12
$\begingroup$

I put
{TensorProduct[{1, 0}, {0, 1}, {0, 1}, {0, 1}, {0, 1}]}/Sqrt[2]//MatrixForm and I got $$ \left( \begin{array}{cc} \left( \begin{array}{cc} \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right) & \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right) \\ \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right) & \left( \begin{array}{cc} 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} \\ \end{array} \right) \\ \end{array} \right) & \left( \begin{array}{cc} \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right) & \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right) \\ \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right) & \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right) \\ \end{array} \right) \\ \end{array} \right) $$ as a result. I now would like to rewrite this as $$ \left( \begin{array}{cc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 \end{array} \right) $$ to calculate eigenvalues of this matrix above.

Could you tell me how?

Improve this question
$\endgroup$

4 Answers 4

Reset to default
10
$\begingroup$ 
mat = {TensorProduct[{1, 0}, {0, 1}, {0, 1}, {0, 1}, {0, 1}]/Sqrt[2]};

FixedPoint[ArrayFlatten, mat] // MatrixForm

$\left( \begin{array}{cccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 \\ \end{array} \right)$

Improve this answer
$\endgroup$
8
$\begingroup$ 
mat = {TensorProduct[{1, 0}, {0, 1}, {0, 1}, {0, 1}, {0, 1}]/ Sqrt[2]};

ArrayFlatten[ArrayFlatten /@ mat] // MatrixForm

$\left( \begin{array}{cccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 \\ \end{array} \right)$

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Yes, it was a typo. Thanks for catching.. $\endgroup$
    – OkkesDulgerci
    Jul 18, 2020 at 3:09
7
$\begingroup$ 
X = {TensorProduct[{1, 0}, {0, 1}, {0, 1}, {0, 1}, {0, 1}]/Sqrt[2]};

Flatten[X, {{1, 3, 5}, {2, 4, 6}}]

(*    {{0, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 1/Sqrt[2], 0, 0, 0, 0}}    *)
Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Side note: to understand this syntax of Flatten, one may read this answer. $\endgroup$
    – xzczd
    Jul 17, 2020 at 12:12
5
$\begingroup$

Another option

(m = {TensorProduct[{1, 0}, {0, 1}, {0, 1}, {0, 1}, {0, 1}]}/Sqrt[2]) // MatrixForm

Mathematica graphics

And now

m1 = ArrayFlatten[m[[1, 1]], 2]
m2 = ArrayFlatten[m[[1, 2]], 2]
Join[m1, m2, 2] // MatrixForm

Mathematica graphics

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.