector $x \in \mathbb{R}^3$ such that there exists $\lambda \in \mathbb{R}$ with the property $Ax = \lambda x$, which means that when we act on $x$ by the linear operator determined by $A$, the image is a vector parallel to $x$. The scalars $\lambda$ for which there exist $x \in \mathbb{R}^3$ that satisfy the equation $Ax = \lambda x$ are called eigenvalues of $A$.

To find eigenvalues of $A$, one has to solve the equation

$$\det(A - \lambda E) = 0 \text{ for } \lambda, \text{ where } E = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$

In our case, if $A = \begin{pmatrix} 2 & 2 & -2 \\ 1 & 3 & 1 \\ 1 & 2 & 1 \end{pmatrix}$, the equation takes the form

$$\left| \begin{array}{ccc} 2 - \lambda & 2 & -2 \\ 1 & 3 - \lambda & 1 \\ 1 & 2 & 1 - \lambda \end{array} \right| = 0,$$

or, after the expansion,

$$- \lambda^ {3} + 6 \lambda^ {2} + 9 \lambda + 4 = 0.$$

First look for rational roots among numbers $\{-1,1,2,-1,4,-4\}$ (recall that if $x_0 = \frac{m}{n}$ is a rational root of the polynomial $a_n x^n + \cdots + a_0$ then $m$ divides $a_0$ and $n$ divides $a_n$).

We see that $\lambda = 4$ and $\lambda = 1$ satisfy the equation.

In order to find the last root, divide $-\lambda^3 + 6\lambda^2 + 9\lambda + 4$ by $(\lambda - 4)(\lambda - 1) = \lambda^2 - 5\lambda + 4$. We obtain that

$$- \lambda^ {3} + 6 \lambda^ {2} + 9 \lambda + 4 = (\lambda - 4) (\lambda - 1) (\lambda - 1),$$

hence the eigenvalues are $\lambda_1 = 4$ and $\lambda_2 = 1$ (the last root is of multiplicity 2).

To find eigenvalues corresponding to $\lambda_1 = 4$, solve the matrix equation

$$A x = 4 x \text{ or } \left( \begin{array}{ccc} -2 & 2 & -2 \\ 1 & -1 & 1 \\ 1 & 2 & -3 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \text{ for } x = \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right),$$

an equivalent system with triangular matrix is $\left( \begin{array}{ccc}1 & -1 & 1\\ 0 & 3 & -4\\ 0 & 0 & 0 \end{array} \right)\left( \begin{array}{c}x_{1}\\ x_{2}\\ x_{3} \end{array} \right) = \left( \begin{array}{c}0\\ 0\\ 0 \end{array} \right),$

from where

$$x_1 - x_2 + x_3 = 0$$ $$3 x_2 = 4 x_3$$

If $3x_2 = 4x_3$, then $x_2 = \frac{4}{3} x_3$.

Next,

if $x_1 - x_2 + x_3 = 0$, then $x_1 = x_2 - x_3 = \frac{4}{3} x_3 - x_3 = \frac{1}{3} x_3$.

We get the general solution (the linear subspace of eigenvectors corresponding to $\lambda_1 = 4$) of the form $\left(\frac{1}{3} x_3, \frac{4}{3} x_3, x_3\right), x_3 \in \mathbb{R}$. Set $x_3 = 3$ to obtain vector $a = (1,4,3)$.

Similarly, for $\lambda_2 = 1$ we solve the system

$$\left( \begin{array}{ccc} 1 & 2 & -2 \\ 1 & 2 & 1 \\ 1 & 2 & 0 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right),$$

which is equivalent to

$$\left( \begin{array}{ccc} 1 & 2 & -2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right),$$

from where

$$x_1 + 2 x_2 - 2 x_3 = 0$$ $$x_3 = 0$$

If $x_3 = 0$, then $x_1 + 2x_2 - 2x_3 = 0$ gives $x_1 + 2x_2 = 0$, hence $x_1 = -2x_2$.

The general solution is $(-2x_2, x_2, 0), x_2 \in \mathbb{R}$. Set $x_2 = 1$ to obtain vector $b = (-2,1,0)$.

Answer: $\lambda_1 = 4$ $a = (1,4,3)$, $\lambda_2 = 1$, $b = (-2,1,0)$.

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