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if $A=\begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix}$, I found eigen values are $\frac{5\pm \sqrt{33}}{2}$ but I tried for eigen vectors which I failed. I tried $$(A-\lambda_i I)x=0$$ then I need help. thanks

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  • $\begingroup$ Your eigenvalues are correct. But for the sake of practice, these numbers don't come out nice. Are you sure you copied the problem right? $\endgroup$
    – imranfat
    May 10, 2016 at 15:02
  • $\begingroup$ @imranfat yes .. then what to do $\endgroup$
    – my stak
    May 10, 2016 at 15:04
  • $\begingroup$ Do you know how to find the eigenvectors of a 2 by 2 when the eigenvalues happen to be integers? Because your question is just hard because of the algebra. As far as the concept is concerned, it is very straight forward... $\endgroup$
    – imranfat
    May 10, 2016 at 15:09

1 Answer 1

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The problem is probably in the algebra, I'll guide you through the calculations. You could avoid these, see below for an extra remark.

For $\lambda = \tfrac{5+\sqrt{33}}{2}$, you need to solve the homogeneous system: $$\left(\begin{array}{cc|c} 1-\tfrac{5+\sqrt{33}}{2} & 2 & 0 \\ 3 & 4-\tfrac{5+\sqrt{33}}{2} & 0 \end{array}\right) \sim \left(\begin{array}{cc|c} \tfrac{-3-\sqrt{33}}{2} & 2 & 0 \\ 3 & \tfrac{3-\sqrt{33}}{2} & 0 \end{array}\right) \sim (*)$$ Now subtract three times the first row divided by $\tfrac{-3-\sqrt{33}}{2}$ from the second row to create a $0$ at the bottom left position. After some calculations $(**)$, you'll find that the second element of this row also becomes $0$, so: $$(*) \sim \left(\begin{array}{cc|c} \tfrac{-3-\sqrt{33}}{2} & 2 & 0 \\ 0 & 0 & 0 \end{array}\right)$$ which means the eigenvector satisfies: $$\tfrac{-3-\sqrt{33}}{2} x + 2 y = 0$$ Let $y=t$, then: $$\frac{-3-\sqrt{33}}{2} x = -2t \Leftrightarrow x = \frac{4}{3+\sqrt{33}}t = \frac{\sqrt{33}-3}{6}t$$ So an eigenvector corresponding to $\lambda = \tfrac{5+\sqrt{33}}{2}$ is (take $t=1$): $$\left(\frac{\sqrt{33}-3}{6} , 1 \right)$$ A similar calculation gives you the eigenvector corresponding to $\lambda = \tfrac{5-\sqrt{33}}{2}$: $$\left(\frac{-\sqrt{33}-3}{6} , 1 \right)$$


$(**)$ $$\begin{array}{rl} \displaystyle \frac{3-\sqrt{33}}{2}-3\frac{2}{\tfrac{-3-\sqrt{33}}{2}} & \displaystyle = \frac{3-\sqrt{33}}{2}-\frac{12}{-3-\sqrt{33}} \\[7pt] & \displaystyle = \frac{3-\sqrt{33}}{2}-\frac{12\color{blue}{(-3+\sqrt{33})}}{(-3-\sqrt{33})\color{blue}{(-3+\sqrt{33})}} \\[9pt] & \displaystyle = \frac{3-\sqrt{33}}{2}-\frac{12(-3+\sqrt{33})}{-24}\\[9pt] & \displaystyle = \frac{3-\sqrt{33}}{2}+\frac{-3+\sqrt{33}}{2} = 0 \end{array}$$


Remark

If you have seen some theory about eigenvalues and eigenvectors, you may know that a 2 by 2-matrix is always diagonalizable if it has two distinct eigenvalues, which is the case here. That means that there will be exactly one eigenvector (up to scaling) per eigenvalue so you expect to get a zero row in the homogeneous system you have to solve. This means you could deduce an eigenvector from the first row straightaway, without the algebraic calculations we did above.

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