I was wondering if its possible to know if a matrix is invertible without row reducing it in attempt to make it I. If the matrix is not nxn, it is never invertible?

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- Oct 3rd 2011, 04:25 PMKumainvertible matrix
I was wondering if its possible to know if a matrix is invertible without row reducing it in attempt to make it I. If the matrix is not nxn, it is never invertible?

- Oct 3rd 2011, 04:28 PMDrexel28Re: invertible matrix
It doesn't make sense to speak of invertibility for non $\displaystyle n\times n$ matrices. In particular, a matrix $\displaystyle A$ which is $\displaystyle n\times m$ can be thought of a linear transformation $\displaystyle A:\mathbb{R}^m\to\mathbb{R}^n$ and so if $\displaystyle A$ is invertible, then the associated map is an isomorphism, which implies that $\displaystyle n=m$.

- Oct 3rd 2011, 04:34 PMKumaRe: invertible matrix
- Oct 3rd 2011, 07:19 PMDrexel28Re: invertible matrix
Well, since you liked the last explanation, perhaps one in the same vein. Pretend for a second that a matrix $\displaystyle A$ really is just a linear transformation $\displaystyle A:\mathbb{R}^n\to\mathbb{R}^n$, then one only has to check any of the following equivalent conditions:

1) $\displaystyle A$ is injective

2) $\displaystyle A$ is surjective

3) $\displaystyle A$ does not have zero as an eigenvalue

4) $\displaystyle \det(A)\ne0$

And there are many more, some of which are very similar to the ones I listed (e.g. 1) is easily seen to be eqiuvalent to the existence of a left inverse). - Oct 3rd 2011, 08:57 PMFernandoRevillaRe: invertible matrix
For $\displaystyle A\in\mathbb{K}^{m\times n}$ we have the concepts of left and right inverse: $\displaystyle B\in \mathbb{K}^{n\times m}$ is a left inverse of $\displaystyle A $ iff $\displaystyle BA = I_n$ and a left-invertible matrix is a matrix with at least one left inverse. Similar considerations for right inverse.