TITLE: Inverse of Matrix of Standard Vectors
QUESTION [1 upvotes]: Having a bit of trouble with the reasoning behind this question:
"Consider a $n\times n$ matrix $A$. Assume that for each standard vector $\vec{e}_i$, there exists another vector $\vec{v_i}$ such that $A\vec{v_i} = \vec{e_i}$ for $i = 1,\cdots,n$. Is the matrix $A$ invertible? If $A$ is invertible, find the inverse matrix $A^{-1}$."
I'm not sure how to proceed here. I know that the inverse of $I_n$ is itself, and if $A$ is $n\times n$, then $AA^{-1}=I_n$. I'm tempted to use an augmented matrix of the form $(A|I)$ to solve for $A^{-1}$ but I'm not sure exactly what that would look like.
Any suggestions? Thanks!

REPLY [1 votes]: You know that $A$ is invertible (you're given this), so you know that $A^{-1}$ exists. So we have that $A^{-1}e_i = v_i$ for each $i$. This is enough to define the matrix, since any vector $u$ can be written as $$u = \sum_{i = 1}^N u_i e_i$$ for some coefficients $(u_i)$. Thus
$$A^{-1}u = \sum_{i=1}^N u_i A^{-1}e_i = \sum_{i=1}^N u_i v_i.$$
(Sorry, I don't know how to make things bold in MathJax! $u_i \in \Bbb R$ and $v_i \in \Bbb R^n$.)
In particular, if we have any matrix $B$, then $B e_i$ is the $i$th column of $B$ (check this yourself if you're not sure. So in our case
$$ A^{-1} = ( v_1 | v_2 | ... | v_n )$$
where the bar | separates the columns. This means that $A^{-1}$ is the matrix that comprises of the columns $v_1 , ..., v_n$ in that order.

In fact, I say at the start "A is invertible (you're given this)", but actually you can see from my construction that the other hypotheses actually say this. Given the initial hypothesis, construct
$$B=(v_1|v_2|...|v_n).$$
Then you can check that $AB=I=BA.$ The easiest way to see this is that
$$ABe_i=Av_i=e_i.$$
So since we're in an $n$ dimensional space - in particular, a finite dimensional space - $B=A^{−1},$ since having a left or right inverse is equivalent to having an inverse.
Alternatively, observe that $BAv_i = Be_i = v_i$, and since the $(e_i)$ form a basis, so do the $v_i$, so we've found that $BA$ and the identity $I$ are equivalent on a basis, and hence equal.
Note that if we had some $(y_i) \subseteq \Bbb R$ such that $\sum_i y_i v_i = 0$, then we would have
$$0 = A \cdot 0 = A(\sum y_i v_i) = \sum y_i A v_i = \sum y_i e_i$$
and we have found linear dependence between the $(e_i)$ - contradiction.

In an infinite dimensional vector space, we have can have a right inverse but not a left (or vice versa). Consider this:
$$\theta : (x_1, x_2, ...) \to (0, x_1, x_2, ...),$$
$$\phi : (x_1, x_2, ...) \to (x_2, x_3, ...).$$
Clearly $\theta \circ \phi$ is the identity map ($x_i \to x_{i+1} \to x_i$), but
$$\phi \circ \theta : (x_1, x_2, x_3, ...) \to (0, x_2, x_3, ...).$$ ($\theta$ and $\phi$ can be represented by matrices as a direct extension from the finite dimensional case.)
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