# Matrix Vector Spaces

1. Apr 29, 2012

### venom192

1. The problem statement, all variables and given/known data

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

2. Relevant equations

No relevant equations.

3. 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.

2. Apr 29, 2012

### micromass

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

3. Apr 30, 2012

### 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, ... vn} and replace vk with c1v1 + ... + cnvn, you still get a basis, as long as ck is not 0. That is easy to prove, if needed.
Take some invertible matrix m1 = c1v1 + ... + cnvn 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 mn = mn-1 + anvn, gives an invertible basis {m1, ..., mn}

4. Apr 30, 2012

### micromass

Staff Emeritus
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.

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