Problem with Eigenvectors

Question

When I want to calculate eigenvectors of the following matrix in Mathematica the only answer it gives me is zero vector, anybody knows how to fix this? here's my matrix : \begin{equation} X=\left(\begin{array}{cccc} 0 & 1 & 0 & 0&0&0\\ 1 & 0 & \sqrt{2} & 0&0&0\\ 0 & \sqrt{2} & 0 & \sqrt{3}&0&0\\ 0 & 0 & \sqrt{3} & 0& \sqrt{4} &0\\0&0&0&\sqrt{4} &0&\sqrt{5} \\ 0&0&0&0&\sqrt{5} &0 \end{array}\right) \end{equation} heres the code:

X = {{0, 1, 0, 0, 0, 0},
   {1, 0, Sqrt[2], 0, 0, 0},
   {0, Sqrt[2], 0, Sqrt[3], 0, 0},
   {0, 0, Sqrt[3], 0, Sqrt[4], 0},
   {0, 0, 0, Sqrt[4], 0, Sqrt[5]},
   {0, 0, 0, 0, Sqrt[5], 0}
   };
Eigenvectors[X]

The point is I don't want to find them numerically I want analytical expression for eigenvectors.

Answer 1

the only answer it gives me is zero vector

It works for me. Please post exact code you used and show the output. Not just Latex. And give which version you used. You might have made a mistake in the input.

On V12:

mat = {{0, 1, 0, 0, 0, 0},
   {1, 0, Sqrt[2], 0, 0, 0},
   {0, Sqrt[2], 0, Sqrt[3], 0, 0},
   {0, 0, Sqrt[3], 0, Sqrt[4], 0},
   {0, 0, 0, Sqrt[4], 0, Sqrt[5]},
   {0, 0, 0, 0, Sqrt[5], 0}
   };

Eigenvectors[mat];
MatrixForm[% // N]

Update

OP wants solution to be analytical solution and not numerical.

Mathematica gives answer using Roots objects.

Using

SetSystemOptions[
  "TypesetOptions" -> "NumericalApproximationForms" -> False];
Eigenvectors[mat]

Gives

TO obtain numerical values, the command N can be applied to the above.

Answer 2

Since you didn't post any code, it's hard to know what you did wrong. Please always post your code!

When in doubt if mathematica is correct about what it's doing, do it by hand..this of course really only works when the question in hand is small like yours.

matA = {{0, 1, 0, 0, 0, 0}, {1, 0, Sqrt[2], 0, 0, 0}, {0, Sqrt[2], 0, Sqrt[3], 0, 0}, {0, 0, Sqrt[3], 0, Sqrt[4], 0}, {0, 0, 0, Sqrt[4],0, Sqrt[5]}, {0, 0, 0, 0, Sqrt[5], 0}};

poly = Det[matA - IdentityMatrix[6] \[Lambda]]

$$\lambda ^6-15 \lambda ^4+45 \lambda ^2-15$$

sol = Solve[poly == 0 , \[Lambda]];

eigen = matA - IdentityMatrix[6] \[Lambda] /. sol[[1]]

Using RowReduce we can find one of our vectors.

sol2 = RowReduce[eigen]
sol2[[All, 6]] // MatrixForm

The Last column being our first vector. The last element is a zero...this should be a one...an artifact of RowReduce I'm not sure why it does that. Regardless, a lot of work when one probably just used Eigenvectors[] the function incorrectly.