# Eigenvector Anomaly

I'm trying to compute the eigenvectors for:

$$M = \left( \begin{array}{ccc} 1 & 4 \\ 4 & 100 \end{array} \right)$$

Both myself and Mathematica report the eigenvalues as:

$$\lambda_1 = \frac{1}{2} \left(101+\sqrt{9865}\right) \approx 100.161 \\ \lambda_2 = \frac{1}{2} \left(101-\sqrt{9865}\right) \approx 0.838647$$

But when I ask for eigenvectors, the answer changes depending on whether I input the numbers as integers or floating-point numbers.

When I ask for Eigenvectors[{{1, 4}, {4, 100}}] I get:

$$v_1 = \left(\frac{1}{8} \left(-99+\sqrt{9865}\right), 1\right) \approx (0.0403383, 1) \\ v_2 = \left(\frac{1}{8} \left(-99-\sqrt{9865}\right), 1\right) \approx (-24.7903, 1)$$

When I ask for Eigenvectors[{{1.0, 4.0}, {4.0, 100.0}}] I get:

$$v_1 \approx (0.0403055, 0.999187) \\ v_2 \approx (-0.999187, 0.0403055)$$

When I calculate by hand, I get a solution which matches the first query.

So, am I going crazy and overlooking some important maths (not unlikely, very tired...) or is this a bug?

Possible related issues?:

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1) We do not discuss Wolfram|Alpha questions/issues here. They're explicitly off-topic. 2) The bugs should be used only if the behaviour has been confirmed by the community as anomalous and a bug. –  rm -rf Mar 21 at 22:31
Normalise/@... on exact case, will get you same results as inexact case. –  rasher Mar 21 at 22:35
This is what Mathematica does. I don't see Alpha referenced int the question. –  Mark Adler Mar 21 at 22:35
This question appears to be off-topic because it appears to be concerned with the behavior of Wolfram|Alpha. –  m_goldberg Mar 21 at 23:21

Eigenvectors for inexact arguments are normalized:

Eigenvectors[{{1, 4}, {4, 100}}]
% // N
Normalize /@ %% // N
Eigenvectors[{{1.0, 4.0}, {4.0, 100.0}}]

(*
{{1/8 (-99+Sqrt[9865]),1},{1/8 (-99-Sqrt[9865]),1}}

{{0.0403383,1.},{-24.7903,1.}}

{{0.0403055,0.999187},{-0.999187,0.0403055}}

{{0.0403055,0.999187},{-0.999187,0.0403055}}
*)

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Eigenvectors are just a direction, so the ones you have are equivalent up to a multiplicative factor. Both answers are correct. The second one is normalized to length 1.

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