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.
 A: Prove that if $Ax=x$ for all $x$ then $AI=I$ where $I$ is the identity matrix. 
A: 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 $AB$.
A: Let $C=A-I$. Then for all $x\in \mathbb R^n$ holds that $Cx=Ax-Ix=x-x=0$. The claim will follow by showing that $C$ must be zero (since then $A-I=0$ which gives $A=I$). Assume to the contrary that $C$ is not the zero matrix, and assume its $(i,j)$ entry is non, zero: $c_{ij}\ne 0$. But direct computation shows that the $j$-th component of the vector $C\cdot e_j$ is $c_{ij}$, and the former is the zero vector. Contradiction. 
A: If $e_k$ denotes the $k$th unit coordinate vector, then for any matrix $M$, $M e_k$ gives the $k$th column of $M$. 
So in the case at hand, $A e_k = e_k$ implies that the $k$th column of $A$ is $e_k$. In other words, $A$ is the identity matrix. 
A: If you write $x$ as $n \times 1$ column matrix, there is a unique linear map $F_{A}:\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ associated with $A$ as follows: for any $x \in \mathbb{R}^{n}$, if we look at this $x$ as a column matrix, obtained $n \times 1$ column matrix $Ax$ has corresponding elements in the $n$-tuple $F_{A}(x) \in \mathbb{R}^{n}$.
What you have written [$Ax = x$ for all $x \in \mathbb{R}^{n}$] means that corresponding $F_{A}$ has to be the identity map. And this is exactly "why" $A = I$.
You have to write out the details in order to see this. Let me know if you find something wrong here.
A: Hint:consider the standard basis $\{e_1,\dots,e_n\}$ of $\mathbb{R}^n$ by the data for any vector is eigen vector with eigen value $1$ so $A(e_1+e_2+\dots+e_n)=(e_1+\dots+e_n)$ and also $Ae_1=e_1,\dots Ae_n=e_n$, can you now do by your hand that $A=I_n$?
A: I am leaving this answer here just for future reference.
As Marso suggested, consider the standard basis $\mathbf{e_1},\mathbf{e_1},...,\mathbf{e_n}\in \mathbb{R}^n$ column vectors of $\mathbb{R}^n$ .
From our hypothesis, it is true that:
$$
\label{eq:1}\tag{1}
\mathbf{Ax}=\mathbf{x},\forall \mathbf{x}\in\mathbb{R}^n
$$
Also, a basic property of the identity matrix, is that it can be expressed as:
$$\label{eq:2}\tag{2}
\mathbf{I}=\sum_{i=1}^{n}\mathbb{e}_i\mathbb{e}_i^T
$$
Hence, using that property we have:
\begin{equation}
\begin{split}
    \mathbf{I}=\sum_{i=1}^{n}\mathbb{e}_i\mathbb{e}_i^T&\Longrightarrow
\mathbf{AI}=\mathbf{A}\sum_{i=1}^{n}\mathbb{e}_i\mathbb{e}_i^T\Longrightarrow\\
&\Longrightarrow\mathbf{A}=\mathbf{A}\sum_{i=1}^{n}\mathbb{e}_i\mathbb{e}_i^T\Longrightarrow\\
&\Longrightarrow\mathbf{A}=\sum_{i=1}^{n}\mathbf{A}\mathbb{e}_i\mathbb{e}_i^T=
\sum_{i=1}^{n}(\mathbf{A}\mathbb{e}_i)\mathbb{e}_i^T\stackrel{(1)}{=}
\sum_{i=1}^{n}\mathbb{e}_i\mathbb{e}_i^T\stackrel{(2)}{=}\mathbf{I}
\end{split}
\end{equation}
