Worked examples — Linear independence — formal definition, testing

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The scenario matrix

Every linear-independence problem lives in one of these cells. We will hit all of them.

Cell	Situation	Right tool	Answer pattern
A	Square, $n$ vectors in ${\mathbb{R}}^n$	determinant	$\det \ne 0$ → indep
B	Square but parallel/multiple (degenerate)	determinant $=0$ or inspection	dependent
C	Non-square, fewer vectors than dimension	row reduce, count pivots	could be either
D	Non-square, more vectors than dimension	count vs dimension	always dependent
E	Set contains the zero vector	inspection	always dependent
F	A vector is a sum of others (hidden)	row reduce reveals it	dependent
G	Real-world word problem	translate → matrix → test	context-dependent
H	Exam twist (parameter $k$: for which $k$ dependent?)	$\det$ as function of $k$	solve $\det =0$

The number of vectors $n$ is just "how many arrows we listed". The dimension is how many coordinates each arrow has (e.g. $(3,1)$ lives in dimension $2$). A pivot is a leading nonzero entry after row reduction — one per genuinely new direction.

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Example 1 — Cell A (square, determinant)

Forecast: They point clearly different ways — neither looks like a stretch of the other. Guess: independent. Look at the figure before reading on.

1. Stack as columns. $A=\left[\begin{array}{cc}3& 1\\ 1& 2\end{array}\right]$. Why this step? The equation $c_1{\mathbf{v}}_1+c_2{\mathbf{v}}_2=0$ is exactly $A\mathbf{c}=0$; the columns of $A$ are our vectors.
2. Two vectors in ${\mathbb{R}}^2$ → square → use the determinant. Why this step? $\det$ only exists for square matrices, and here $A$ is $2\times 2$. Geometrically $\det$ is the signed area of the parallelogram the two vectors make (blue in the figure). Zero area means they lie on one line (dependent); nonzero area means they open up a genuine 2D patch.
3. Compute. $\det A=3\cdot 2-1\cdot 1=6-1=5\ne 0$.

Answer: independent.

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Example 2 — Cell B (square, degenerate: parallel)

Forecast: Notice $-6=-3\cdot 2$ and $3=-3\cdot (-1)$. Smells like a multiple. Guess: dependent.

1. Spot the scalar multiple. ${\mathbf{v}}_2=-3{\mathbf{v}}_1$. Why this step? If one vector is a scalar multiple of another, they lie on the same line through the origin (see figure — both arrows and their extension share one dashed line). That is the picture of dependence.
2. Write the nontrivial combo. $3{\mathbf{v}}_1+1\cdot {\mathbf{v}}_2=3(2,-1)+(-6,3)=(0,0)=0$, with $c_1=3\ne 0$. Why this step? Dependence needs at least one nonzero coefficient producing $0$; we just exhibited one.
3. Confirm with determinant. $\det \left[\begin{array}{cc}2& -6\\ -1& 3\end{array}\right]=2\cdot 3-(-6)(-1)=6-6=0$.

Answer: dependent.

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Example 3 — Cell C (non-square, fewer than dimension)

Forecast: Only $2$ vectors in a $3$-dimensional space, and neither is a multiple of the other. Guess: independent (but $\det$ is off-limits — the matrix is $3\times 2$).

1. Why no determinant. $A=\left[\begin{array}{cc}1& 0\\ 0& 3\\ 2& 1\end{array}\right]$ is $3\times 2$ — not square, so $\det$ does not exist. Why this step? The determinant is defined only for square matrices; reaching for it here is a classic error. Use rank via row reduction instead.
2. Row reduce. $\left[\begin{array}{cc}1& 0\\ 0& 3\\ 2& 1\end{array}\right]\stackrel{R_3-2R_1}{\to }\left[\begin{array}{cc}1& 0\\ 0& 3\\ 0& 1\end{array}\right]\stackrel{R_3-\frac{1}{3}{R}_2}{\to }\left[\begin{array}{cc}1& 0\\ 0& 3\\ 0& 0\end{array}\right].$ Why this step? Row operations don't change dependence relations among columns; we just want to count pivots (leading entries).
3. Count pivots. Pivots in column 1 (the $1$) and column 2 (the $3$): 2 pivots for 2 columns. Why this step? Pivots $=$ number of columns ⇒ $\mathrm{rank}=2=n$ ⇒ no free variable ⇒ only trivial solution.

Answer: independent.

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Example 4 — Cell D (more vectors than dimension)

Forecast: $3$ arrows squeezed into a flat $2$-dimensional plane. Guess: dependent — there's simply no room for $3$ independent directions.

1. Count vectors vs dimension. $n=3$ vectors, dimension $=2$. Matrix $A$ is $2\times 3$. Why this step? $\mathrm{rank}(A)\le \min (\text{rows},\text{cols})=2<3$, so at least one column has no pivot ⇒ a free variable exists ⇒ nontrivial solution guaranteed. We already know the answer before computing.
2. Find an explicit relation (for satisfaction). Note ${\mathbf{v}}_3={\mathbf{v}}_1+{\mathbf{v}}_2$? Check: $(1,1)+(2,3)=(3,4)={\mathbf{v}}_3$. ✓ Why this step? Producing the actual combo confirms dependence concretely: ${\mathbf{v}}_1+{\mathbf{v}}_2-{\mathbf{v}}_3=0$, with coefficients $(1,1,-1)\ne 0$.

Answer: dependent.

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Example 5 — Cell E (contains the zero vector)

Forecast: One of them is $0$. Guess: dependent — and this holds no matter what the other vectors are.

1. Isolate the zero vector. Call them ${\mathbf{v}}_1,{\mathbf{v}}_2=0,{\mathbf{v}}_3$. Why this step? Any set containing $0$ is instantly dependent — we don't even need the other entries.
2. Build a nontrivial combo. $0\cdot {\mathbf{v}}_1+1\cdot {\mathbf{v}}_2+0\cdot {\mathbf{v}}_3=1\cdot 0=0$. Why this step? Here $c_2=1\ne 0$, yet the sum is $0$. That single nonzero coefficient is enough for dependence.

Answer: dependent.

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Example 6 — Cell F (hidden sum, must row reduce)

Forecast: No two are visible multiples. Tempting to shout "independent", but the trap is a hidden combination. Row reduce to be safe.

1. Stack columns and reduce. $A=\left[\begin{array}{ccc}1& 2& 4\\ 2& 1& 5\\ 1& 0& 2\end{array}\right]\stackrel{\begin{array}{c}{R}_2-2R_1\\ R_3-R_1\end{array}}{\to }\left[\begin{array}{ccc}1& 2& 4\\ 0& -3& -3\\ 0& -2& -2\end{array}\right]\stackrel{R_3-\frac{2}{3}{R}_2}{\to }\left[\begin{array}{ccc}1& 2& 4\\ 0& -3& -3\\ 0& 0& 0\end{array}\right].$ Why this step? A zero row appears — only 2 pivots for 3 columns, so $\mathrm{rank}=2<3$ ⇒ free variable ⇒ dependent.
2. Recover the relation. Guess ${\mathbf{v}}_3=2{\mathbf{v}}_1+{\mathbf{v}}_2$: $2(1,2,1)+(2,1,0)=(4,5,2)={\mathbf{v}}_3$. ✓ Why this step? The dependence was ${\mathbf{v}}_3=2{\mathbf{v}}_1+{\mathbf{v}}_2$, invisible at a glance — exactly why row reduction beats eyeballing.

Answer: dependent.

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Example 7 — Cell G (real-world word problem)

Forecast: "Reproduced by mixing others" is literally the definition of a redundant vector. Scan: ${\mathbf{s}}_3$ looks like ${\mathbf{s}}_1+{\mathbf{s}}_2$. Guess: dependent (one stock is wasteful).

1. Translate the story. "Mix solution X from the others" = "X is a linear combination of the others" = dependence. Why this step? Mixing with amounts $a,b$ means $a{\mathbf{s}}_1+b{\mathbf{s}}_2$; asking if this equals ${\mathbf{s}}_3$ is asking about a dependency.
2. Test the guess. ${\mathbf{s}}_1+{\mathbf{s}}_2=(2+0,0+4,1+2)=(2,4,3)={\mathbf{s}}_3$. ✓ Why this step? One explicit nontrivial combo (${\mathbf{s}}_1+{\mathbf{s}}_2-{\mathbf{s}}_3=0$) proves dependence — no need to reduce.

Answer: dependent — ${\mathbf{s}}_3$ is exactly $1$ part ${\mathbf{s}}_1$ + $1$ part ${\mathbf{s}}_2$, so it is a redundant stock.

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Example 8 — Cell H (exam twist: a parameter)

Forecast: Square $3\times 3$ case, so dependence ⇔ $\det =0$. There will be one special value of $k$. Guess: solve $\det (k)=0$.

1. Set up the determinant as a function of $k$. $A=\left[\begin{array}{ccc}1& 0& 1\\ 2& 1& 3\\ 3& 4& k\end{array}\right].$ Why this step? Square matrix ⇒ determinant test is valid, and $\det$ becomes a formula in $k$; the dependent values are its roots.
2. Expand along row 1. $\det A=1\cdot (1\cdot k-3\cdot 4)-0+1\cdot (2\cdot 4-1\cdot 3)=(k-12)+(8-3)=k-7.$ Why this step? Cofactor expansion turns a $3\times 3$ determinant into small $2\times 2$ ones we can evaluate.
3. Solve $\det =0$. $k-7=0\Rightarrow k=7$. Why this step? $\det =0$ is exactly the dependence condition (columns collapse to $<3$ directions).

Answer: the set is dependent only when $k=7$; independent for every other $k$.

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Case-coverage checklist

Recall Did we hit every cell?

A (square/det) → Ex 1 ::: ✓ B (parallel degenerate) → Ex 2 ::: ✓ C (non-square, fewer) → Ex 3 ::: ✓ D (more than dimension) → Ex 4 ::: ✓ E (contains zero vector) → Ex 5 ::: ✓ F (hidden sum) → Ex 6 ::: ✓ G (word problem) → Ex 7 ::: ✓ H (parameter twist) → Ex 8 ::: ✓

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Connections

• Determinant — the $\det =0$ test powering Cells A, B, H.
• Rank of a matrix — pivot-counting for the non-square Cells C, D, F.
• Homogeneous systems and null space — every test is really "does $A\mathbf{c}=0$ have a nontrivial solution?".
• Basis and dimension — Cell D is the "too many for the dimension" fact.
• Span and spanning sets — the word problem (Cell G) asks whether a stock lies in the span of the others.
• Invertible matrix theorem — for the square cells, independent ⇔ invertible.