Given an $n \times n$ matrix $A$, if $Ax = x$ for all $x \in \Bbb R^n$, prove that $A$ is the identity matrix. How can I prove that this statement is true? I found this in an old textbook I was flipping through and was wondering how I could construct a proof for it.

Let $B$ be any $n\times n$ matrix, and let $b_1,\dots,b_n$ be the columns of $A$. By hypothesis $Ab_k=b_k$ for $k=1,\dots,n$, and it follows immediately that $AB=B$. (Technically you also have to show that a left identity in the ring of $n\times n$ matrices is a two-sided identity.)
Added: The key point is that if $A$ and $B$ are any $n\times n$ matrices, the $k$-th column of $AB$ is $Ab_k$, where $b_k$ is the $k$-th column of $B$. For example, let
$$A=\pmatrix{0&1&2\\3&-1&1\\1&1&2}$$ and $$B=\pmatrix{1&2&3\\2&1&4\\0&2&1}\;.$$
Then $$AB=\pmatrix{2&5&6\\1&7&6\\3&7&9}\;,\tag{1}$$ and for example
$$A\pmatrix{2\\1\\2}=\pmatrix{5\\7\\7}\;.\tag{2}$$
The calculation in $(2)$ is identical to the part of the calculation in $(1)$ that produces the second column of $AB$.
