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How can I prove the following statement?

Every $m \times n$ matrix of rank $r$ reduces to $(m \times r)$ times $(r \times n)$:

$A = $ (pivot columns of $A$) (first $r$ rows of $R$) = (COL)(ROW).

[ Source: Gilbert Strang, Introduction to Linear Algebra, question $56$, section $3.2$. ]

I think that the $R$ is reduced row echelon form. I believe there is a brief elegant proof.

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  • $\begingroup$ How do you define the rank of a matrix? $\endgroup$ – Brian May 5 at 12:44
  • $\begingroup$ @Brian The dimension of column space. $\endgroup$ – Tengerye May 6 at 1:32
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Let $\{v_1,\dotsc, v_r\}$ be a basis of $\operatorname{Col}(A)$ where $A$ is $m\times n$. Put these basis vectors into the columns of a matrix $V$, so that $$ V=\left[\begin{array}{ccc}v_1 & \cdots & v_r\end{array}\right] $$ Then, for every $b\in\operatorname{Col}(A)$ there exists a $x_b\in\Bbb R^r$ such that $Vx_b=b$.

Now, let $\{a_1,\dotsc, a_n\}$ be the columns of $A$ so that $$ A=\left[\begin{array}{ccc}a_1 & \cdots & a_r\end{array}\right] $$ Each column $a_k$ of $A$ is in $\operatorname{Col}(A)$. So, let $$ W=\left[\begin{array}{ccc}x_{a_1} & \cdots & x_{a_r}\end{array}\right] $$ What happens when we compute $VW$?

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  • $\begingroup$ How shall we connect 𝑊 with reduced row echelon form 𝑅 please? @Brain $\endgroup$ – Tengerye May 16 at 9:24
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Suppose $\mathbf{A}=[\mathbf{p_1}, \mathbf{f_2}, \cdots, \mathbf{p_n}]$ where $\mathbf{p_i}$ is the column of $\mathbf{A}$ indexed $i$ and it is pivot column; while $\mathbf{f_i}$ is free column with similar notation. Therefore, $\mathbf{L}=[\mathbf{p_1}, \cdots, \mathbf{p_n}]$ have columns of pivot columns of $\mathbf{A}$. Let $\mathbf{R}=[\mathbf{p_1^R}, \mathbf{f_2^R}, \cdots, \mathbf{p_n^R}]$ be the reduced row echelon form of $\mathbf{A}$. $\mathbf{R^c}$ be the first $r$ rows of $\mathbf{R}$.

Matrix $\mathbf{A}$ is of $m$ by $n$ with rank $r$. Because $\mathbf{R}$ is $rref(\mathbf{A})$, the last $r$ rows of $R$ are null row vectors. Thus, $\mathbf{L}\mathbf{R^c}=[\mathbf{p_1}, \cdots, \mathbf{p_n}, \mathbf{f_2}, \cdots]\mathbf{R}$ where the last $n-r$ columns of $[\mathbf{p_1}, \cdots, \mathbf{p_n}, \mathbf{f_2}, \cdots]$ are free columns and first $n$ columns are pivot columns.

Now we need to prove $\mathbf{A}=[\mathbf{p_1}, \cdots, \mathbf{p_n}, \mathbf{f_2}, \cdots]R$. Every entry of $\mathbf{p_i^R}$ is 0 except 1 at the $i^{th}$ row, $[\mathbf{p_1}, \cdots, \mathbf{p_n}, \mathbf{f_2}, \cdots] \mathbf{p_i^R} = \mathbf{p_i}$. The pivot columns of $\mathbf{A}$ remain. For $\mathbf{f_j^R}=[c_0, \cdots, c_n, 0, \cdots]^{T}$, we have $c_0\mathbf{p_0^R}+\cdots+c_n\mathbf{p_n^R}=\mathbf{f_j^R}$. Because $\mathbf{R}$ has the same row space and nullspace of $\mathbf{A}$, $c_0\mathbf{p_0}+\cdots+c_n\mathbf{p_n}=\mathbf{f_j}$, i.e. $[\mathbf{p_1}, \cdots, \mathbf{p_n}, \mathbf{f_2}, \cdots]\mathbf{f_j^R}=\mathbf{f_j}$.


Welcome to alternative proofs.

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