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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.

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up vote 4 down vote accepted

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


The calculation in $(2)$ is identical to the part of the calculation in $(1)$ that produces the second column of $AB$.

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Sorry, do you mind elaborating just a little more? – user1903336 Feb 6 '13 at 5:55
@user1903336: Just look at how matrix multiplication works: the $k$-th column of $AB$ is simply $Ab_k$. – Brian M. Scott Feb 6 '13 at 5:56
Why is B the columns of A? – user1903336 Feb 6 '13 at 5:58
@user1903336: $B$ is any $n\times n$ matrix. – Brian M. Scott Feb 6 '13 at 5:59
@user1903336: Because one can prove that the identity is unique, so any matrix that acts like the identity must be the identity. – Brian M. Scott Feb 6 '13 at 6:12

Prove that if $Ax=x$ for all $x$ then $AI=I$ where $I$ is the identity matrix.

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I can understand that, but from what the book says, $x$ is an n x 1 matrix. – user1903336 Feb 6 '13 at 6:02
@user1903336: Yes: your hypothesis is that $Ax=x$ for all $n\times 1$ matrices $x$. Gerry has in mind the same idea that I gave in more detail. – Brian M. Scott Feb 6 '13 at 6:18
I'm suggesting you can prove that if $Ax=x$ for all vectors $x$ then $AI=I$ where $I$ is the identity matrix. – Gerry Myerson Feb 6 '13 at 6:25

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.

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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.

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This is basically equivalent to two of the answers already posted. – Cameron Williams Jun 9 '14 at 4:03
No it's basically not. In that there is nothing extraneous in this answer. – Zarrax Jun 9 '14 at 4:04

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.

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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 $A=I_n$?

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