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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!

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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.

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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.

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