# Thread: Proving matrix inverse properties

1. ## Proving matrix inverse properties

Assume that A inR^(nxn) and without using the invertible matrix theorem, prove the following:

3.1.
Spanning Sets. If A is an n x n matrix and A^(-1) exists, then the columns of A span R^n.
3.2.
Pivot Structure. If A is an n x n matrix and Ax = b has a solution for each b inR^n, then A is invertible.

3.3.
Linear Independence. If the matrix A is invertible, then the columns of A^(1)are linearly independent.

I'm not sure how to do any of these. I don't know where to begin or anything. If someone could guide me through them that would be very helpful. Thank you

2. Originally Posted by SpiffyEh
Assume that A inR^(nxn) and without using the invertible matrix theorem, prove the following:

3.1.
Spanning Sets. If A is an n x n matrix and A^(-1) exists, then the columns of A span R^n.
3.2.
Pivot Structure. If A is an n x n matrix and Ax = b has a solution for each b inR^n, then A is invertible.

3.3.
Linear Independence. If the matrix A is invertible, then the columns of A^(1)are linearly independent.

I'm not sure how to do any of these. I don't know where to begin or anything. If someone could guide me through them that would be very helpful. Thank you
3.1: I suggest that you try to prove the converse: that if the columns of A do not span $\mathbb{R}^n$, then A cannot be invertible: because, in that case, there would exist a vector $y\in \mathbb{R}^n\backslash \mathrm{span}(A)$ for which there is no $x\in \mathbb{R}^n$ with $Ax = y$, which contradicts invertibility of A.

3.2 To say that Ax=b has a solution for each b in $\mathbb{R}^n$ amounts to the same thing as saying that the span of A is the whole of $\mathbb{R}^n$. So this follows directly from 3.1.

3.3. If the columns of $A^{-1}$ are not linearly independent, then the columns of $A^{-1}$ cannot span the whole of $\mathbb{R}^n$, hence by 3.1 $A^{-1}$ is not invertible, thus contradicting the invertibility of $A$ itself.