# Invertibility of Ax = e_i

The question above has a typo where it should be i = 1,...,n.

Why is the matrix A invertible? I understand how to calculate it's inverse by multiplying both sides by A's inverse, but don't understand why A is invertible to begin with

My initial thoughts are that Av_i equaling the standard unit vectors somehow implies that the columns are linearly independent, but I don't really know how to put it into words. Any help would be great!

I'll give two approaches: the first is a little more intuitive than the second, which is clearer to beginners.

To see that $$A$$ is invertible, we first guess that it is invertible. (This is not mathematically incorrect unless we arrive at a contradiction). Then we define a transformation $$B$$, given by $$Be_i = v_i$$, where $$v_i$$ are the vectors which exist so that $$Av_i = e_i$$. Then, extend linearly: if $$x = \sum x_ie_i$$, then $$Bx = \sum x_iv_i$$.

Now, you can check that $$AB= I$$, since $$(AB)e_i = (A)(Be_i) = Av_i= e_i$$ for all $$i = 1,2, \ldots , n$$.

Now, (this is not true for infinite dimensions!) since we are in a finite dimensional space, $$AB = I$$ implies $$BA = I$$, hence $$A$$ is invertible, with inverse $$B$$.

There's another approach, but they all boil down to the same thing: We will show that $$A$$ is surjective. Let $$x$$ be a vector, then $$x = \sum x_ie_i$$ for some scalars $$x_i$$, and hence we can see that $$A(\sum x_iv_i) = \sum x_i A(v_i) = \sum x_ie_i = x$$. So $$A$$ is surjective, hence injective (rank-nullity theorem), hence is invertible.

Note that the converse is also true : if $$A$$ is invertible, then it is surjective, so every vector, let alone unit vectors, will have a pre-image.

HINT

Remember that a matrix is invertible if and only if its kernel has dimension zero and that the dimension of the kernel is $n$ minus the dimension of the range. So if you can prove the range has dimension $n$ you are done.