Question

I am asked the question "Is it true that there is a basis for R^(3x3) consisting only of non singular matrices.

I know that in order to be a basis, the candidates must be linearly independent, and that the span of s must =V

I'm also pretty sure that since this is R(3x3), the dimension of the basis must be 9 (not completely sure of this though, if this is incorrect, please tell me!)

How would I go about answering this? I am a little stumped.

Answer 1

Also, I know that there is a basis, the standard basis vectors, that will be a basis for R(3x3), but they are all singular...

Answer 2

Yes it is true

Answer 3

Figured so, but how can we prove it?

Answer 4

It is reversible meaning that we set it up the matrices as an identity matrix

Answer 5

and that means it's a pivot point of every column

Answer 6

So the identity matrix is a basis for R(3x3), but the determinant of I is 1, so that means it is non singular.

Answer 7

But we wold need to prove that the columns of I are linearly independent, and that they span v, right?

Answer 8

Yeah but you already said "non singular matrices" which means that the matrix has an inverse

Answer 9

and invertible means that it is linearly independent...

Answer 10

I think that's one of the axioms that as long as you can prove one ( one of them being if matrix A is invertible) then all of the other ones are true by default (another one includes that the columns of matrix A are lin inde)

Answer 11

Thanks!

Answer 12

I thought about it. That can't be right. If all of the bases are I, then you can never get to an answer such as [1,1,0... ... ...] because there will always be 0's added in the position 1,3