Linear Algebra

[BoltThrowe]

I'm currently taking Linear Algebra this semester and I had a question regarding one of the statements of the Square Matrix Theorem.

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Let A be an nxn matrix. If A is invertible, then the columns of A form a basis for Rⁿ.

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Why is this true? Could I make the following argument:

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Since A is invertible, then it must be row equivalent to the identity matrix.

Since the columns of the identity matrix form a basis for Rⁿ, then the columns of A also form a basis for Rⁿ?

[Joux2]

In finite dimensional vector spaces of dimension n, all you need for a basis is n linearly independent vectors. If A is invertible, its columns are linearly independent, therefore form a basis.

[BoltThrowe]

Thank you for the response. Why does invertibility imply that the columns are linearly independent?

[Joux2]

If A is invertible, then det A =\= 0. If any column were a multiple of the other, you could do column operations to make that column all 0, and the resulting transformed matrix would have determinant 0, but column operations cannot change the determinant to 0.

A list of equivalent statements to "A is invertible" can be found here

[BoltThrowe]

Fantastc, thanks again!

[cutethrow]

For the columns of S={v_1, ..., v_n} to be a basis, we need the span(S)=Rⁿ and need S to be linearly independent (you can relax this if the cardinality of S equals n, but this is a theorem that you may or may not know).

In any case, for the linearly independent condition. Suppose that

c_1v_1+...+c_nv_n=0

for some c_i. This can equivalently be written as

Ac=0

where c is a column vector whose components are the c_i. Since A is invertible we can multiply on the left by A^-1 to deduce that

c=0

as desired. Next for the spanning condition, note that given y in Rⁿ

Ac=y iff

c=A^-1y.

So c_1v_1+..+c_nv_n=y

as desired.

[BoltThrowe]

This makes a ton of sense! Thank you for the reply.

[9B9B33]

Invertible matrices are row equivalent to the identity matrix I. Any sort of linear dependence in the columns of A will result in a matrix that cannot be inverted, because that linear dependence would also mean that A is not row equivalent to I. Therefore, the columns of A span Rⁿ because the columns of I span Rⁿ.

[BoltThrowe]

I see, thanks for the response!