Linear Algebra · Book or PDF

Essential Student Algebra: Volume Four, Linear Algebra

T. S. Blyth & E. F. Robertson · First-year university · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 5 · pp. 68–78

Chapter 5: The Rational and Classical Forms

This section develops duality for vector spaces. A dual space collects all the linear ways of turning vectors into scalars. We study its bases, how it changes when the basis of $V$ changes, and how a linear map produces a transpose map between dual spaces. These ideas explain why a matrix and its transpose have the same rank.

Linear maps and the dual space

For vector spaces $V$ and $W$ over a field $F$ , the set $\mathrm{Lin}(V,W)$ of linear maps from $V$ to $W$ is itself a vector space over $F$ . Operations are done pointwise: $(f+g)(x)=f(x)+g(x)$ and $(\lambda f)(x)=\lambda f(x)$ . Both results are again linear, so the vector space axioms hold. (p. 68)

The most important case is $W=F$ . The dual space is $V^d=\mathrm{Lin}(V,F)$ . Its elements are called linear functionals (or linear forms). Each one sends a vector to a scalar. For $V=\mathbb{R}^3$ , the map $(x,y,z)\mapsto 2x-y+5z$ is a linear functional. (p. 68)

Other natural examples are:

the $i$ -th coordinate projection on $\mathbb{R}^n$ , which sends $(x_1,\dots,x_n)$ to $x_i$ ;
the trace on square matrices, which adds the diagonal entries;
integration of polynomials over a fixed interval, which gives a real number.

Each of these is linear and takes values in the field. (p. 68)

The dual basis

Let $\{v_1,\dots,v_n\}$ be a basis of $V$ . Define $v_i^d$ by $v_i^d(v_j)=\delta_{ij}$ . So $v_i^d$ gives 1 on $v_i$ and 0 on every other basis vector. A linear map is determined by its values on a basis, so this defines $v_i^d$ completely. These maps form a basis of $V^d$ , called the dual basis. (p. 69)

Independence: if $\sum\lambda_i v_i^d=0$ , apply it to $v_j$ . The result is $\lambda_j=0$ .
Spanning: any functional $f$ satisfies $f=\sum f(v_i)\,v_i^d$ .

(p. 69)

The dual basis has $n$ elements when $V$ has a basis of $n$ elements. Hence, for finite-dimensional $V$ , $\dim V^d=\dim V$ . For example, $\mathbb{R}^3$ and its dual both have dimension 3, and $\mathbb{R}_n[X]$ and its dual both have dimension $n+1$ . (p. 69)

Two expansion formulas follow: $x=\sum_i v_i^d(x)\,v_i,\qquad f=\sum_i f(v_i)\,v_i^d.$ The first says $v_i^d(x)$ is the $i$ -th coordinate of $x$ , so the dual basis maps are called coordinate forms. The second says the coefficients of $f$ are just its values on the basis vectors. (pp. 69–70)

Common mistake. It is wrong to think $v_i^d$ depends only on $v_i$ . It must give 0 on all the other basis vectors, so changing any other basis vector changes it. In $\mathbb{R}^2$ with the standard basis, $e_1^d(x,y)=x$ . With the basis $\{(1,0),(1,1)\}$ , the first vector is the same, but the first dual functional becomes $(x,y)\mapsto x-y$ . (pp. 69–70)

Computing dual bases

Standard basis. The dual basis of $\{e_1,\dots,e_n\}$ in $\mathbb{R}^n$ is the set of coordinate projections $\{p_1,\dots,p_n\}$ , because $p_i$ gives 1 on $e_i$ and 0 on the other $e_j$ . (p. 70)

A direct example. For the basis $\{(1,2),(2,3)\}$ of $\mathbb{R}^2$ , write $v_1^d(x,y)=ax+by$ . We need $v_1^d(1,2)=1$ and $v_1^d(2,3)=0$ . This gives $a+2b=1$ and $2a+3b=0$ , so $a=-3$ and $b=2$ . The same method gives $v_1^d(x,y)=-3x+2y$ and $v_2^d(x,y)=2x-y$ . (p. 70)

Evaluation functionals. Take $n+1$ distinct reals $t_1,\dots,t_{n+1}$ and define $\phi_i(p)=p(t_i)$ on $\mathbb{R}_n[X]$ . These form a basis of the dual space. Since the dual has dimension $n+1$ , independence is enough. If $\sum\lambda_i\phi_i=0$ , apply it to $1,X,\dots,X^n$ . This gives a linear system whose matrix has entries $t_i^{k}$ . Its determinant is the Vandermonde product $\prod_{j<i}(t_i-t_j)$ , which is nonzero because the $t$ 's are distinct. So all $\lambda_i=0$ . (pp. 70–71)

Lagrange polynomials. The polynomials $L_j=\prod_{i\ne j}\frac{X-t_i}{t_j-t_i}$ form the basis of $\mathbb{R}_n[X]$ whose dual basis is $\{\phi_j\}$ . We need $L_j(t_i)=\delta_{ij}$ . At $t_j$ every factor equals 1. At $t_i$ with $i\ne j$ , the factor $X-t_i$ vanishes. (p. 71)

Changing the basis

If $P$ is the transition matrix from one basis of $V$ to another, the transition matrix between the corresponding dual bases is $(P^{-1})^t$ . The dual basis transforms by the inverse transpose, not by $P$ . To prove it, express the new dual vectors in terms of the old ones using a matrix $Q$ , then check that $Q^tP=I$ . So $Q=(P^{-1})^t$ . (pp. 71–72)

This gives a practical method for a basis $B$ of $\mathbb{R}^n$ :

Form $P$ with the basis vectors as columns.
Compute $P^{-1}$ .
The $i$ -th row $[a_{i1},\dots,a_{in}]$ of $P^{-1}$ is the functional $(x_1,\dots,x_n)\mapsto\sum_j a_{ij}x_j$ .

For $B=\{(1,2),(2,3)\}$ , the inverse has rows $[-3,2]$ and $[2,-1]$ . This matches the functionals found above. (pp. 72–73)

Pairing and the bidual

Write $\langle x,y^d\rangle=y^d(x)$ . This pairing is linear in each slot: $\langle x+y,z^d\rangle=\langle x,z^d\rangle+\langle y,z^d\rangle,\quad \langle x,y^d+z^d\rangle=\langle x,y^d\rangle+\langle x,z^d\rangle,$ and $\langle\lambda x,y^d\rangle=\lambda\langle x,y^d\rangle=\langle x,\lambda y^d\rangle$ . The notation treats vectors and functionals symmetrically. (p. 73)

The bidual is $V^{dd}=(V^d)^d$ . The canonical map $\alpha_V:V\to V^{dd}$ sends $x$ to $\hat{x}$ , where $\hat{x}(y^d)=y^d(x)$ . In words, $x$ acts on functionals by letting them evaluate at $x$ . Each $\hat{x}$ is linear on $V^d$ , and $x\mapsto\hat{x}$ is linear on $V$ . (p. 73)

If $V$ is finite-dimensional, $\alpha_V$ is an isomorphism. If $\hat{x}=0$ , then $v_i^d(x)=0$ for every $i$ , so all coordinates of $x$ vanish and $x=0$ . So the map is injective. Because $\dim V=\dim V^d=\dim V^{dd}$ , it is also surjective. (pp. 73–74)

We therefore identify $V$ with $V^{dd}$ . The reason is that $\alpha_V$ is natural: it uses only evaluation and no choice of basis. Many isomorphisms between $V$ and $V^d$ need a basis, so they cannot serve for this identification. (p. 74)

The transpose of a linear map

For linear $f:V\to W$ , the transpose is $f^t:W^d\to V^d$ given by $f^t(y^d)=y^d\circ f$ . A functional on $W$ is pulled back to a functional on $V$ . The map $f^t$ is linear because $(y^d+z^d)\circ f=y^d\circ f+z^d\circ f$ , and similarly for scalars. (p. 75)

The key identity is $\langle f(x),y^d\rangle=\langle x,f^t(y^d)\rangle$ . Both sides equal $y^d(f(x))$ . Proofs about transposes rest on it. (p. 75)

If $A$ is the matrix of $f$ relative to chosen bases, then $A^t$ is the matrix of $f^t$ relative to the dual bases. The $(i,j)$ entry of the matrix of $f^t$ is $\langle a_i,f^t(b_j^d)\rangle=\langle f(a_i),b_j^d\rangle$ , which is the $(j,i)$ entry of $A$ . (p. 75)

Properties of transposition:

$(\mathrm{id}_V)^t=\mathrm{id}_{V^d}$ ;
$(f+g)^t=f^t+g^t$ ;
$(g\circ f)^t=f^t\circ g^t$ .

The order reverses because $y^d\mapsto y^d\circ g\circ f$ applies $g^t$ first and then $f^t$ . This mirrors $(AB)^t=B^tA^t$ . (p. 76)

If $f$ is an isomorphism, so is $f^t$ , and $(f^t)^{-1}=(f^{-1})^t$ . Apply the composition rule to $f^{-1}\circ f=\mathrm{id}_V$ and $f\circ f^{-1}=\mathrm{id}_W$ . (p. 76)

Finally, $f^{tt}\circ\alpha_V=\alpha_W\circ f$ , because $f^{tt}(\hat{x})(y^d)=\hat{x}(f^t(y^d))=\langle f(x),y^d\rangle=\widehat{f(x)}(y^d)$ . After the identifications, $f^{tt}=f$ , matching $(A^t)^t=A$ . (p. 76)

Annihilators and rank

A functional $y^d$ annihilates $x$ if $\langle x,y^d\rangle=0$ . The annihilator of a subset $E\subseteq V$ is $E^\circ=\{y^d\in V^d : y^d(x)=0\text{ for all }x\in E\}.$ It is always a subspace of $V^d$ . Also $\{0\}^\circ=V^d$ and $V^\circ=\{0\}$ . (p. 77)

For a subspace $W$ of finite-dimensional $V$ , $\dim W^\circ=\dim V-\dim W$ . To see this, extend a basis $a_1,\dots,a_m$ of $W$ to a basis of $V$ . Then $a_{m+1}^d,\dots,a_n^d$ form a basis of $W^\circ$ . For example, a plane in $\mathbb{R}^3$ has a 1-dimensional annihilator. Also $W\subseteq W^{\circ\circ}$ and both have dimension $m$ , so $W^{\circ\circ}=W$ . (p. 77)

For a linear map $f$ between finite-dimensional spaces (the dimensions of $V$ and $W$ need not be equal): $(\mathrm{Im}\,f)^\circ=\mathrm{Ker}\,f^t,\qquad (\mathrm{Ker}\,f)^\circ=\mathrm{Im}\,f^t.$ The first holds because $y^d$ kills the image of $f$ exactly when $\langle x,f^t(y^d)\rangle=0$ for all $x$ , that is, when $f^t(y^d)=0$ . Applying this to $f^t$ and using $f^{tt}=f$ gives the second. Comparing dimensions with the annihilator formula shows that $f$ and $f^t$ have the same rank. (p. 78)

This gives a matrix result: the row rank and column rank of any matrix over a field are equal. If $A$ represents $f$ , then $A^t$ represents $f^t$ . The column rank of $A$ is $\dim\mathrm{Im}\,f$ . The row rank of $A$ is the column rank of $A^t$ , which is $\dim\mathrm{Im}\,f^t$ . (p. 78)

Check yourself

Find the dual basis of $\{(1,0),(1,1)\}$ in $\mathbb{R}^2$ and explain why $e_1^d$ changed from the standard case.
Why does independence alone prove that the evaluation functionals $\phi_1,\dots,\phi_{n+1}$ form a basis of $\mathbb{R}_n[X]^d$ ?
If $P$ is the transition matrix between two bases of $V$ , what is the transition matrix between the dual bases, and how can you use it to find a dual basis in $\mathbb{R}^n$ ?
Why is the canonical map $V\to V^{dd}$ an isomorphism for finite-dimensional $V$ , and why does it not depend on a basis?
A subspace $W$ of $\mathbb{R}^5$ has dimension 2. What is $\dim W^\circ$ , and how does the transpose help show that row rank equals column rank?