A set of vectors $\{v_1, \ldots, v_n\}$ is LINEARLY INDEPENDENT iff
A$\sum a_i v_i = 0$ forces all $a_i = 0$
Btheir dot products are all zero
Cno vector equals zero
D$\sum v_i \neq 0$
Answer & Solution
Correct answer: A. $\sum a_i v_i = 0$ forces all $a_i = 0$
1. DEFINITION: vectors are LINEARLY INDEPENDENT iff the only way to write $\sum a_i v_i = 0$ is with all coefficients $a_i = 0$.
2. Equivalently: no vector in the set is a linear combination of the others.
3. If some non-trivial $\sum a_i v_i = 0$ exists, the vectors are LINEARLY DEPENDENT.
4. Option B describes ORTHOGONALITY (different property). Options C, D are too weak.
_Source: Sergei Treil, "Linear Algebra Done Wrong", §2.2 (Linear independence)._
Related questions
Let $A$ be a $3 \times 3$ matrix with eigenvalues 1, 2, 3. Then $\det(A)$ equalsIf $A$ is invertible, the eigenvalues of $A^{-1}$ areThe TRACE of $A = \begin{pmatrix} 2 & 5 \\ 7 & 3 \end{pmatrix}$ isGRAM-SCHMIDT process converts a basis intoTwo vectors $u, v$ are ORTHOGONAL iffAn ORTHOGONAL matrix $Q$ satisfiesFor a SYMMETRIC real matrix $A$, all eigenvalues areCAYLEY-HAMILTON THEOREM states that for any $n \times n$ matrix $A$ with characteristic po