USTHLinear Algebra · MAT1.002

Modules › Vector spaces

Vector spaces

Vector-space axioms, subspaces, span, independence, basis, dimension, rank, coordinates and change of basis.

Lecture 4 · Exercise Sheet 4 · Vector spaces
Textbook sections: Elementary Linear Algebra ch. 4 §4.1–4.7 · Strang ch. 3 §3.1–3.6
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Vector spaces

Linear Algebra · Module 4
Lecture 4 · Exercise Sheet 4
Chapter: Vector spaces
Books: Elementary Linear Algebra ch. 4 §4.1–4.7 · Strang ch. 3 §3.1–3.6
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After this module you can

  • Test whether a set is a vector space or a subspace.
  • Decide independence/spanning, find a basis and the dimension; compute the bases of null, column and row spaces.
  • Compute coordinates and transition matrices.
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Vector spaces and subspaces

  • A vector space satisfies 10 axioms: closure under $+$ and scalar multiplication, commutativity, associativity, a zero vector, additive inverses, two distributive laws, $(cd)u=c(du)$, $1u=u$.
  • Subspace test (inside a known space): $\mathbf0\in W$, closed under $+$, closed under scalar multiples.
  • $\{b_1=0\}$, planes through $0$ and spans are subspaces; $\{b_1=1\}$, $\{b_2b_3=0\}$ (a union of two planes), $\{b_1\ge0\}$ are not.
  • Non-examples among the sheet's sets: $\{(x,y):x\ge0\}$ (not closed under negative scalars), singular matrices (not closed under $+$), polynomials of degree exactly $2$ (no zero, not closed).
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Span, independence, basis, dimension

NotionTest in $\mathbb{R}^n$
linearly independent$c_1v_1+\cdots+c_kv_k=0$ only for $c=0$: matrix with the $v_i$ as columns has rank $k$
spanevery vector is a combination: rank $=n$ (for $n$ vectors in $\mathbb{R}^n$: $\det\ne0$)
basisindependent AND spanning; all bases have $\dim V$ vectors
dimensions$\dim\mathbb{R}^n=n$, $\dim M_{2\times2}=4$, $\dim P_n=n+1$
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The fundamental subspaces

  • Row-reduce $A$. Column space: pivot columns of $A$ (original columns!). Row space: non-zero rows of the RREF. Null space: one vector per free variable.
  • Rank–nullity: $\operatorname{rank}A+\operatorname{nullity}A=n$ (number of columns).
  • $A\mathbf x=\mathbf b$ consistent: $\mathbf x=\mathbf x_p+\mathbf x_h$ with $\mathbf x_h\in\ker A$.
  • Example: $\begin{bmatrix}1&3&2&22&13\\1&0&2&-2&1\\3&6&5&42&27\end{bmatrix}$ has rank $3$, nullity $2$.
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Coordinates and change of basis

  • Basis $B=\{b_1,\dots,b_n\}$, $P_B=[b_1\ \cdots\ b_n]$ (columns). $\mathbf x=P_B[\mathbf x]_B$, so $[\mathbf x]_B=P_B^{-1}\mathbf x$.
  • Transition matrix $B\to B'$: $P=P_{B'}^{-1}P_B$ with $[\mathbf x]_{B'}=P[\mathbf x]_B$ (row-reduce $[B'\mid B]\to[I\mid P]$).
  • Example: $B=\{(1,1),(-1,1)\}$, $[\mathbf x]_B=(3,5)^T$: $\mathbf x=3(1,1)+5(-1,1)=(-2,8)$.
  • If $v_1,v_2,v_3$ are independent, $\{v_1-v_2,v_2-v_3,v_3-v_1\}$ is dependent (the sum is $0$).
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Typical mistakes

Column space basisTake the pivot columns of the ORIGINAL matrix, not of the RREF.
Subspace testDo not forget the zero vector and both closure conditions.
Dimension count$n$ vectors in an $n$-dimensional space: independent $\iff$ basis.
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Practice next

  • Chapter quizzes Ch. 4 (3 sets)
  • Practice bank, filtered to this chapter, then written mode
  • Lab: Rank, null space, column and row space
  • Then the mock exams (Exams page)

Worked examples from the course problems

Worked example 1 · Vector space axioms (Exercise Sheet 4, Ex. 3(a))
Why is $\mathbb{R}^3$ with usual addition and $c\cdot(x,y,z)=(cx,y,z)$ NOT a vector space? Select the property that fails.
Show the answer and the solution

Answer: $(c+d)u=cu+du$ fails

Test the axioms on examples: $(c+d)u=cu+du$ fails. The other listed properties do hold for this set.
Worked example 2 · Coordinates and change of basis (Exercise Sheet 4, Ex. 16(a))
Find $[x]_{B'}$ given $B=\{(1,1),(1,-1)\}$, $B'=\{(0,1),(1,2)\}$, $[x]_B=(3,-3)^T$.
Show the answer and the solution

Answer: $(6, 0)$

First $x=P_B[x]_B=(0, 6)$ (standard coordinates), then solve $P_{B'}[x]_{B'}=x$: $[x]_{B'}=(6, 0)$.
Worked example 3 · Vector arithmetic (Exercise Sheet 4, Ex. 2)
Solve $2x-u+3v+2w=0$ for $x$, where $u=(1,-1,2)$, $v=(0,2,3)$, $w=(0,1,1)$.
Show the answer and the solution

Answer: $(\frac{1}{2}, - \frac{9}{2}, - \frac{9}{2})$

$x=\tfrac12(u-3v-2w)=\tfrac12(1,-1-6-2,2-9-2)=(\frac{1}{2}, - \frac{9}{2}, - \frac{9}{2})$.
Worked example 4 · Linear combinations (Exercise Sheet 4, Ex. 5)
Write $v=(3,0,-6)$ as $c_1u_1+c_2u_2+c_3u_3$ with $u_1=(1,-1,2)$, $u_2=(2,4,-2)$, $u_3=(1,2,-4)$. What is $c_1$?
Show the answer and the solution

Answer: $2$

Solve $[u_1\ u_2\ u_3]c=v$: $\det[u_1\ u_2\ u_3]=-18\neq0$, so the solution is unique: $c=(2, -1, 3)$.

Practice

Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Rank, null space, column and row space

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