Matrix Vector Spaces: Invertible Basis?

[venom192]

Homework Statement

Does the vector space of all square matrices have a basis of invertible matrices?

Homework Equations

No relevant equations.

The Attempt at a Solution

I know that the 2x2 case has an invertible basis, but I don't know how to generalize it for the vector space of all nxn matrices. Any hint would be appreciated.

 

[micromass]

Can you find a basis of the square matrices (without the requirement that every element of the basis should be invertible)??

 

[hamsterman]

Here's what I'm thinking.
Take a basis of vector space of square matrices V, such that each matrix has a single 1 and all other elements are 0.
If you take a basis {v₁, ... v_n} and replace vk with c₁v₁ + ... + c_nv_n, you still get a basis, as long as c_k is not 0. That is easy to prove, if needed.
Take some invertible matrix m₁ = c₁v₁ + ... + c_nv_n in V, such that c₁ is not 0 and replace v₁ with it. I'm going to build an invertible matrix m₂ = m₁ + a₂v₂. Since det is a linear function, we can write it as kx + b, where x is some element of the matrix. The thing to prove is that if kx+b ≠ 0, then there exists 'a' ≠ 0 such that k(x+a) + b ≠ 0. The condition implies that k and b can't be both 0, so (due to linearity) there exists at most 1 point where k(x+a)+b = 0. We can always choose a different one, thus we can always build m₂. Likewise m_n = m_n-1 + a_nv_n, gives an invertible basis {m₁, ..., m_n}

 

[micromass]

Well first of all, the determinant is not a linear function.

But you have a nice idea. Just take a matrix with 0 everywhere but on one place a 1. This is obviously not invertible. But can you modify it so that it becomes invertible? Try to do some stuff to the diagonal.