0
$\begingroup$

I'm confused by 4 in Theorem 4. What if the row has a row of all 0s? There is no pivot position there, but it would still work for echelon form, and if the augmented matrix can be put into echelon form it has a solution, but that would violate 1 in Theorem 4. Could someone clarify my current misunderstanding?

Theorem 4: Let $A$ be an m x n matrix. Then the following statements are logically equivalent. That is, for a particular $A$, either they are all true statements or they are all false.

  1. For each $b$ in $\mathbb{R}^m$,the equation $Ax=b$ has a solution.

  2. Each $b$ in $\mathbb{R}^m$ is a linear combination of the columns of $A$.

  3. The columns of $A$ span $\mathbb{R}^m$.

  4. $A$ has a pivot position in every row.

  • 0
    echelon form of augmented matrix is no guarantee for the existence of solutions. For example, $\begin{bmatrix} 0&0&0 &| & 1\end{bmatrix}$ is augmented and in echelon form but no solution. – 2017-02-20
  • 0
    Sorry, I think I was being unclear. I meant the whole row is 0: [0 0 0 | 0] – 2017-02-20
  • 0
    if the whole row is zero row, then the equation has only solution for $Ax=0$ and not for every right hand side $b$, so it does not violate the theorem. – 2017-02-20
  • 0
    What about the matrix where b = [ 1 2 0], with the last row of the agumented matrix of A and b being all 0s? – 2017-02-20
  • 0
    Yes, for that $b$ the system would have a solution. But the claim is that it has solutions **for all** $b$, not only for some suitably chosen $b$. So, what if $b=\begin{bmatrix}0&1&2\end{bmatrix}$? – 2017-02-20
  • 0
    I see how that $b$ would violate rules 1 and 2. Matrix A's last column is all 0s, so I'd see how it would violate rule 4. But how does rule 3 get into this mix? – 2017-02-20
  • 1
    If every column has a zero in its last entry, then everything in the span of the columns must have a zero as its last entry, so 3 is violated. – 2017-02-20

1 Answers 1

0

Quoting your comment above:

Sorry, I think I was being unclear. I meant the whole row is $0$: $\begin{bmatrix}0&0&0&|&0\end{bmatrix}$.

That's precisely your misunderstanding. You can't choose what's after the vertical bar in the augmented matrix, because in the statement of this theorem $A$ is treated as only the coefficient matrix of a system. And pretty much the whole idea of the theorem is to examine what may happen if we augment different right-hand side columns to this coefficient matrix, i.e. if we consider $\begin{bmatrix}A&|&b\end{bmatrix}$ for all possible column-vectors $b$. If, for example, the last row of $A$ consists of zeroes only, then the last row of such an augments matrix is $\begin{bmatrix}0&0&0&|&?\end{bmatrix}$, where the question mark can be anything — and if it's nonzero, then the system doesn't have a solution.