I'm actually working with the Jordan Canonical form I should solve this simple equation:

$Av_2 = v_1 + 2v_2$


$(A - 2I)v_2 = v_1$

now I plug in my known matrix and vector and after subtracting the identity the system looks like that:

$\begin{bmatrix}1&4&3\\-1&-2&-1\\1&2&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}1\\-1\\1\end{bmatrix}$

how do I solve this? Can I simplify it to row echelon form?


Yes, this is just a regular linear system that can be converted to row echelon form.

Notice that the second and third rows are negative of each other.

You can perform the row operations $R_3+ R_2$ and $R_2 + R_1$ to convert it to row echelon form.

Remark: There are infinitely many solutions to the system.

  • $\begingroup$ So then the solution would be $x_1 -x_3 = 1 $ and $x_2 + x_3 =-1$ ? but how does the vector look like? $\endgroup$ – Leroy Sep 3 '17 at 4:20
  • $\begingroup$ If $x_3= t$, can you solve for $x_1$ and $x_2$ in terms of $t$? $\endgroup$ – Siong Thye Goh Sep 3 '17 at 4:21
  • $\begingroup$ $x_1 = 1 + t$ and $x_2 = -1 - t$ ? $\endgroup$ – Leroy Sep 3 '17 at 4:28
  • $\begingroup$ awesome. $(x_1, x_2, x_3) = (1, -1, 0) + t (1, -1, 1)$ $\endgroup$ – Siong Thye Goh Sep 3 '17 at 4:29
  • $\begingroup$ How can I add that to a basis of eigenvectors? $\endgroup$ – Leroy Sep 3 '17 at 4:32

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