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I am trying to solve a problem related to the Eigenvalue and Eigenvector.

The $A$ ($2\times2$ matrix) is: $$A = \left(\begin{array}{cc} 44.8 & 42\\ 42 & 44.8 \end{array}\right).$$

I found the answers successfully.

The lambdas were $86.8$ and $2.8$ and the eigenvectors were $V_1 = K_1 (1, 1)$ and $V_2 = K_2 (1, -1),$ where $K_1$ and $K_2$ are constants.

I confirmed them with the results obtained using https://www.easycalculation.com/matrix/eigenvectors-eigenvalues.php.

However, I checked them with the results obtained using http://www.wolframalpha.com/widgets/view.jsp?id=3f00f874e9837b0ec850a34c85432d66.

The wolfram gave me different but unique results.

The $V_1$ is $(0.707, 0.707)$ and the $V_2$ is $(-0.707, 0.707)$.

Does anyone tell me how can I calculate in order to obtain the same results as the one in wolfram?

Thank you very much!

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  • $\begingroup$ “K1 and K2 are constants.” $\endgroup$ – amd Sep 18 '16 at 7:51
  • $\begingroup$ I know that K1 and K2 are constants and they can be assigned into different values in order to obtain the same values for both V1 and V2 as in wolfarm. However, the normalization made on the V1 = K1 (1, 1) and V2 = K2 (1, -1) can't obtain the V1 = (0.707, 0.707) and V2 = (-0.707, 0.707) (i.e. V1 = (1/sqrt(1^2+1^2), 1/sqrt(1^2+1^2)) and V2 = (1/sqrt(1^2+(-1)^2), -1/sqrt(1^2+(-1)^2))). $\endgroup$ – chrisych Sep 20 '16 at 11:29
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Any multiple of an eigenvector is also an eigenvector. Different tools have different conventions for normalising them.

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