Actually matrix $A$:

$$A = \begin{pmatrix} 2&6&-15\\1&1&-5\\1&2&-6 \end{pmatrix}$$

So eigenvalue:

$$\det|A-\lambda I| = \begin{vmatrix} 2-\lambda&6&-15\\1&1-\lambda&-5\\1&2&-6-\lambda \end{vmatrix} = -\lambda^3-3\lambda^2-3\lambda-1 = -(\lambda+1)^3 \tag{1}$$

Solving cubic equation we have $\lambda_1= -1$

Eigenvalue has been found.

Next step is root subspace:

First of all substitute $\lambda_1$ into the original matrix $A$ we have:

$$A_{\varphi} = \begin{pmatrix} 3&6&-15\\1&2&-5\\1&2&-5 \end{pmatrix}$$

Now as $(1)$ has multiplicity $3$ we have to find $(A_{\varphi})^3$ therefore we have:

$$(A_{\varphi})^3 = \begin{pmatrix} 3&6&-15\\1&2&-5\\\color{red}1&\color{red}2&\color{red}-\color{red}5 \end{pmatrix}^3$$

But as we have two the same lines we finally got non-square matrix, and as far as I remember it is not possible to raise it to the thrid power, so how should I proceed?

And how would I proceed if only we had square matrix? (In theory).

  • 1
    $\begingroup$ What is the rank of your $A_{\varphi}$ matrix ? $\endgroup$ – Widawensen Jun 27 '17 at 15:25
  • $\begingroup$ @Widawensen, exactly it equals $2$ $\endgroup$ – M.Mass Jun 27 '17 at 15:26
  • $\begingroup$ No, the rank is 1. 2 and 3 columns are the first column multiplied by 2 and - 5. $\endgroup$ – Widawensen Jun 27 '17 at 15:27
  • $\begingroup$ @Widawensen, ah indeed, I was looking at only lines, not columns $\endgroup$ – M.Mass Jun 27 '17 at 15:29
  • $\begingroup$ @Widawensen, but how should I interpretate the answer then? what will be root subspace? $\endgroup$ – M.Mass Jun 27 '17 at 15:33

Root subspace for $\lambda=-1$

$=\{(x,y,z)\in R^3 \;\;: x+2y-5z=0\}.$ the three equations reduce to one.

By Cayley Hamilton Theorem $$(A_{\varphi})^3=(A+I)^3=0$$

  • $\begingroup$ And if matrix were square I had to solve system like this? $A^3\cdot x = 0$ $\endgroup$ – M.Mass Jun 27 '17 at 15:47
  • $\begingroup$ Hmm, then I have only one more question, is there even any difference between tasks type "find root subspace" and "find eigenvector(s)" ? We perform nearly the same steps for each type, I am a bit confised. $\endgroup$ – M.Mass Jun 27 '17 at 15:54
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    $\begingroup$ @M.Mass the root subspace of an eigenvalue lambda is SPAN of eigenvectors X satisfying AX=lambda.X $\endgroup$ – hamam_Abdallah Jun 27 '17 at 15:59

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