How to find the rank of a matrix using row echelon form?

**Practical Definition:** The rank of a matrix is the **number of pivots** in its row echelon form. It equals the number of linearly independent rows (or columns). **Answering your questions:** 1. Row echelon form and RREF give the **same** rank, because the number of non-zero rows is invariant under further row operations. Either form works for computing rank. 2. The zero matrix has rank $0$. An $n \times n$ identity matrix has rank $n$. **3. Computing rank of your matrix:** $$A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix}$$ Row reduce: $R_2 \rightarrow R_2 - 2R_1$: $$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 3 & 6 & 9 \end{pmatrix}$$ $R_3 \rightarrow R_3 - 3R_1$: $$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ There is only **one** non-zero row, so $\text{rank}(A) = 1$. Notice that all rows are multiples of $(1, 2, 3)$, so the row space is one-dimensional. **4. Full rank:** An $m \times n$ matrix has **full rank** if $\text{rank}(A) = \min(m, n)$. - A square $n \times n$ matrix has full rank iff it is invertible ($\det(A) \ eq 0$) - For a rectangular matrix, full row rank means $\text{rank} = m$, full column rank means $\text{rank} = n$ **Rank-Nullity Theorem:** For an $m \times n$ matrix $A$: $$\text{rank}(A) + \text{nullity}(A) = n$$ where nullity is the dimension of the null space (solutions to $A\mathbf{x} = \mathbf{0}$).