Linear dependence and independence

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2013, 2015, 2023)
• Priority tier: T3
• Marks (count): 10 (2), 8 (1)
• Average solve time: ~6 min
• Difficulty mix: easy 2, medium 1
• Section: A | Dominant type: proof

Why This Chapter Matters

Linear dependence and independence is the most fundamental concept in linear algebra and one of the fastest questions to solve when you know the method. UPSC asks it in three distinct flavours — testing independence over different fields ($\mathbb R$ vs $\mathbb C$), deciding membership in a span, and row-reducing to find the rank and an explicit dependence relation. All three take under 8 minutes with the right technique. At 8–10 marks each, these are “safe” marks: straightforward once the method is automatic, but a mark trap for students who try to compute determinants of non-square matrices or who skip exhibiting the explicit dependence relation.

Minimum Theory

Linear dependence and independence. Let $V$ be a vector space over a field $\mathbb F$. Vectors $v_1,\ldots,v_k\in V$ are linearly dependent over $\mathbb F$ if there exist scalars $c_1,\ldots,c_k\in\mathbb F$, not all zero, with $\sum c_i v_i=0$. They are linearly independent if the only solution to $\sum c_i v_i=0$ is $c_1=\cdots=c_k=0$.

Span and membership. $w\in\operatorname{span}\{v_1,\ldots,v_k\}$ iff the system $c_1 v_1+\cdots+c_k v_k=w$ is consistent. Test consistency by row-reducing the augmented matrix $[v_1|\cdots|v_k|w]$.

Row-rank test. Stack the vectors as rows of a matrix. The vectors are linearly independent iff the matrix has full row-rank. Row-reduce to echelon form and count non-zero rows. If any row reduces to zero, the corresponding vector is in the span of the others — read the dependence relation from the row operations.

Field dependence/independence. The same set of vectors can be independent over $\mathbb R$ but dependent over $\mathbb C$, because over $\mathbb C$ there are more (complex) scalars available. To show independence over $\mathbb R$: let $c_i\in\mathbb R$ and use the real/imaginary-part trick (a real linear combination of complex vectors is zero iff its real and imaginary parts are separately zero — each complex component gives two real equations). To show dependence over $\mathbb C$: explicitly find complex $c_i$ (not all zero) with $\sum c_i v_i=0$.

Question Archetypes

field-dependent-independence (1 question(s); 2013)

Recognition Cues

• “Show the vectors are independent over $\mathbb R$ but dependent over $\mathbb C$.”
• The vectors are in $\mathbb C^n$ (complex entries) and the question explicitly names both fields.
• Dependence over $\mathbb C$ is shown by finding specific complex scalars; independence over $\mathbb R$ uses the real/imaginary-part argument.

Solution Template

1. Dependence over $\mathbb C$: Try $c=1$ (or another fixed value) and solve for the remaining complex scalars. Show the system has a non-trivial solution.
2. Independence over $\mathbb R$: Let $a,b,c\in\mathbb R$ with $aV_1+bV_2+cV_3=0$. Look for a component that forces immediate conclusions. If any component gives $\alpha+\beta i=0$ with $\alpha,\beta\in\mathbb R$ real-valued expressions, then $\alpha=0$ and $\beta=0$ separately. Drive all scalars to zero.
3. Conclude each part explicitly.

Worked Example

2013 Paper 1, 2013-P1-Q2c-ii (8 marks)

Show that the vectors $X_1=(1,1+i,i)$, $X_2=(i,-i,1-i)$, $X_3=(0,1-2i,2-i)$ in $\mathbb C^3$ are linearly independent over $\mathbb R$ but linearly dependent over $\mathbb C$.

Part 1 — Dependence over $\mathbb C$.

Try $aX_1+bX_2=X_3$ (i.e. $a,b,1$ are the coefficients with the third scalar set to $-1$):

• Coord 1: $a+bi=0\Rightarrow a=-bi$.
• Coord 2: $a(1+i)+b(-i)=1-2i$. Substitute $a=-bi$: $-bi-bi^2-bi=1-2i\Rightarrow b-2bi=1-2i\Rightarrow b=1$.
• Then $a=-i$.
• Check Coord 3: $(-i)(i)+(1)(1-i)=1+1-i=2-i$ ✓.

The relation $-iX_1+X_2-X_3=0$ (or equivalently $iX_1-X_2+X_3=0$) has non-zero complex coefficients, so $\{X_1,X_2,X_3\}$ is linearly dependent over $\mathbb C$. $\blacksquare$

$\boxed{\;iX_1-X_2+X_3=0.\;}$

Part 2 — Independence over $\mathbb R$.

Let $a,b,c\in\mathbb R$ satisfy $aX_1+bX_2+cX_3=0$.

• Coord 1: $a+bi=0$. Since $a,b\in\mathbb R$: real part $a=0$; imaginary part $b=0$.
• Coord 3 (with $a=b=0$): $c(2-i)=0$. Since $2-i\neq0$: $c=0$.

So $a=b=c=0$, and $\{X_1,X_2,X_3\}$ is linearly independent over $\mathbb R$. $\blacksquare$

Common Traps

• “Independent over $\mathbb R$” does not mean the vectors are real. The vectors are in $\mathbb C^3$; only the scalars are restricted to $\mathbb R$. As a real vector space $\mathbb C^3$ has dimension 6, so three vectors are generically independent.
• Matching real and imaginary parts doubles the equations. For $\alpha+\beta i=0$ with $\alpha,\beta\in\mathbb R$, both $\alpha=0$ and $\beta=0$ must hold. This is the key that forces independence over $\mathbb R$.
• Show dependence over $\mathbb C$ by an explicit relation — don’t just say “there are more scalars available.” Write the coefficients explicitly and verify all components.
• Sign: $iX_1-X_2+X_3=0$ and $-iX_1+X_2-X_3=0$ are the same relation; either form earns full marks.

membership-in-span (1 question(s); 2023)

Recognition Cues

• “Does $w\in\operatorname{span}\{V_1,V_2,V_3\}$?” or “Is $w$ expressible as a linear combination of $V_1,V_2,V_3$?”
• You have $k$ vectors in $\mathbb R^n$ with $n>k$ (over-determined system).
• Answer: yes if consistent, no if inconsistent.

Solution Template

1. Set up the system. Write $aV_1+bV_2+cV_3=w$ as a system of $n$ equations in $k$ unknowns.
2. Reduce. Use two equations to express all unknowns in terms of one free parameter.
3. Substitute into a third equation. Check for consistency. If consistent: $w\in\operatorname{span}$. If inconsistent: $w\notin\operatorname{span}$.
4. Conclude explicitly.

Worked Example

2023 Paper 1, 2023-P1-Q1a (10 marks)

Let $V_1=(2,-1,3,2)$, $V_2=(-1,1,1,-3)$, $V_3=(1,1,9,-5)$. Does $(3,-1,0,-1)\in\operatorname{span}\{V_1,V_2,V_3\}$?

Step 1 — System. Seek $a,b,c\in\mathbb R$ with $aV_1+bV_2+cV_3=(3,-1,0,-1)$: $2a-b+c=3,\qquad -a+b+c=-1,\qquad 3a+b+9c=0,\qquad 2a-3b-5c=-1.$

Step 2 — Reduce. Add Eq 1 and Eq 2: $a+2c=2\Rightarrow a=2-2c$. Substitute into Eq 2: $b=1-3c$.

Step 3 — Check Eq 3. $3(2-2c)+(1-3c)+9c=6-6c+1-3c+9c=7$. For consistency, need $7=0$. Contradiction.

Conclusion. $\boxed{\;(3,-1,0,-1)\notin\operatorname{span}\{V_1,V_2,V_3\}.\;}$

(Also check Eq 4 for completeness: $2(2-2c)-3(1-3c)-5c=4-4c-3+9c-5c=1\neq -1$ — also inconsistent.)

Common Traps

• Do not compute a determinant of a non-square matrix. The matrix $[V_1|V_2|V_3]$ is $4\times 3$ — it has no determinant. Use elimination instead.
• Use only 2 equations to solve for $a,b$ in terms of $c$, then test the remaining equations. Don’t try to solve all 4 simultaneously from scratch.
• If the system is consistent, find explicit values of $a,b,c$ and write $w=aV_1+bV_2+cV_3$ explicitly as the conclusion.

test-independence (1 question(s); 2015)

Recognition Cues

• “Are $V_1,\ldots,V_k$ linearly independent? Justify.” — especially when phrased as a leading statement to be verified.
• Vectors in $\mathbb R^n$; decision by row-rank.
• Full marks require: row reduction, rank determination, and (if dependent) an explicit linear relation.

Solution Template

1. Form the matrix $M$ with the vectors as rows (or columns).
2. Row-reduce to echelon form, recording the operations.
3. Identify rank. Count non-zero rows in echelon form.
4. If rank $<k$: the vectors are dependent. Read off the dependence relation from the row operations (or solve $\sum c_i V_i=0$ directly with the reduced system).
5. State conclusion clearly: true/false + the explicit relation.

Worked Example

2015 Paper 1, 2015-P1-Q1a (10 marks)

The vectors $V_1=(1,1,2,4)$, $V_2=(2,-1,-5,2)$, $V_3=(1,-1,-4,0)$, $V_4=(2,1,1,6)$ in $\mathbb R^4$ are linearly independent. Is it true? Justify.

Step 1 — Matrix. $M=\begin{pmatrix}1&1&2&4\\2&-1&-5&2\\1&-1&-4&0\\2&1&1&6\end{pmatrix}.$

Step 2 — Row-reduce.

$R_2\to R_2-2R_1$, $R_3\to R_3-R_1$, $R_4\to R_4-2R_1$: $\begin{pmatrix}1&1&2&4\\0&-3&-9&-6\\0&-2&-6&-4\\0&-1&-3&-2\end{pmatrix}.$

Observe: $R_2=3R_4$ and $R_3=2R_4$. So $R_2\to R_2-3R_4$ and $R_3\to R_3-2R_4$: $\begin{pmatrix}1&1&2&4\\0&0&0&0\\0&0&0&0\\0&-1&-3&-2\end{pmatrix}.$

Step 3 — Rank. Two non-zero rows. Rank $=2<4$.

Conclusion. The statement is false — the vectors are linearly dependent.

Step 4 — Explicit relation. Solve $\alpha V_1+\beta V_2+\gamma V_3+\delta V_4=0$. From the row operations: $R_2-3R_4=0$ means $-3V_2+(-1)(-3)V_4=0$… more directly, try $\alpha=0$: then $2\beta+\gamma+2\delta=0$, $-\beta-\gamma+\delta=0$, $2\beta+6\delta=0\Rightarrow\beta=-3\delta$. Then $\gamma=4\delta$. Take $\delta=1$: $\boxed{\;V_4=3V_2-4V_3.\;}$

Verify: $3V_2-4V_3=3(2,-1,-5,2)-4(1,-1,-4,0)=(6-4,-3+4,-15+16,6-0)=(2,1,1,6)=V_4$ ✓.

Common Traps

• The question is a leading statement — “they are linearly independent. Is it true?” One must check and potentially refute. The answer here is “false.”
• Rank is 2, not 3. After one round of elimination, three rows $R_2,R_3,R_4$ are proportional to $(0,-1,-3,-2)$ — all three collapse. Reporting “rank 3” would be wrong.
• Exhibit the explicit relation. Simply saying $\det M=0$ is insufficient; the relation $V_4=3V_2-4V_3$ (or equivalent) is required for full marks.
• Row-reduce carefully when rows become proportional. The proportionality $R_2=3R_4$ is visible before computing $R_3$; noticing it early saves time.

Marks-Aware Writing

8-mark questions (field independence, 2013): Two parts, roughly 4 marks each. For independence over $\mathbb R$: write the linear combination equation, use one component to force real and imaginary parts to zero simultaneously, drive other scalars to zero. One display equation suffices. For dependence over $\mathbb C$: write the explicit relation with the non-zero complex coefficients and verify by component check.

10-mark questions (span/row-rank): For membership-in-span: the system setup + two-step reduction + consistency check + conclusion covers all marks. For test-independence: matrix + row operations + rank determination + explicit dependence relation. In both cases, the final boxed conclusion sentence (“the vectors are/are not independent”, ”$w$ is/is not in the span”) is required.

Practice Set

• 2014-P1-Q1a (10 m) — — Hint: row-reduce; the four vectors span only a 3D subspace; exhibit a relation.
• 2022-P1-Q1a (10 m) — — Hint: linear dependence test; check rank by row reduction.
• 2017-P1-Q3b (10 m) — — Hint: span/independence in a polynomial or function space; use evaluation.
• 2025-P1-Q1a (10 m) — — Hint: row-reduce and report rank; look for a zero row early.