# How do I compute the eigenvectors and general eigenvector of this matrix for $\lambda = 3$?

Let $$A =\begin{pmatrix} 177& 548& 271& -548& -356\\ 19& 63& 14& -79& -23\\ 8& 24& 17& -20& -20\\ 42& 132& 55& -141& -76\\ 56& 176& 80& -184& -105\end{pmatrix}$$

We know that $$\lambda = 3$$ is an eigenvalue with algebraic multiplicity 4. Find the eigenvectors and generalized eigenvectors.

It was easy to find the eigenvectors. Finding the eigenvectors for $$\lambda = 3$$ involves finding the kernel of the matrix $$(A -\lambda I)$$, which involves finding the reduced-row echelon form. By gaussian elimination we get.

$$\left[\begin{matrix} 1 & 0 & 0 & 19/5 & -6 \\ 0 & 1 & 0 & -14/5 & 7/4 \\ 0 & 0 & 1 & 6/5 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}\right] \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix}\right] = \left[\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}\right].$$

The eigenvector of which is $$\left[\begin{matrix} x_1 \\ x_2 \\ \frac{14}{29}x_1 + \frac{220}{203}x_2 \\ \frac{5}{29}x_1 + \frac{120}{203}x_2 \\ \frac{8}{29}x_1 + \frac{76}{203}x_2 \end{matrix}\right] = \left[\begin{matrix} x_1 \\ 0 \\ \frac{14}{29}x_1 \\ \frac{5}{29}x_1 \\ \frac{8}{29}x_1 \end{matrix}\right] + \left[\begin{matrix} 0 \\ x_2 \\ \frac{220}{203}x_2 \\ \frac{120}{203}x_2 \\ \frac{76}{203}x_2 \end{matrix}\right].$$ However, this is not enough, so we need to find the eigenvector of $$(A - 3I)^2$$. But using gaussian elimination gives

$$\left[\begin{matrix} 1 & 8/3 & 0 & 11/3 & -4/3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}\right] \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix}\right] = \left[\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}\right].$$

Which doesn't give us the right number of generalized eigenvectors (we need two more). What went wrong, and how do I make it right?

• I see four zero rows in the reduced matrix, so its kernel is four-dimensional. – amd Apr 13 at 3:44
• @ErotemeObelus: I get the eigenvectors $(24,-7,4,0,4),(-19,14,-6,5,0)$ and the two generalized eigenvectors $(22/5,-33/20,3/5,0,0),(-8,3,-1,0,0)$. Those last two were derived using the RREF of $[A - 3I]v_3 = v_1$ and $[A - 3I]v_4 = v_2$ – Moo Apr 13 at 3:56
• @Moo Let's suppose that I'm dealing with a hypothetical matrix $C$. C has eigenvectors v and w but the eigenvalue of v, w has algebraic multiplicity three. Which vector do I use to calculate the generalized eigenvector? – ErotemeObelus Apr 13 at 4:46
• @Moo how do I determine "which one works?" – ErotemeObelus Apr 13 at 5:15
• math.upenn.edu/~moose/240S2013/slides7-31.pdf – Moo Apr 13 at 12:13