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I'm studying for my exam of linear algebra.. I want to prove the following corollary:

Given $A \in{R^{n\times n}}$, there is a solution $x$ to $Ax = y$ for all $y$, if and only if $A$ has rank $m$ (full row rank).

I know that the rank of a matrix is the maximum number of columns (rows respectively) that are linearly independent and is defined by:

$\operatorname{Img} (A) = \operatorname{Rg} (A):= y \in{C^m}:y = Ax, x \in{C^n}$

My problem is that I can not find a way to relate the two concepts in order to reach a formal proof. Any help?

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sorry for the notation used but do not know how to express formulas in LaTeX notation in this page – franvergara66 Aug 26 '12 at 17:44
use $ to embed the inline math. Use $$ to embed the display math. – user21436 Aug 26 '12 at 17:49
And use > to quote material. See my edit. – celtschk Aug 26 '12 at 17:54
If you linear map is injective then it is onto, since you are considering maps from n-dimensional to n-dimensional spaces. If it is injective then linearly independent set is mapped to a linearly independent set also. – Sigur Aug 26 '12 at 17:56
There is a possible typo here. The matrix is $n\times n$ but the question states the full row rank $ = m$?? Is $m$ a typo? – user2468 Aug 26 '12 at 18:07
up vote 2 down vote accepted

Write the system $Ax = y$ as $$ \pmatrix{\mid & \mid & & \mid \\ a_1 & a_2 & \dots & a_n \\ \mid & \mid & & \mid } \pmatrix{x_1 \\ x_2 \\ \vdots \\ x_n} = \mathbf{y} \tag{1} $$ or $$ x_1 \pmatrix{\mid \\ a_1 \\ \mid } + x_2 \pmatrix{\mid \\ a_2 \\ \mid } + \cdots + x_n \pmatrix{\mid \\ a_n \\ \mid } = \mathbf{y} \tag{2} $$ For every $\mathbf{y} \in \Bbb{R}^n$, equation $(2)$ has a unique solution $x_1, x_2, \ldots, x_n$ iff the vectors $\{ a_1, a_2, \ldots, a_n \}$ form a basis of $\Bbb{R}^n$, which is true since $A$ is $n\times n$ and full rank.

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Remark 1: The proof depends on the definition of basis. If $\{ a_1, a_2, \ldots, a_n \}$ is a basis of $\Bbb{R}^n$ then for every vector $y \in \Bbb{R}^n$ there are unique "coordinates" for $y$ in terms of $\{ a_1, a_2, \ldots, a_n \}$. Let me know if you have questions. – user2468 Aug 26 '12 at 18:19
Remark 2: However, you will need to re-write this proof as $\Rightarrow$ and $\Leftarrow$ parts, which is easy to do, once you know that full rank $\Leftrightarrow$ linear independence $\Leftrightarrow$ basis $\Leftrightarrow$ unique coordinates. – user2468 Aug 26 '12 at 18:22

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