given matrix is $$\begin{bmatrix} 2&1\\ 3&2\end{bmatrix}$$ The eigen vectors of this matrix are

A) $\begin{bmatrix}\sqrt 3\\3\end{bmatrix}$, $\begin{bmatrix}\sqrt 3\\-3\end{bmatrix}\qquad$ B) $\begin{bmatrix}i\sqrt 3\\3\end{bmatrix}$, $\begin{bmatrix}i\sqrt 3\\-3\end{bmatrix}\qquad $ C) $\begin{bmatrix}\sqrt 3\\3\end{bmatrix}$, $\begin{bmatrix}-\sqrt 3\\-3\end{bmatrix}\qquad $ D) $\begin{bmatrix}\sqrt 3\\1\end{bmatrix}$, $\begin{bmatrix}\sqrt 3\\-1\end{bmatrix}\qquad $

my try: assuming $\lambda$ eigen value, $$\left|\begin{matrix} 2-\lambda & 1\\ 3 & 2-\lambda \end{matrix}\right|=0\implies \lambda= 2\pm\sqrt 3$$ then eigen vectors for $\lambda=2-\sqrt3$ are
$$\begin{bmatrix} \sqrt 3 & 1\\ 3 & \sqrt 3 \end{bmatrix}\cdot \begin{bmatrix} x_1\\ x_2\end{bmatrix}=\begin{bmatrix} 0\\ 0\end{bmatrix} $$ $$x_1\sqrt3 +x_2=0\ \ or\ \ 3x_1+x_2\sqrt 3=0$$ so eigen vectors are $\begin{bmatrix}\sqrt 3\\1\end{bmatrix}$ or $\begin{bmatrix}\sqrt 3\\3\end{bmatrix}\qquad$

similarly for $\lambda =2+\sqrt 3$, eigen vectors are so vectors are $\begin{bmatrix}\sqrt 3\\-1\end{bmatrix}$ or $\begin{bmatrix}\sqrt 3\\-3\end{bmatrix}\qquad$

i know answer is option A but i don't understand the reason why don't we take $\begin{bmatrix}\sqrt 3\\1\end{bmatrix}$ & $\begin{bmatrix}\sqrt 3\\-1\end{bmatrix}$ ? please help me. thanks

  • $\begingroup$ oops, i have not shown that step. anyway i have computed eigen values so please give answer to my question $\endgroup$ – Bhaskara-III Sep 27 '16 at 19:14
  • $\begingroup$ what do you mean by RREF? $\endgroup$ – Bhaskara-III Sep 27 '16 at 19:22
  • $\begingroup$ How are you ending up with two linearly independent choices of eigenvector for each eigenvalue? Only one of them is actually a solution for the equations that you’ve set up. $\endgroup$ – amd Sep 27 '16 at 19:24
  • $\begingroup$ The vector $\begin{bmatrix}\sqrt 3\\1\end{bmatrix}$ does not solve the equations $x_1\sqrt3 +x_2=0$ or $3x_1+x_2\sqrt 3=0$. Likewise for the vector $\begin{bmatrix}\sqrt 3\\-1\end{bmatrix}$. $\endgroup$ – Did Sep 27 '16 at 19:24
  • $\begingroup$ @Did vector $\begin{bmatrix}\sqrt 3\\3\end{bmatrix}\qquad$ also doesn't satisfy $x_1\sqrt 3+x_2=0$ or $3x_1+x_2\sqrt3=0$ so why do we take vector $\begin{bmatrix}\sqrt 3\\3\end{bmatrix}\qquad$?. then we should only take vector $\begin{bmatrix}\sqrt 3\\-3\end{bmatrix}\qquad$ because it satisfies both $x_1\sqrt 3+x_2=0$ & $3x_1+x_2\sqrt3=0$ $\endgroup$ – Bhaskara-III Sep 27 '16 at 19:35

From $x_1\sqrt 3 + x_2=0$ you should get that the eigenvector is $\begin {bmatrix} 1\\-\sqrt 3 \end {bmatrix}$ Neither of the vectors you propose satisfiesthe equation. Similarly, from $-x_1\sqrt 3 + x_2=0$ you should find $\begin {bmatrix} 1\\ \sqrt 3 \end {bmatrix}$. Multiplying both by $\sqrt 3$ gives the choices in A. You should only get one eigenvalue per (nondegenerate) eigenvalue.


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