Finding eigenvector of a non-integer eigenvalue I have a quick question I want to find the eigenvectors of the following $2 \times 2$ Matrix:
$$A = \begin{pmatrix}1& 1.5\\ 1.5 &9\end{pmatrix}$$
I have found the eigenvalues, $\lambda_1 = 9.27$  and $\lambda_2 = 0.73$ (exact $\lambda_{1,2} = 5 \pm \sqrt{18.25})$
However, how do I find the eigenvectors with length 1 belonging to the eigenvalues if I can't use a program/programmed calculator, but have to do it by row echelon?
Thanks in advance!
 A: Using SymPy Live:
>>> A = Matrix([[1, 1.5], [1.5, 9]])
>>> A
[ 1   1.5]
[        ]
[1.5   9 ]

Computing the characteristic polynomial and the eigenvalues:
>>> A.charpoly()
PurePoly(1.0*_lambda**2 - 10.0*_lambda + 6.75, _lambda, domain='RR')
>>> A.eigenvals()
     ____           ____        
   \/ 73          \/ 73         
{- ------ + 5: 1, ------ + 5: 1}
     2              2           

Plugging one of the eigenvalues in $\lambda \mathrm I_2 - \mathrm A$:
>>> ((5 - sqrt(73)/2) * eye(2) - A)
[    ____                  ]
[  \/ 73                   ]
[- ------ + 4      -1.5    ]
[    2                     ]
[                          ]
[                  ____    ]
[                \/ 73     ]
[    -1.5      - ------ - 4]
[                  2       ]
>>> ((5 - sqrt(73)/2) * eye(2) - A).det()
0

Hence,
$$\left( 4 - \frac{\sqrt{73}}{2} \right) x - \frac 32 y = 0$$
Choosing $y = 1$, we obtain
$$x = \frac{3}{8 - \sqrt{73}} = \cdots = -\frac{8+\sqrt{73}}{3}$$
Verifying,
>>> float(-(8 + sqrt(73)) / 3)
-5.51466791511
>>> A.eigenvects()
[(0.727998127341234, 1, [[-5.51466791510584]]), (9.27200187265877, 1, [[0.18133458177251]])]
                         [                 ]                           [                ]   
                         [       1.0       ]                           [      1.0       ]   

It is correct. The other eigenvector can be found in a similar manner.
