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Given a matrix $A_1$ as part of the equation $A\vec{x}=\vec{b}$:

$$ \begin{bmatrix} P & f & f & f\\ 0 & P & f & f\\ 0 & 0 & P & f \end{bmatrix} $$

What do we know based on the fact that there is a pivot in every row?


Given a matrix $A_2$ as part of the equation $A\vec{x}=\vec{b}$:

$$ \begin{bmatrix} P & f & f\\ 0 & P & f\\ 0 & 0 & P\\ 0 & 0 & 0 \end{bmatrix} $$

What do we know based on the fact that there is a pivot in every column?


My understanding is that a pivot in every row (as in $A_1$) tells us that the columns of $A_1$ span $\mathbb{R}^m$. And that a pivot in every column (as in $A_2$) tells us that the columns are linearly independent. Are these understandings correct?

I'm sure that we know a lot about a matrix given the conditions listed above, but I'm just looking for the most obvious or helpful information.

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1 Answer 1

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A pivot in every row means that the linear system $Ax=b$ has at least one solution, for every $b$.

If every column has a pivot, then the linear system $Ax=b$ has at most one solution.

If both hold (which can happen only if $A$ is a square matrix), we get that the system $Ax=b$ has unique solution for every $b$.

A pivot in every row is equivalent to $A$ having a right inverse, and equivalent to the columns of $A$ spanning $\mathbb{R}^m$ ($m$ is the number of rows).

A pivot in every column is equivalent to $A$ having a left inverse, and equivalent to the columns of $A$ being linearly independent.

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  • $\begingroup$ This is exactly what I was looking for. Thank you. $\endgroup$ Oct 30, 2018 at 15:23
  • $\begingroup$ why does having a pivot in every row necessarily mean Ax=b has at least one solution? Even if there weren't pivots in every row, couldn't we still have solutions—for eg, if A = [4 5 6 ; 0 0 0] and b= [5 ; 0] then we have 4*x_1 + 5*x_2 + 6*x_3 = 5, which does give at least one solution (x_2 and x_3 in particular are free variables), but A doesn't have a pivot in every row. $\endgroup$
    – space
    Jun 29, 2021 at 13:20
  • $\begingroup$ also, regarding your fourth point, you mean col(A) won't be necessarily independent, right? $\endgroup$
    – space
    Jun 29, 2021 at 13:40
  • $\begingroup$ @space What do you mean by $\operatorname{col}(A)$ being independent? $\endgroup$
    – egreg
    Jun 29, 2021 at 13:44
  • $\begingroup$ @egreg—I meant the columns of A being independent. $\endgroup$
    – space
    Jun 30, 2021 at 15:27

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