How to find the rank of a matrix using row echelon form?
I'm studying linear algebra and I need to understand how to compute the rank of a matrix. I know the definition:
The rank of a matrix is the dimension of its column space (or row space).
But practically, I've been told to convert the matrix to row echelon form and count the number of non-zero rows. However, I'm confused about:
- Does the rank depend on whether I use row echelon form or reduced row echelon form?
- What is the rank of a zero matrix? An identity matrix?
- How do I find the rank of:
- What does it mean for a matrix to have full rank?
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1 Answer
**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 $n \ imes n$ 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:
$R_2 \
ightarrow R_2 - 2R_1$:
$$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 3 & 6 & 9 \end{pmatrix}$$
$R_3 \
ightarrow R_3 - 3R_1$:
$$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
There is only **one** non-zero row, so $\ ext{rank}(A) = 1$.
Notice that all rows are multiples of $(1, 2, 3)$, so the row space is one-dimensional.
**4. Full rank:**
An $m \ imes n$ matrix has **full rank** if $\ ext{rank}(A) = \min(m, n)$.
- A square $n \ imes n$ matrix has full rank iff it is invertible ($\det(A) \
eq 0$)
- For a rectangular matrix, full row rank means $\ ext{rank} = m$, full column rank means $\ ext{rank} = n$
**Rank-Nullity Theorem:**
For an $m \ imes n$ matrix $A$:
$$\ ext{rank}(A) + \ ext{nullity}(A) = n$$
where nullity is the dimension of the null space (solutions to $A\mathbf{x} = \mathbf{0}$).
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