Recipe

Test independence by row-reducing the coefficient matrix

Use when: the vectors live in \mathbb{R}^n (or any setting where 'stack them as columns' makes sense) and you'd rather row-reduce a matrix than solve the dependence equation by hand substitution

Goal: a pivot-in-every-column verdict of independent, or a free column that hands you the dependence relation's coefficients directly

Taught in Independence, dependence, and homogeneous systems · extends Test whether a set of vectors is linearly independent

1. 1

   Put \mathbf v_1,\dots,\mathbf v_k into the columns of a matrix A, in order. The dependence equation c_1\mathbf v_1+\cdots+c_k\mathbf v_k=\mathbf 0 from the last section's recipe is now exactly A\mathbf x=\mathbf 0, with \mathbf x=(c_1,\dots,c_k).

   Example

   For \mathbf v_1=(1,1,0), \mathbf v_2=(2,0,1), \mathbf v_3=(0,2,-1): A=\begin{bmatrix}1&2&0\\1&0&2\\0&1&-1\end{bmatrix}

   ⚡ Shortcut

   Before reducing anything, count columns against rows: if A has more columns than rows (k>n vectors in \mathbb{R}^n), the vectors are automatically dependent — a matrix has at most one pivot per row, so k>n columns can't all get one. Skip straight to Step 4 to find the actual relation; no elimination is needed just to know a nontrivial one exists.

   Why: this is the identical dependence equation from Step 1 of the earlier recipe, just written with a matrix instead of listed scalar equations — same unknowns, same equation, better bookkeeping once k gets large.

2. 2

   Run ordinary row reduction on A itself, exactly as in any homogeneous system — there's no augmented column to track since the right-hand side is \mathbf 0 throughout.

   Example

   R_2 \to R_2 - R_1 turns row 2 into [0,-2,2]; then R_3 \to R_3+\tfrac12 R_2 turns row 3 into [0,0,0]. Normalizing row 2 and clearing column 2 out of row 1 gives \operatorname{rref}(A) = \begin{bmatrix}1&0&2\\0&1&-1\\0&0&0\end{bmatrix}

   Why: solving A\mathbf x=\mathbf 0 is exactly the null-space recipe's Step 1 — the coefficient vector \mathbf x this method wants is found the same way any homogeneous system's solutions are.

3. 3

   Read the RREF: a pivot in every column means the trivial solution is the only one; a column with no pivot means that variable is free and a nontrivial solution exists.

   Example

   In \begin{bmatrix}1&0&2\\0&1&-1\\0&0&0\end{bmatrix}, columns 1 and 2 have pivots, but column 3 does not — so c_3 is free.

   • every column has a pivot — stop; \mathbf v_1,\dots,\mathbf v_k are linearly independent.
   • at least one column has no pivot — continue to Step 4; that free column is about to hand you the dependence relation.

   Why: this is the same trivial-vs-nontrivial fork the null-space recipe uses, applied to the coefficients c_i instead of a generic \mathbf x.

4. 4

   Set the free variable to 1 (or any convenient nonzero number), back-substitute using the RREF rows, and the resulting vector \mathbf x=(c_1,\dots,c_k) IS the dependence relation — no separate substitution chain required.

   Example

   Row 1 gives c_1+2c_3=0 \Rightarrow c_1=-2c_3. Row 2 gives c_2-c_3=0 \Rightarrow c_2=c_3. Set c_3=1: c_1=-2, c_2=1. \boxed{-2\mathbf v_1+\mathbf v_2+\mathbf v_3=\mathbf 0} Check: -2(1,1,0)+(2,0,1)+(0,2,-1) = (-2+2+0,\ -2+0+2,\ 0+1-1) = (0,0,0) ✓.

   Why: the free column's back-substituted values are precisely a null-space vector of A, and that vector's entries are exactly the coefficients the original dependence equation asked for — no extra work translates one into the other.