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Hey we just learned about eigen vectors and i'm not really what it is and how to work with it I know how to this thing with lambda for the diagonals. $$\begin{bmatrix} -8&33&38&173&-30 \\ 0&0&-1&-4&0 \\ 0&0&-5&-25&1 \\0&0&1&5&0 \\ 4&-16&-19&-86&15 \\ \end{bmatrix}$$

1) For this 5x5 vector we were asked to find the "characteristic polynomial of R" and hence find the eigenvalues of the system. (What is the character polynomial is that the thing you get using the lambda method?).

2) For each of the eigenvalues find how many linearly independent eigenvectors can be found.

3) is it possible to make it look like this: "here are eigenvalues: a, b, c and eigenvectors: $v_{1}= \begin{bmatrix} d \\ e \\ f \end{bmatrix} v_{2}= \begin{bmatrix} g\\ h \\i \end{bmatrix} v_{3}= \begin{bmatrix} j\\ k \\l \end{bmatrix}$"

Q1) so i plugged it in wolfram...

Characteristi polyonomial = $-x^5+7 x^4-19 x^3+25 x^2-16 x+4$

Eigen values $\lambda_{1}=2$ $\lambda_{2}=2$ $\lambda_{3}=1$ $\lambda_{4}=1$ $\lambda_{5}=1$

Eigen vectors

$v_1 = (-31, -2, -12, 4, 16)$

$v_2 = (0, 0, 0, 0, 0)$

$v_3 = (-1, 0, -4, 1, 1)$

$v_3 = (-1, 0, -4, 1, 1)$

$v_4 = (0, 0, 0, 0, 0)$

$v_5 = (0, 0, 0, 0, 0)$

q2) I get lost from. How would i check for linear independence?

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  • 2
    $\begingroup$ You cannot have zero eigenvectors, so you have a deficient matrix. You must find three generalized eigenvectors. Here is how you can check linear independence. $\endgroup$ – Amzoti Mar 19 '15 at 4:36
  • $\begingroup$ I really just need help with 3... I'm saying it is not because determinant=0 and vectors algebraic multiples which make them dependent on other vectors. $\endgroup$ – Ivan Mar 19 '15 at 20:43

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