# What does a having pivot in every row tell us? What about a pivot in every column?

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.

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.

• This is exactly what I was looking for. Thank you. Oct 30, 2018 at 15:23
• 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. Jun 29, 2021 at 13:20
• also, regarding your fourth point, you mean col(A) won't be necessarily independent, right? Jun 29, 2021 at 13:40
• @space What do you mean by $\operatorname{col}(A)$ being independent? Jun 29, 2021 at 13:44
• @egreg—I meant the columns of A being independent. Jun 30, 2021 at 15:27