Dual Spaces and Bilinear Forms: Examples and Counterexamples

The matrix columns give $T(e_1)=e_1$ and $T(e_2)=2e_1+e_2$. Therefore $T^{\ast }({\epsilon }_1)$ takes values $1,2$ on $e_1,e_2$, while $T^{\ast }({\epsilon }_2)$ takes values $0,1$.

By the dual-basis expansion [L2], these four values give $T^{\ast }({\epsilon }_1)={\epsilon }_1+2{\epsilon }_2$ and $T^{\ast }({\epsilon }_2)={\epsilon }_2$.

The coordinate columns in step 2.1 give $[T^{\ast }]=\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)$, which independently agrees with [L3].

By [L3], a general functional is $f=a{x}^{\ast }+b{y}^{\ast }+c{z}^{\ast }$. If $f\in U^{\circ }$, evaluation at $(1,0,0)$ and $(0,1,0)$ gives $a=b=0$.

Conversely every $c{z}^{\ast }$ vanishes on $(x,y,0)$, so [L1] gives $U^{\circ }=\mathrm{span}\{z^{\ast }\}$.

As a check, $\dim U=2$ and [L2] gives $\dim U^{\circ }=3-2=1$, agreeing with step 1.2.

By [L1], the monomials $1,x,x^2,\dots$ form an infinite Hamel basis. Let ${\delta }_n$ extract the coefficient of $x^n$, and put $\mathrm{\Phi }=\mathrm{span}\{{\delta }_n:n\ge 0\}$. Define $\varphi ({\sum }_n{a}_n{x}^n)={\sum }_n{a}_n$; finite support makes this a functional. Every member of $\mathrm{\Phi }$ vanishes on all but finitely many monomials, whereas $\varphi (x^n)=1$ for every $n$, so $\varphi \mathrm{\notin }\mathrm{\Phi }$.

Apply the separation statement in [L3] inside $V^{\ast }$ to choose $L\in V^{\ast \ast }$ with $L{\mathrm{\mid }}_{\mathrm{\Phi }}=0$ and $L(\varphi )=1$. If $L=J_V(p)$, then $0=L({\delta }_n)={\delta }_n(p)$ for every $n$, so all coefficients of $p$ vanish and $p=0$; this would give $L=0$, contradicting $L(\varphi )=1$.

Thus $L$ is outside the image, while injectivity follows from [L2]. This explicitly realizes the finite-dimensional boundary in [L2].

If $v\mathrm{\ne }0$, some coordinate of its finite basis expansion is nonzero, so the corresponding $b^{\ast }\in \mathrm{\Phi }$ does not kill $v$. Hence [L1] gives ${}^{\circ }\mathrm{\Phi }=\{0\}$.

By [L3], every nonzero $v$ is detected by some member of $V^{\ast }$, so ${}^{\circ }(V^{\ast })=\{0\}$.

Yet [L2] gives $\mathrm{\Phi }\mathrm{\ne }{V}^{\ast }$, while steps 1.1 and 1.2 give equal preannihilators. These are the required distinct subspaces.

By [L1], $B(e_1,e_2)=1$ but $B(e_2,e_1)=0$, so symmetry fails by [L2].

Also $B(e_1+e_2,e_1+e_2)=1$, so alternation fails by [L2].

The same form therefore witnesses both failures.

Expanding gives $2(x+y{)}^2+3y^2=2x^2+4xy+5y^2=q(x,y)$. With $u=x+y$, $v=y$, old coordinates are $(x,y)=(u-v,v)$, so $P=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)$ is invertible.

Direct multiplication gives $P^{\mathsf{T}}AP=\left(\begin{array}{cc}2& 0\\ 0& 3\end{array}\right)$. Both diagonal entries are positive, and positive rescaling changes this matrix to $I_2$.

By [L2], the inertia is $(2,0,0)$, in agreement with the diagonalization promised by [L1].

The matrix has zero diagonal and is skew-symmetric, so $B(v,v)=0$ by direct expansion. Also $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$.

Multiplication by $A$ kills exactly the vectors $(0,0,z)$, so the radical is $\mathrm{span}\{e_3\}$; the first two columns are independent, giving rank $2$.

Thus $(e_1,e_2)$ is one symplectic pair followed by the radical basis $(e_3)$, exactly the normal form of [L2].

For $v=(x,y)$, $B(v,v)=xy+yx=2xy=0$, so $B$ is alternating and hence symmetric by [L1] and [L2]. Its matrix has determinant $-1=1$ in ${\mathbb{F}}_2$, so it is nondegenerate by [L3].

If $(u,v)$ were an orthogonal basis, alternation would give $B(u,u)=B(v,v)=0$ and orthogonality would give $B(u,v)=B(v,u)=0$. The matrix in that basis would be zero, contradicting nondegeneracy.

Thus this symmetric form has no orthogonal basis, and the characteristic-not-$2$ hypothesis in the diagonalization theorem is essential.

Both $q_0$ and $q_1$ have the required degree-two homogeneity. The polar form of $q_0$ is zero, while for $u=(x,y)$ and $v=(s,t)$, $b_{q_1}(u,v)=(x+s{)}^2-x^2-s^2=2xs=0$ by [L1]. Thus both are quadratic forms with zero polar form.

They are distinct because $q_0(1,0)=0$ while $q_1(1,0)=1$.

Hence polarization is not injective in characteristic $2$.

By [L2], $\det A=(-1)(-1)(1)=1>0$.

But $e_1^{\mathsf{T}}A{e}_1=-1<0$, so [L1] shows that $A$ is not positive definite. Its inertia is $(1,2,0)$.

Thus a positive determinant alone does not imply positive definiteness.

The leading principal minors are ${\mathrm{\Delta }}_1=2$, ${\mathrm{\Delta }}_2=\det \left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)=3$, and ${\mathrm{\Delta }}_3=2(4-1)-(2)=4$. All three are positive.

Independently, for $(x,y,z)\mathrm{\ne }0$, one has $\left(\begin{array}{ccc}x& y& z\end{array}\right)A\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=2{(x+\frac{y}{2})}^2+\frac{3}{2}{(y+\frac{2z}{3})}^2+\frac{4}{3}{z}^2>0$, because the three squared linear forms vanish simultaneously only at $(0,0,0)$.

Hence [L1] proves that $A$ is positive definite.

Both calculations agree, and each boundary minor $k=1,2,3$ has been evaluated explicitly.

Since $P^{\mathsf{T}}AP=[2][1][2]=[4]=C$ and $P$ is invertible, [L1] makes $A$ and $C$ congruent.

By [L3], $\mathrm{tr}A=1$, $\mathrm{tr}C=4$, $\det A=1$, and $\det C=4$. Thus neither trace nor determinant is preserved.

Both matrices nevertheless have rank $1$ and are nondegenerate, in agreement with [L2]; the counterexample isolates exactly the two false invariance claims.