Set of vectors with each subset forming a basis

Constantinos

Hey!

Let M and N be two natural numbers and N>M. I want to build a set A with N vectors of size M such that each subset S of A, where |S| = M, contains linearly independent vectors.

Another way to put it is that every S should be a basis for R^M.

Any ideas? Thanks!

Vargo

Do you want an explicit construction or a proof that such a set exists?

HallsofIvy

For example, if M= 2, you can take i= <1, 0>, j= <0, 1>, and k= i+ j= <1, 1>. Then any subset of order 2, {i, j}, {i, k}, and {j, k}, is a basis.

For M= 3, start with i= <1, 0, 0>, j=<0, 1, 0>, and k= <0, 0, 1> and add l= i+ j+ k.

Can you continue that?

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Constantinos	Set of vectors with each subset forming a basis Hey! Let M and N be two natural numbers and N>M. I want to build a set A with N vectors of size M such that each subset S of A, where \|S\| = M, contains linearly independent vectors. Another way to put it is that every S should be a basis for R^M. Any ideas? Thanks!

Vargo	Do you want an explicit construction or a proof that such a set exists?

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	For example, if M= 2, you can take i= <1, 0>, j= <0, 1>, and k= i+ j= <1, 1>. Then any subset of order 2, {i, j}, {i, k}, and {j, k}, is a basis. For M= 3, start with i= <1, 0, 0>, j=<0, 1, 0>, and k= <0, 0, 1> and add l= i+ j+ k. Can you continue that?