Exercises — Linear independence — formal definition, testing

• ${\mathbf{v}}_1,\dots ,{\mathbf{v}}_n$ — our vectors (arrows / lists of numbers).
• $c_1,\dots ,c_n$ — unknown scalars (plain numbers) we multiply vectors by.
• $0$ — the zero vector (the arrow of length zero; the list of all zeros).
• $A\mathbf{c}=0$ — stack vectors as columns of matrix $A$; then $A\mathbf{c}=c_1{\mathbf{v}}_1+\cdots +c_n{\mathbf{v}}_n$. Asking "when is this $0$?" is the independence test.
• $\mathrm{rank}(A)$ — number of pivots after row reduction = number of genuinely independent directions.
• $\det A$ — a single number, defined only for square $A$; zero means the columns are squashed flat (dependent).

Independent. $(1,0)$ points purely East, $(0,1)$ purely North — neither is a scalar multiple of the other, so neither is redundant. Algebraically $c_1(1,0)+c_2(0,1)=(c_1,c_2)=0$ forces $c_1=c_2=0$: only the trivial (boring) solution.

Dependent. Look: $(2,4,6)=2\cdot (1,2,3)$. That is a visible scalar multiple. So $1\cdot (2,4,6)-2\cdot (1,2,3)=0$ is a nontrivial combo (the coefficient $1$ is not zero). Dependent.

Dependent. Any set containing $0$ is dependent, because $1\cdot (0,0)+0\cdot (1,5)+0\cdot (2,9)=0$ has coefficient $c_1=1\ne 0$ — a nontrivial way to hit zero.

$A=\left[\begin{array}{cc}1& 3\\ 2& 5\end{array}\right],\det A=1\cdot 5-3\cdot 2=5-6=-1.$ $\det A=-1\ne 0$, so independent. Why determinant here? For two vectors in the plane, $\det A$ is the signed area of the parallelogram they span (Figure below). Zero area means they lie on one line (parallel = dependent); nonzero area means they open up a genuine 2D region.

$\det \left[\begin{array}{cc}2& -4\\ 1& -2\end{array}\right]=2\cdot (-2)-(-4)\cdot 1=-4+4=0.$ $\det A=0\Rightarrow$ dependent. Indeed $(-4,-2)=-2\cdot (2,1)$: the two arrows lie on the same line, so the parallelogram is squashed to a line segment — zero area (see figure).

$A=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right]\stackrel{R_3-R_1}{\to }\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 1\end{array}\right]\stackrel{R_3-R_2}{\to }\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right].$ Two pivots (columns 1 and 2), for two columns. Every column has a pivot $\Rightarrow$ no free variable $\Rightarrow$ only $\mathbf{c}=0$. Independent.

$A=\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 2& 1& 5\end{array}\right]\stackrel{R_3-2R_1}{\to }\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 0& 1& 1\end{array}\right]\stackrel{R_3-R_2}{\to }\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 0& 0& 0\end{array}\right].$ Two pivots for three columns → a free variable → dependent. Find the combo: solve $A\mathbf{c}=0$. From the reduced rows, $c_1+2c_3=0$ and $c_2+c_3=0$. Let $c_3=1$: then $c_1=-2,c_2=-1$. Check: $-2(1,0,2)-1(0,1,1)+1(2,1,5)=(-2+2,-1+1,-4-1+5)=(0,0,0).✓$ So ${\mathbf{v}}_3=2{\mathbf{v}}_1+{\mathbf{v}}_2$ — the redundant passenger.

Dependent $⟺\det A=0$: $\det \left[\begin{array}{cc}1& 3\\ 2& k\end{array}\right]=k-6=0\Rightarrow k=6.$ At $k=6$, ${\mathbf{v}}_2=(3,6)=3(1,2)=3{\mathbf{v}}_1$ — parallel, dependent. For every other $k$, independent.

Stack the $4$ vectors as columns: $A$ is $3\times 4$. Row reduction produces at most one pivot per row, and there are only $3$ rows, so $\mathrm{rank}(A)\le 3$. But independence of $4$ columns needs a pivot in each of the $4$ columns, i.e. $\mathrm{rank}=4$. Since $3<4$, that is impossible: a free variable always exists → dependent. Maximum independent count $=$ number of rows $=$ dimension $=3$. (This is why $\dim {\mathbb{R}}^3=3$; see Basis and dimension.)

Square case, so use the determinant. With columns ${\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3$: $A=\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 1& t\end{array}\right].$ Expand along the first row: $\det A=1\cdot (0\cdot t-1\cdot 1)-1\cdot (1\cdot t-1\cdot 0)+0=(-1)-(t)=-1-t.$ $\det A\ne 0⟺-1-t\ne 0⟺t\ne -1$. So they form a basis for all $t\ne -1$. At $t=-1$, $\det A=0$: dependent, not a basis. Why this is the whole answer: for $n$ vectors in ${\mathbb{R}}^n$, invertible $⟺$ independent $⟺$ spanning $⟺$ basis. One determinant settles everything.

(a) No. Independence of $5$ columns needs $5$ pivots, but rank $=3$ gives only $3$ pivots → $2$ free variables → nontrivial solutions → dependent. (b) A maximal independent subset has size $=\mathrm{rank}=3$ columns (the pivot columns). (c) By the rank–nullity idea, $\dim (\text{null space})=(\#\text{columns})-\mathrm{rank}=5-3=2$. Two free variables ⇔ a $2$-dimensional space of nontrivial combinations. (See Homogeneous systems and null space.)

Row reduce. $\stackrel{R_2-2R_1,R_3-R_1}{\to }\left[\begin{array}{cccc}1& 2& 3& 1\\ 0& 0& 0& -2\\ 0& 0& 0& 4\end{array}\right]\stackrel{R_3+2R_2}{\to }\left[\begin{array}{cccc}1& 2& 3& 1\\ 0& 0& 0& -2\\ 0& 0& 0& 0\end{array}\right].$ Pivots sit in column 1 and column 4. (a) $\mathrm{rank}(A)=2$. (b) Dependent ($2$ pivots $<4$ columns). (c) Pivot columns give a maximal independent subset: $\{{\mathbf{v}}_1=(1,2,1),{\mathbf{v}}_4=(1,0,5)\}$. (d) Non-pivot columns ${\mathbf{v}}_2,{\mathbf{v}}_3$ are combinations of ${\mathbf{v}}_1$. Directly: ${\mathbf{v}}_2=(2,4,2)=2{\mathbf{v}}_1$ and ${\mathbf{v}}_3=(3,6,3)=3{\mathbf{v}}_1$ (each uses $0\cdot {\mathbf{v}}_4$). Check: $2\cdot (1,2,1)=(2,4,2)={\mathbf{v}}_2.✓$

Suppose $a({\mathbf{v}}_1+{\mathbf{v}}_2)+b({\mathbf{v}}_2+{\mathbf{v}}_3)+c({\mathbf{v}}_1+{\mathbf{v}}_3)=0$. Group by original vector: $(a+c){\mathbf{v}}_1+(a+b){\mathbf{v}}_2+(b+c){\mathbf{v}}_3=0.$ Since $\{{\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3\}$ is independent, every coefficient must be $0$: $a+c=0,a+b=0,b+c=0.$ From the first two, $c=-a$ and $b=-a$. Put into the third: $(-a)+(-a)=-2a=0\Rightarrow a=0$, hence $b=c=0$. Only the trivial solution → the new set is independent. $■$ (The system $a+c=a+b=b+c=0$ has coefficient matrix with determinant $\det \left[\begin{array}{ccc}1& 0& 1\\ 1& 1& 0\\ 0& 1& 1\end{array}\right]=2\ne 0$, confirming only the trivial solution.)

Coordinate vectors (in the order $[\text{const},x,x^2]$): $p_1=(1,1,0),p_2=(0,1,1),p_3=(1,0,1)$. Same setup as L5.2's implicit matrix. Determinant: $\det \left[\begin{array}{ccc}1& 0& 1\\ 1& 1& 0\\ 0& 1& 1\end{array}\right]=1(1\cdot 1-0\cdot 1)-0+1(1\cdot 1-1\cdot 0)=1+1=2\ne 0.$ $\det \ne 0\Rightarrow$ independent. Linear independence is a coordinate statement — it works identically for polynomials once you list their coefficients as a vector.