Linear Algebra THEOREM 4

Question

I am a university student studying Linear Algebra. I read the book Linear Algebra by David C. Lay and wonder about d. of Theorem 4

Let $A$ be an $m \times n$ matrix. The following statements are equivalent.
a. For each $\mathbf{b}$ in $\mathbb{R}^m$, the equation $A \mathbf{x}=\mathbf{b}$ has a solution.
b. Each $\mathbf{b}$ in $\mathbb{R}^m$ is a linear combination of the columns of $A$.
c. The columns of $A$ span $\mathbb{R}^m$.
d. $A$ has a pivot position in every row.

By THEOREM $4$, $A$ has a pivot position in every row.

After reading this, I think $A$ is a $2 \times 3$ matrix like this.

$$ \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right) $$

It doesn't have a pivot position in the $2$th row.

Is the THEOREM 4 wrong?

Thanks for your help.

Sources: Lay, David C., Linear algebra and its applications. Pearson Education India, 2003.

Answer 1

I think you are misreading theorem 4... It claims that the statements are equivalent, meaning that they are all true or all false. In your example, since the matrix does not have a pivot position in every row, it follows that:

1. The system $Ax =b$ does not have a solution for every $b$;
2. There are vectors in $\mathbb{R}^m$ that are not linear combinations of the columns of $A$.
3. The columns of $A$ do not span $\mathbb{R}^m$.