How to determine if a set of vectors is linearly independent?

Here are the practical methods ranked by efficiency. **Method 1: Row Reduction (Most General)** Place the vectors as **columns** of a matrix $A$ and row-reduce to reduced row echelon form (RREF). - If every column has a leading 1 (pivot), the vectors are independent. - If any column lacks a pivot, the vectors are dependent, and the free columns correspond to vectors that can be expressed as combinations of earlier vectors. This works for any set of vectors in $\mathbb{R}^n$. **Method 2: Determinant (Square Matrices Only)** If you have exactly $n$ vectors in $\mathbb{R}^n$, form an $n \times n$ matrix $A$ with the vectors as columns. Then: - $\det(A) \ eq 0$ $\iff$ vectors are linearly independent $\iff$ $A$ is invertible - $\det(A) = 0$ $\iff$ vectors are linearly dependent **Example:** Check if $\mathbf{v}_1 = (1,2,3)$, $\mathbf{v}_2 = (4,5,6)$, $\mathbf{v}_3 = (7,8,10)$ are independent. \begin{align*} \det\begin{pmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 10 \end{pmatrix} &= 1(5\cdot10 - 8\cdot6) - 4(2\cdot10 - 8\cdot3) + 7(2\cdot6 - 5\cdot3) \\ &= 1(50 - 48) - 4(20 - 24) + 7(12 - 15) \\ &= 1(2) - 4(-4) + 7(-3) = 2 + 16 - 21 = -3 \ eq 0 \end{align*} Since $\det(A) = -3 \ eq 0$, the vectors are linearly independent. **Method 3: Inspection (Simple Cases)** - Two vectors: independent iff neither is a scalar multiple of the other. - A set containing the zero vector is always dependent. - More vectors than the dimension of the space is always dependent (e.g., 4 vectors in $\mathbb{R}^3$).