Question

2 1 1,1 2 1,0 0 1 find eigen values and vectors
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Answer 1

Answer:

eigen value are 1,1,3

eigen vector at 3 is 1,1,0

eigen vector at 1 is 1,-1,0 and -1,1,0

Answer 2

Answer:

Eigenvalues $\lambda_{1}=3, \lambda_{2}=1, \lambda_{3}=1$,

Eigenvectors $\left\{\left[\begin{array}{c}-1 \\ 1 \\ 0\end{array}\right],\left[\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right]\right\}$

Step-by-step explanation:

Given: Matrix

To find: Eigen values and vectors

Solution:

Start from forming a new matrix by subtracting $\lambda$ from the diagonal entries of the given matrix:

$\left[\begin{array}{ccc}2-\lambda & 1 & 1 \\1 & 2-\lambda & 1 \\0 & 0 & 1-\lambda\end{array}\right]$

The determinant of the obtained matrix is $-(\lambda-3)(\lambda-1)^{2}$(for steps, see determinant calculator).

Solve the equation $-(\lambda-3)(\lambda-1)^{2}=0$

The roots are $\lambda_{1}=3, \lambda_{2}=1, \lambda_{3}=1$ (for steps, see equation solver).

These are the eigenvalues.

Next, find the eigenvectors.

$\lambda=3$

$\left[\begin{array}{ccc}2-\lambda & 1 & 1 \\1 & 2-\lambda & 1 \\0 & 0 & 1-\lambda\end{array}\right]=\left[\begin{array}{ccc}-1 & 1 & 1 \\1 & -1 & 1 \\0 & 0 & -2\end{array}\right]$

The null space of this matrix is $\left\{\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right]\right\}$ (for steps, see null space calculator).

This is the eigenvector.

$- \lambda=1$

$\left[\begin{array}{ccc}2-\lambda & 1 & 1 \\1 & 2-\lambda & 1 \\0 & 0 & 1-\lambda\end{array}\right]=\left[\begin{array}{lll}1 & 1 & 1 \\1 & 1 & 1 \\0 & 0 & 0\end{array}\right]$

The null space of this matrix is $\left\{\left[\begin{array}{c}-1 \\ 1 \\ 0\end{array}\right],\left[\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right]\right\}$(for steps, see $\underline{\text { null space calculator) }} .$

These are the eigenvectors.