# Find eigenvalue and root subspace of given matrix $A$

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).

• What is the rank of your $A_{\varphi}$ matrix ? – Widawensen Jun 27 '17 at 15:25
• @Widawensen, exactly it equals $2$ – M.Mass Jun 27 '17 at 15:26
• No, the rank is 1. 2 and 3 columns are the first column multiplied by 2 and - 5. – Widawensen Jun 27 '17 at 15:27
• @Widawensen, ah indeed, I was looking at only lines, not columns – M.Mass Jun 27 '17 at 15:29
• @Widawensen, but how should I interpretate the answer then? what will be root subspace? – 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 $$(A_{\varphi})^3=(A+I)^3=0$$
• And if matrix were square I had to solve system like this? $A^3\cdot x = 0$ – M.Mass Jun 27 '17 at 15:47