Attempt these from memory before reading.
Define linear independence. What is a basis, and what two things does it have to do? State the dimension formula and the rank-nullity theorem. How are they related? Does R 3 \mathbb{R}^3 R 3 What is the rank of a matrix, in terms of its rows and its columns? Check your answers against the text. Prompt 4 is the one that makes dimension a theorem rather than a definition: that you cannot have both is the Steinitz exchange lemma.
The naive idea is that dimension is the number of coordinates. But “the number of coordinates” presupposes a basis, and the whole point of the concept is that every basis gives the same number.
Suppose someone hands you { ( 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) } \{(1,0), (0,1), (1,1)\} {( 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 )} R 2 \mathbb{R}^2 R 2 { ( 1 , 0 ) , ( 0 , 1 ) } \{(1,0), (0,1)\} {( 1 , 0 ) , ( 0 , 1 )}
The something deeper is the Steinitz exchange lemma. An independent set is never larger than a spanning set. That single result forces all bases of a space to have the same size, and that is what makes “dimension” a property of the space rather than of a choice of basis. It is the first genuinely structural theorem in the subject, and it is why every later result — rank- nullity, the dimension formula, the existence of eigenbases — can even be stated.
Rank-nullity is the same idea applied to a map. A linear map T : V → W T: V \to W T : V → W
dim V = dim ker T + dim im T . \dim V = \dim\ker T + \dim\operatorname{im} T . dim V = dim ker T + dim im T . It reads like arithmetic and it is a theorem about how maps behave, and it is the reason a system of equations is either inconsistent, underdetermined, or has a unique solution — with no fourth option.
A set of vectors { v 1 , v 2 , … , v k } ⊆ V \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\} \subseteq V { v 1 , v 2 , … , v k } ⊆ V linearly Independent if the equation
α 1 v 1 + α 2 v 2 + ⋯ + α k v k = 0 \alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 + \cdots + \alpha_k \mathbf{v}_k = \mathbf{0} α 1 v 1 + α 2 v 2 + ⋯ + α k v k = 0 Implies α 1 = α 2 = ⋯ = α k = 0 \alpha_1 = \alpha_2 = \cdots = \alpha_k = 0 α 1 = α 2 = ⋯ = α k = 0 linearly dependent .
Proposition 2.1 (Equivalent formulations). The following are equivalent for vectors v 1 , … , v k ∈ V \mathbf{v}_1, \ldots, \mathbf{v}_k \in V v 1 , … , v k ∈ V
{ v 1 , … , v k } \{\mathbf{v}_1, \ldots, \mathbf{v}_k\} { v 1 , … , v k } No v j \mathbf{v}_j v j If ∑ i = 1 k α i v i = ∑ i = 1 k β i v i \sum_{i=1}^k \alpha_i \mathbf{v}_i = \sum_{i=1}^k \beta_i \mathbf{v}_i ∑ i = 1 k α i v i = ∑ i = 1 k β i v i α i = β i \alpha_i = \beta_i α i = β i i i i Proof. (1 ⇒ \Rightarrow ⇒ v j = ∑ i ≠ j α i v i \mathbf{v}_j = \sum_{i \neq j} \alpha_i \mathbf{v}_i v j = ∑ i = j α i v i ∑ i ≠ j α i v i − v j = 0 \sum_{i \neq j} \alpha_i \mathbf{v}_i - \mathbf{v}_j = \mathbf{0} ∑ i = j α i v i − v j = 0
(2 ⇒ \Rightarrow ⇒ ∑ ( α i − β i ) v i = 0 \sum (\alpha_i - \beta_i)\mathbf{v}_i = \mathbf{0} ∑ ( α i − β i ) v i = 0 α i = β i \alpha_i = \beta_i α i = β i i i i
(3 ⇒ \Rightarrow ⇒ ∑ α i v i = 0 = ∑ 0 ⋅ v i \sum \alpha_i \mathbf{v}_i = \mathbf{0} = \sum 0 \cdot \mathbf{v}_i ∑ α i v i = 0 = ∑ 0 ⋅ v i α i = 0 \alpha_i = 0 α i = 0 i i i ■ \blacksquare ■
The span of a set S ⊆ V S \subseteq V S ⊆ V s p a n ( S ) \mathrm{span}(S) span ( S ) S S S
s p a n ( S ) = { ∑ i = 1 k α i v i : k ∈ N , α i ∈ F , v i ∈ S } \mathrm{span}(S) = \left\{ \sum_{i=1}^k \alpha_i \mathbf{v}_i : k \in \mathbb{N},\, \alpha_i \in F,\, \mathbf{v}_i \in S \right\} span ( S ) = { i = 1 ∑ k α i v i : k ∈ N , α i ∈ F , v i ∈ S } Proposition 2.2. s p a n ( S ) \mathrm{span}(S) span ( S ) V V V s p a n ( S ) \mathrm{span}(S) span ( S ) S S S W W W S ⊆ W S \subseteq W S ⊆ W s p a n ( S ) ⊆ W \mathrm{span}(S) \subseteq W span ( S ) ⊆ W
Proof. s p a n ( S ) \mathrm{span}(S) span ( S ) 0 = 0 ⋅ v \mathbf{0} = 0 \cdot \mathbf{v} 0 = 0 ⋅ v v ∈ S \mathbf{v} \in S v ∈ S W W W S S S S S S s p a n ( S ) ⊆ W \mathrm{span}(S) \subseteq W span ( S ) ⊆ W ■ \blacksquare ■
A set B ⊆ V B \subseteq V B ⊆ V basis for V V V
B B B s p a n ( B ) = V \mathrm{span}(B) = V span ( B ) = V Theorem 2.1. Every vector space has a basis. All bases of a finite-dimensional vector space have The same number of elements.
The dimension of V V V dim ( V ) \dim(V) dim ( V ) V V V
Lemma 2.3 (Steinitz Exchange Lemma). Let { u 1 , … , u k } \{\mathbf{u}_1, \ldots, \mathbf{u}_k\} { u 1 , … , u k } V V V { w 1 , … , w m } \{\mathbf{w}_1, \ldots, \mathbf{w}_m\} { w 1 , … , w m } V V V k ≤ m k \leq m k ≤ m w j \mathbf{w}_j w j
{ u 1 , … , u k , w k + 1 , … , w m } \{\mathbf{u}_1, \ldots, \mathbf{u}_k, \mathbf{w}_{k+1}, \ldots, \mathbf{w}_m\} { u 1 , … , u k , w k + 1 , … , w m } Also spans V V V
Proof. We proceed by induction on k k k k = 0 k = 0 k = 0
Assume the result holds for k − 1 k - 1 k − 1 { u 1 , … , u k } \{\mathbf{u}_1, \ldots, \mathbf{u}_k\} { u 1 , … , u k } u k ≠ 0 \mathbf{u}_k \neq \mathbf{0} u k = 0 u k ∈ s p a n { w 1 , … , w m } \mathbf{u}_k \in \mathrm{span}\{\mathbf{w}_1, \ldots, \mathbf{w}_m\} u k ∈ span { w 1 , … , w m } w j \mathbf{w}_j w j V V V u k = ∑ j = 1 m α j w j \mathbf{u}_k = \sum_{j=1}^m \alpha_j \mathbf{w}_j u k = ∑ j = 1 m α j w j α j ∈ F \alpha_j \in F α j ∈ F α j \alpha_j α j
After relabelling, assume α 1 ≠ 0 \alpha_1 \neq 0 α 1 = 0 w 1 = α 1 − 1 ( u k − ∑ j = 2 m α j w j ) \mathbf{w}_1 = \alpha_1^{-1}(\mathbf{u}_k - \sum_{j=2}^m \alpha_j \mathbf{w}_j) w 1 = α 1 − 1 ( u k − ∑ j = 2 m α j w j ) w 1 ∈ s p a n { u k , w 2 , … , w m } \mathbf{w}_1 \in \mathrm{span}\{\mathbf{u}_k, \mathbf{w}_2, \ldots, \mathbf{w}_m\} w 1 ∈ span { u k , w 2 , … , w m }
s p a n { w 1 , … , w m } = s p a n { u k , w 2 , … , w m } = V \mathrm{span}\{\mathbf{w}_1, \ldots, \mathbf{w}_m\} = \mathrm{span}\{\mathbf{u}_k, \mathbf{w}_2, \ldots, \mathbf{w}_m\} = V span { w 1 , … , w m } = span { u k , w 2 , … , w m } = V Now { u 1 , … , u k − 1 } \{\mathbf{u}_1, \ldots, \mathbf{u}_{k-1}\} { u 1 , … , u k − 1 } { u k , w 2 , … , w m } \{\mathbf{u}_k, \mathbf{w}_2, \ldots, \mathbf{w}_m\} { u k , w 2 , … , w m } V V V k − 1 ≤ m − 1 k - 1 \leq m - 1 k − 1 ≤ m − 1 k ≤ m k \leq m k ≤ m { u 1 , … , u k − 1 , w k , … , w m } \{\mathbf{u}_1, \ldots, \mathbf{u}_{k-1}, \mathbf{w}_k, \ldots, \mathbf{w}_m\} { u 1 , … , u k − 1 , w k , … , w m } V V V u k \mathbf{u}_k u k { u 1 , … , u k , w k + 1 , … , w m } \{\mathbf{u}_1, \ldots, \mathbf{u}_k, \mathbf{w}_{k+1}, \ldots, \mathbf{w}_m\} { u 1 , … , u k , w k + 1 , … , w m } V V V ■ \blacksquare ■
Theorem 2.4 (Dimension is Well-Defined). If V V V V V V
Proof. Let B 1 \mathcal{B}_1 B 1 B 2 \mathcal{B}_2 B 2 ∣ B 1 ∣ = k \lvert\mathcal{B}_1\rvert = k ∣ B 1 ∣ = k ∣ B 2 ∣ = m \lvert\mathcal{B}_2\rvert = m ∣ B 2 ∣ = m B 1 \mathcal{B}_1 B 1 B 2 \mathcal{B}_2 B 2 k ≤ m k \leq m k ≤ m m ≤ k m \leq k m ≤ k k = m k = m k = m ■ \blacksquare ■
Theorem 2.5 (Dimension Formula). If U U U W W W V V V
dim ( U + W ) = dim ( U ) + dim ( W ) − dim ( U ∩ W ) \dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W) dim ( U + W ) = dim ( U ) + dim ( W ) − dim ( U ∩ W ) Theorem 2.6 (Rank-Nullity Theorem). Let A ∈ M m × n ( F ) A \in \mathcal{M}_{m \times n}(F) A ∈ M m × n ( F )
r a n k ( A ) + n u l l i t y ( A ) = n \mathrm{rank}(A) + \mathrm{nullity}(A) = n rank ( A ) + nullity ( A ) = n Where r a n k ( A ) = dim ( c o l ( A ) ) \mathrm{rank}(A) = \dim(\mathrm{col}(A)) rank ( A ) = dim ( col ( A )) n u l l i t y ( A ) = dim ( n u l l ( A ) ) \mathrm{nullity}(A) = \dim(\mathrm{null}(A)) nullity ( A ) = dim ( null ( A ))
Proof. Let { v 1 , … , v k } \{\mathbf{v}_1, \ldots, \mathbf{v}_k\} { v 1 , … , v k } n u l l ( A ) \mathrm{null}(A) null ( A ) k = n u l l i t y ( A ) k = \mathrm{nullity}(A) k = nullity ( A ) { v 1 , … , v k , v k + 1 , … , v n } \{\mathbf{v}_1, \ldots, \mathbf{v}_k, \mathbf{v}_{k+1}, \ldots, \mathbf{v}_n\} { v 1 , … , v k , v k + 1 , … , v n } F n F^n F n
We claim that { A v k + 1 , … , A v n } \{A\mathbf{v}_{k+1}, \ldots, A\mathbf{v}_n\} { A v k + 1 , … , A v n } c o l ( A ) \mathrm{col}(A) col ( A )
Spanning: For any y ∈ c o l ( A ) \mathbf{y} \in \mathrm{col}(A) y ∈ col ( A ) x ∈ F n \mathbf{x} \in F^n x ∈ F n y = A x \mathbf{y} = A\mathbf{x} y = A x x = ∑ i = 1 n α i v i \mathbf{x} = \sum_{i=1}^n \alpha_i \mathbf{v}_i x = ∑ i = 1 n α i v i
y = A ( ∑ i = 1 n α i v i ) = ∑ i = 1 n α i A v i = ∑ i = k + 1 n α i A v i \mathbf{y} = A\left(\sum_{i=1}^n \alpha_i \mathbf{v}_i\right) = \sum_{i=1}^n \alpha_i A\mathbf{v}_i = \sum_{i=k+1}^n \alpha_i A\mathbf{v}_i y = A ( i = 1 ∑ n α i v i ) = i = 1 ∑ n α i A v i = i = k + 1 ∑ n α i A v i Since A v i = 0 A\mathbf{v}_i = \mathbf{0} A v i = 0 i ≤ k i \leq k i ≤ k
Linear independence: If ∑ i = k + 1 n α i A v i = 0 \sum_{i=k+1}^n \alpha_i A\mathbf{v}_i = \mathbf{0} ∑ i = k + 1 n α i A v i = 0 A ( ∑ i = k + 1 n α i v i ) = 0 A\left(\sum_{i=k+1}^n \alpha_i \mathbf{v}_i\right) = \mathbf{0} A ( ∑ i = k + 1 n α i v i ) = 0 ∑ i = k + 1 n α i v i ∈ n u l l ( A ) \sum_{i=k+1}^n \alpha_i \mathbf{v}_i \in \mathrm{null}(A) ∑ i = k + 1 n α i v i ∈ null ( A ) { v 1 , … , v k } \{\mathbf{v}_1, \ldots, \mathbf{v}_k\} { v 1 , … , v k } ∑ i = k + 1 n α i v i = ∑ i = 1 k β i v i \sum_{i=k+1}^n \alpha_i \mathbf{v}_i = \sum_{i=1}^k \beta_i \mathbf{v}_i ∑ i = k + 1 n α i v i = ∑ i = 1 k β i v i β i \beta_i β i ∑ i = 1 n ( − β i ) v i + ∑ i = k + 1 n α i v i = 0 \sum_{i=1}^n (-\beta_i)\mathbf{v}_i + \sum_{i=k+1}^n \alpha_i \mathbf{v}_i = \mathbf{0} ∑ i = 1 n ( − β i ) v i + ∑ i = k + 1 n α i v i = 0 α i = 0 \alpha_i = 0 α i = 0 i ≥ k + 1 i \geq k + 1 i ≥ k + 1
Therefore r a n k ( A ) = n − k = n − n u l l i t y ( A ) \mathrm{rank}(A) = n - k = n - \mathrm{nullity}(A) rank ( A ) = n − k = n − nullity ( A ) ■ \blacksquare ■
Problem. Find a basis for and the dimension of the subspace W = s p a n { ( 1 , 2 , − 1 , 0 ) , ( 3 , 1 , 0 , 2 ) , ( − 1 , 3 , − 2 , − 2 ) } W = \mathrm{span}\{(1, 2, -1, 0), (3, 1, 0, 2), (-1, 3, -2, -2)\} W = span {( 1 , 2 , − 1 , 0 ) , ( 3 , 1 , 0 , 2 ) , ( − 1 , 3 , − 2 , − 2 )} R 4 \mathbb{R}^4 R 4
Solution Form the matrix whose rows are the given vectors and row-reduce:
( 1 2 − 1 0 3 1 0 2 − 1 3 − 2 − 2 ) → R 2 − 3 R 1 ( 1 2 − 1 0 0 − 5 3 2 − 1 3 − 2 − 2 ) \begin{pmatrix} 1 & 2 & -1 & 0 \\ 3 & 1 & 0 & 2 \\ -1 & 3 & -2 & -2 \end{pmatrix} \xrightarrow{R_2 - 3R_1} \begin{pmatrix} 1 & 2 & -1 & 0 \\ 0 & -5 & 3 & 2 \\ -1 & 3 & -2 & -2 \end{pmatrix} 1 3 − 1 2 1 3 − 1 0 − 2 0 2 − 2 R 2 − 3 R 1 1 0 − 1 2 − 5 3 − 1 3 − 2 0 2 − 2 → R 3 + R 1 ( 1 2 − 1 0 0 − 5 3 2 0 5 − 3 − 2 ) → R 3 + R 2 ( 1 2 − 1 0 0 − 5 3 2 0 0 0 0 ) \xrightarrow{R_3 + R_1} \begin{pmatrix} 1 & 2 & -1 & 0 \\ 0 & -5 & 3 & 2 \\ 0 & 5 & -3 & -2 \end{pmatrix} \xrightarrow{R_3 + R_2} \begin{pmatrix} 1 & 2 & -1 & 0 \\ 0 & -5 & 3 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix} R 3 + R 1 1 0 0 2 − 5 5 − 1 3 − 3 0 2 − 2 R 3 + R 2 1 0 0 2 − 5 0 − 1 3 0 0 2 0 The row echelon form has two non-zero rows, so dim ( W ) = 2 \dim(W) = 2 dim ( W ) = 2 { ( 1 , 2 , − 1 , 0 ) , ( 0 , − 5 , 3 , 2 ) } \{(1, 2, -1, 0), (0, -5, 3, 2)\} {( 1 , 2 , − 1 , 0 ) , ( 0 , − 5 , 3 , 2 )} ■ \blacksquare ■
Problem. Find a basis for the null space of
A = ( 1 2 1 − 1 2 4 0 1 0 0 1 3 ) A = \begin{pmatrix} 1 & 2 & 1 & -1 \\ 2 & 4 & 0 & 1 \\ 0 & 0 & 1 & 3 \end{pmatrix} A = 1 2 0 2 4 0 1 0 1 − 1 1 3 Solution Row-reduce A A A
( 1 2 1 − 1 2 4 0 1 0 0 1 3 ) → R 2 − 2 R 1 ( 1 2 1 − 1 0 0 − 2 3 0 0 1 3 ) → R 3 + R 2 / 2 ( 1 2 1 − 1 0 0 − 2 3 0 0 0 9 / 2 ) \begin{pmatrix} 1 & 2 & 1 & -1 \\ 2 & 4 & 0 & 1 \\ 0 & 0 & 1 & 3 \end{pmatrix} \xrightarrow{R_2 - 2R_1} \begin{pmatrix} 1 & 2 & 1 & -1 \\ 0 & 0 & -2 & 3 \\ 0 & 0 & 1 & 3 \end{pmatrix} \xrightarrow{R_3 + R_2/2} \begin{pmatrix} 1 & 2 & 1 & -1 \\ 0 & 0 & -2 & 3 \\ 0 & 0 & 0 & 9/2 \end{pmatrix} 1 2 0 2 4 0 1 0 1 − 1 1 3 R 2 − 2 R 1 1 0 0 2 0 0 1 − 2 1 − 1 3 3 R 3 + R 2 /2 1 0 0 2 0 0 1 − 2 0 − 1 3 9/2 This has pivots in columns 1, 3, and 4. The free variable is x 2 x_2 x 2 x 2 = t x_2 = t x 2 = t x 4 = 0 x_4 = 0 x 4 = 0 x 3 = 0 x_3 = 0 x 3 = 0 x 1 = − 2 t x_1 = -2t x 1 = − 2 t { t ( − 2 , 1 , 0 , 0 ) : t ∈ R } \{t(-2, 1, 0, 0) : t \in \mathbb{R}\} { t ( − 2 , 1 , 0 , 0 ) : t ∈ R } { ( − 2 , 1 , 0 , 0 ) } \{(-2, 1, 0, 0)\} {( − 2 , 1 , 0 , 0 )} ■ \blacksquare ■
Problem. Determine whether the vectors v 1 = ( 1 , 2 , 3 ) \mathbf{v}_1 = (1, 2, 3) v 1 = ( 1 , 2 , 3 ) v 2 = ( 4 , 5 , 6 ) \mathbf{v}_2 = (4, 5, 6) v 2 = ( 4 , 5 , 6 ) v 3 = ( 7 , 8 , 9 ) \mathbf{v}_3 = (7, 8, 9) v 3 = ( 7 , 8 , 9 ) R 3 \mathbb{R}^3 R 3
Solution Form the matrix A = [ v 1 ∣ v 2 ∣ v 3 ] A = [\mathbf{v}_1 \mid \mathbf{v}_2 \mid \mathbf{v}_3] A = [ v 1 ∣ v 2 ∣ v 3 ]
det ( A ) = 1 ( 45 − 48 ) − 2 ( 36 − 42 ) + 3 ( 32 − 35 ) = − 3 + 12 − 9 = 0 \det(A) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = -3 + 12 - 9 = 0 det ( A ) = 1 ( 45 − 48 ) − 2 ( 36 − 42 ) + 3 ( 32 − 35 ) = − 3 + 12 − 9 = 0 Since det ( A ) = 0 \det(A) = 0 det ( A ) = 0 { v 1 , v 2 , v 3 } \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\} { v 1 , v 2 , v 3 } v 3 − 2 v 2 + v 1 = 0 \mathbf{v}_3 - 2\mathbf{v}_2 + \mathbf{v}_1 = \mathbf{0} v 3 − 2 v 2 + v 1 = 0
■ \blacksquare ■
whose Columns are those vectors. If det ≠ 0 \det \neq 0 det = 0 det = 0 \det = 0 det = 0
Problem. Let V = P 3 ( R ) V = \mathcal{P}_3(\mathbb{R}) V = P 3 ( R ) W = { p ∈ P 3 : p ( 1 ) = p ( − 1 ) = 0 } W = \{p \in \mathcal{P}_3 : p(1) = p(-1) = 0\} W = { p ∈ P 3 : p ( 1 ) = p ( − 1 ) = 0 }
Solution Write p ( x ) = a x 3 + b x 2 + c x + d p(x) = ax^3 + bx^2 + cx + d p ( x ) = a x 3 + b x 2 + c x + d
p ( 1 ) = a + b + c + d = 0 p(1) = a + b + c + d = 0 p ( 1 ) = a + b + c + d = 0 p ( − 1 ) = − a + b − c + d = 0 p(-1) = -a + b - c + d = 0 p ( − 1 ) = − a + b − c + d = 0
Adding: 2 b + 2 d = 0 2b + 2d = 0 2 b + 2 d = 0 d = − b d = -b d = − b 2 a + 2 c = 0 2a + 2c = 0 2 a + 2 c = 0 c = − a c = -a c = − a
Therefore p ( x ) = a x 3 + b x 2 − a x − b = a ( x 3 − x ) + b ( x 2 − 1 ) p(x) = ax^3 + bx^2 - ax - b = a(x^3 - x) + b(x^2 - 1) p ( x ) = a x 3 + b x 2 − a x − b = a ( x 3 − x ) + b ( x 2 − 1 )
A basis for W W W { x 3 − x , x 2 − 1 } \{x^3 - x, x^2 - 1\} { x 3 − x , x 2 − 1 } dim ( W ) = 2 \dim(W) = 2 dim ( W ) = 2
If you get this wrong, revise: Section 2.7 (Worked Examples).
Linear independence captures the idea that no vector in the set is “redundant.” If you have three vectors in R 3 \mathbb{R}^3 R 3 R 3 \mathbb{R}^3 R 3
Span answers the question: “What is the largest subspace I can reach using these vectors?” If you think of each vector as an arrow, the span is the set of all destinations you can reach by traveling along these arrows (and their negatives), possibly stretching them. For two non-parallel vectors in R 3 \mathbb{R}^3 R 3
Basis combines both ideas: a basis is a set that spans the entire space with no redundancy. It is a “minimal spanning set” and a “maximal linearly independent set” simultaneously.
Dimension counts the number of independent directions. R 3 \mathbb{R}^3 R 3 e 1 , e 2 , e 3 \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 e 1 , e 2 , e 3
The key theorem (Steinitz exchange) says that any linearly independent set can be extended to a basis by swapping in vectors from any spanning set. This is analogous to the following physical process: given a set of non-redundant directions and a set of directions that cover all of space, you can systematically replace redundant spanning directions with your independent ones.
Problem. Find a basis for the row space of A = ( 1 2 3 4 5 6 7 8 9 ) A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} A = 1 4 7 2 5 8 3 6 9
Solution Row-reduce A A A
( 1 2 3 4 5 6 7 8 9 ) → R 2 − 4 R 1 , R 3 − 7 R 1 ( 1 2 3 0 − 3 − 6 0 − 6 − 12 ) \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} \xrightarrow{R_2 - 4R_1, R_3 - 7R_1} \begin{pmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{pmatrix} 1 4 7 2 5 8 3 6 9 R 2 − 4 R 1 , R 3 − 7 R 1 1 0 0 2 − 3 − 6 3 − 6 − 12 → R 3 − 2 R 2 ( 1 2 3 0 − 3 − 6 0 0 0 ) \xrightarrow{R_3 - 2R_2} \begin{pmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & 0 & 0 \end{pmatrix} R 3 − 2 R 2 1 0 0 2 − 3 0 3 − 6 0 The non-zero rows form a basis for the row space: { ( 1 , 2 , 3 ) , ( 0 , − 3 , − 6 ) } \{(1, 2, 3), (0, -3, -6)\} {( 1 , 2 , 3 ) , ( 0 , − 3 , − 6 )} { ( 1 , 2 , 3 ) , ( 0 , 1 , 2 ) } \{(1, 2, 3), (0, 1, 2)\} {( 1 , 2 , 3 ) , ( 0 , 1 , 2 )}
Note that r a n k ( A ) = 2 < 3 = n \mathrm{rank}(A) = 2 < 3 = n rank ( A ) = 2 < 3 = n A A A 3 − 2 = 1 3 - 2 = 1 3 − 2 = 1 { ( 1 , − 2 , 1 ) T } \{(1, -2, 1)^T\} {( 1 , − 2 , 1 ) T }
Connection to determinants: det ( A ) = 0 \det(A) = 0 det ( A ) = 0 3 = 2 ( 2 ) − 1 3 = 2(2) - 1 3 = 2 ( 2 ) − 1 6 = 2 ( 5 ) − 4 6 = 2(5) - 4 6 = 2 ( 5 ) − 4 9 = 2 ( 8 ) − 7 9 = 2(8) - 7 9 = 2 ( 8 ) − 7 ■ \blacksquare ■
Linear independence of infinitely many vectors. The definition only directly applies to finite subsets. A set S S S S S S Dimension and spanning. A set of n n n R n \mathbb{R}^n R n R n \mathbb{R}^n R n n n n R n \mathbb{R}^n R n R n \mathbb{R}^n R n The empty set spans { 0 } \{\mathbf{0}\} { 0 } The span of the empty set is the trivial subspace, and the empty set is a basis for { 0 } \{\mathbf{0}\} { 0 } Row space and column space have the same dimension (rank), but are different subspaces. The row space is a subspace of R n \mathbb{R}^n R n A A A R m \mathbb{R}^m R m A A A A basis is not unique. Any set of n n n R n \mathbb{R}^n R n Claim True? Counterexample Any set of vectors that spans is a basis no it may not be independent; { ( 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) } \{(1,0),(0,1),(1,1)\} {( 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 )} R 2 \mathbb{R}^2 R 2 Any independent set is a basis no it may not span; a single vector in R 2 \mathbb{R}^2 R 2 A subset of a basis is a basis no it is independent but does not span Adding a vector to a basis gives a basis no if it is already in the span the set becomes dependent Rank equals the number of non-zero rows no that is the number of non-zero rows in row echelon form ; before elimination non-zero rows can be dependent Rank-nullity applies to any function no it needs linearity; a general function has neither a kernel that is a subspace nor an image that is one An infinite-dimensional space can have a finite basis no the existence of a finite basis is the definition of finite-dimensional Two subspaces of dimensions p p p q q q p + q p+q p + q no they may overlap; dim ( U + W ) = dim U + dim V − dim ( U ∩ V ) \dim(U+W) = \dim U + \dim V - \dim(U\cap V) dim ( U + W ) = dim U + dim V − dim ( U ∩ V )
The first four rows are all the same trap from different angles: a basis needs two properties, and each is easy to satisfy on its own. The definition is deliberately two clauses long.
The last row is the one that catches people who remember the dimension formula but forget the intersection term, and it is worth writing out the two-plane example in R 3 \mathbb{R}^3 R 3
Infinite dimensions. The exchange lemma needs a finiteness argument, and in infinite dimensions the statement “every independent set can be extended to a basis” is equivalent to the axiom of choice. Hamel bases exist but are unconstructible, which is why analysis uses Schauder bases instead.Module theory. Drop the field and keep the ring, and bases stop existing: Z \mathbb{Z} Z Matroid theory. Independence axiomatised. Every matrix gives a matroid on its columns, and the theory abstracts exactly the properties that make exchange arguments work.Dimension in quantum mechanics. Hilbert spaces can be infinite- dimensional, and the finite-dimensional intuition about bases and dimension carries over only with care — separability, orthonormal bases and the spectral theorem are the replacements.Mix these with Vectors and Vector Spaces , Eigenvalues and Eigenvectors , Systems of Linear Equations , Measure Theory and Representation Theory .
(Vectors.) Determine whether { ( 1 , 1 , 0 ) , ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) } \{(1,1,0), (0,1,1), (1,0,1)\} {( 1 , 1 , 0 ) , ( 0 , 1 , 1 ) , ( 1 , 0 , 1 )} R 3 \mathbb{R}^3 R 3 (Eigenvalues.) Show that eigenvectors belonging to distinct eigenvalues are linearly independent, by induction on the number of eigenvalues. This is why an n × n n \times n n × n n n n (Systems of equations.) A consistent system A x = b Ax = b A x = b A A A r r r n n n (Measure theory.) State what goes wrong with dimension for the space of functions spanned by { 1 , x , x 2 , … } \{1, x, x^2, \ldots\} { 1 , x , x 2 , … } (Representation theory.) A representation of a group on a vector space has a dimension. Explain why irreducible representations have dimensions that are constrained by the group’s order, referencing the sum-of-squares theorem. (Research.) Prove that any two bases of a vector space have the same cardinality in the infinite-dimensional case, and explain where the proof requires the axiom of choice. Vectors and Vector Spaces Matrices Systems of Linear Equations 