Inner Product Spaces and Orthogonality: Examples and Counterexamples

Finite-sum algebra gives linearity, conjugate symmetry, and $⟨x,x{⟩}_a={\sum }_k{a}_k\mathrm{\mid }{x}_k{\mathrm{\mid }}^2\ge 0$. Because every $a_k>0$, this sum vanishes exactly when every coordinate of $x$ vanishes, so [L1] holds. For $n=0$, [L4] leaves only the zero vector.

For weights $(1,4)$, [L2] gives $\parallel (1,1){\parallel }_a^2=1+4=5$. Put $u=(1,1)$, $v=(1,0)$ and $W=\mathrm{span}(u)$. Then $e=\parallel u{\parallel }_a^{-1}u$ is an orthonormal basis of $W$, so [L3] gives $P_{W}v=⟨v,e{⟩}_{a}e=⟨v,u{⟩}_a⟨u,u{⟩}_a^{-1}u$. Here $⟨v,u{⟩}_a=1$ and $⟨u,u{⟩}_a=5$, so $P_{W}v=(1\mathrm{/}5)u$. With both weights $1$ the same computation uses $⟨v,u⟩=1$ and $⟨u,u⟩=2$ and gives $(1\mathrm{/}2)u$.

Entrywise operations and finite-sum algebra give linearity in $A$. Applying conjugation termwise and using [L3] gives conjugate symmetry.

On the diagonal, [L3] gives the displayed sum of squared moduli. It is nonnegative and vanishes exactly when every entry vanishes, which by [L1] means $A=0$. Hence [L2] is satisfied.

If $m=0$ or $n=0$, the matrix space contains only its zero matrix and [L4] makes the formula zero, so definiteness remains valid.

By [L3], the displayed formula is bilinear and symmetric. Also $⟨p,p⟩={\sum }_{k<4}p(t_k{)}^2\ge 0$.

If $⟨p,p⟩=0$, every real square in the sum is zero, so $p(t_k)=0$ for four distinct $t_k$. If $p$ were nonzero, [L2] would give at most three roots because $\mathrm{deg}p\le 3$. Thus $p=0$, proving definiteness and hence [L1].

By [L1], $\parallel v_0\parallel =\sqrt{3}$, giving $e_0$. The second residual is $u_1=v_1-⟨v_1,e_0⟩e_0=(-1,0,1)$, whose norm is $\sqrt{2}$, giving $e_1$.

The third residual is $u_2=v_2-⟨v_2,e_0⟩e_0-⟨v_2,e_1⟩e_1=(1\mathrm{/}2,-1,1\mathrm{/}2)$. Its norm is $\sqrt{6}\mathrm{/}2$, so normalisation gives the displayed $e_2$.

Direct dot products give $⟨e_i,e_j⟩={\delta }_{ij}$. Moreover $e_0$ is a nonzero multiple of $v_0$, $u_1=v_1-v_0$, and $u_2$ differs from $v_2$ by a combination of the first two residual directions, verifying all three successive-span equalities from [L2].

From [L1], $\parallel v_0\parallel =\sqrt{2}$, so $e_0=v_0\mathrm{/}\sqrt{2}$. The projection coefficient is $⟨v_1,e_0⟩=(1-i)\mathrm{/}\sqrt{2}$.

Subtraction gives $u_1=v_1-⟨v_1,e_0⟩e_0=((1+i)\mathrm{/}2,(1-i)\mathrm{/}2)$. By [L3], $\parallel u_1{\parallel }^2=1$, so $e_1=u_1$.

Direct use of [L1] gives $⟨e_1,e_0⟩=0$ and, by conjugate symmetry, $⟨e_0,e_1⟩=0$. Both vectors have norm one, so the displayed list is orthonormal.

The columns are $a_0=(1,1)$ and $a_1=(1,0)$. By [L2]–[L3], $q_0=(1,1)\mathrm{/}\sqrt{2}$, while $a_1-⟨a_1,q_0⟩q_0=(1\mathrm{/}2,-1\mathrm{/}2)$ normalises to $q_1=(1,-1)\mathrm{/}\sqrt{2}$.

The coefficients are $R_{00}=\sqrt{2}$, $R_{01}=1\mathrm{/}\sqrt{2}$, and $R_{11}=1\mathrm{/}\sqrt{2}$, with $R_{10}=0$. Thus $R$ is upper triangular with positive diagonal.

Direct multiplication gives $QR=A$, and $Q^{\mathsf{T}}Q=I$. This verifies the displayed QR factorisation.

By [L2], the two coefficients are $1$ and $2$, while $\parallel v{\parallel }^2=1^2+2^2+3^2=14$. Hence the Bessel sum is $5$ and the gap is $9$.

The projection onto $\mathrm{span}(e_0,e_1)$ is $(1,2,0)$, leaving $3e_2$ of squared norm $9$. Since $v$ is outside the span, strictness also follows from [L1].

The vectors $u_0=(1,-1,0)\mathrm{/}\sqrt{2}$ and $u_1=(1,1,-2)\mathrm{/}\sqrt{6}$ form an orthonormal basis of $W$ by [L3]. Their coefficients against $v$ are $-1\mathrm{/}\sqrt{2}$ and $-3\mathrm{/}\sqrt{6}$.

Substitution in [L1] gives $P_{W}v=(-1,0,1)$. The residual is $(2,2,2)$, which is orthogonal to $W$ and has squared norm $12$.

For $q\in W$, the residual and $P_{W}v-q$ are orthogonal, so Pythagoras gives the displayed equality. It is minimised uniquely at $q=P_{W}v$, in agreement with [L2].

By [L2], $⟨z,(1-i,2i,3)⟩=(1+i)z_0-2i{z}_1+3z_2=f(z)$. Uniqueness in [L1] identifies the displayed vector as the representative.

If the functional is multiplied by $i$, its representative is $-iw$, since $⟨z,-iw⟩=i⟨z,w⟩$. This explicitly exhibits the conjugate-linear dependence asserted by [L1].

Transpose $A$ and conjugate each entry. This gives the displayed $2\times 3$ matrix $A^{\ast }$, exactly as [L1] prescribes.

For generic $x\in {\mathbb{C}}^2$ and $y\in {\mathbb{C}}^3$, expansion with [L2] gives $⟨Ax,y⟩=x^{\mathsf{T}}{A}^{\mathsf{T}}\bar{y}=⟨x,A^{\ast }y⟩$. Thus the displayed matrix also verifies the adjoint identity directly.

Expanding the determinant of $\left(\begin{array}{cc}⟨u,u⟩& ⟨u,v⟩\\ ⟨v,u⟩& ⟨v,v⟩\end{array}\right)$ and using conjugate symmetry gives the displayed formula. Its nonnegativity and equality case agree with [L1] and [L2].

For $(1,0),(1,2)$, the Gram matrix is $\left(\begin{array}{cc}1& 1\\ 1& 5\end{array}\right)$, whose determinant is $4$; this is positive, and $(1,0),(1,2)$ are independent, as [L1] requires. For $(u,2u)$, the two rows are dependent and the determinant is $0$.

The matrix of $B$ is $\mathrm{diag}(1,-1)$, whose determinant is $-1$, so [L1] makes $B$ nondegenerate. It is indefinite because $B((1,0),(1,0))=1$ and $B((0,1),(0,1))=-1$.

Yet $B(v,v)=1-1=0$. Therefore $B(v,av)=0$ for every $av\in W$, so the nonzero vector $v$ lies in both $W$ and $W^{{\perp }_B}$. Thus their sum is not direct.

This does not contradict [L2]: positivity, not merely nondegeneracy, is the hypothesis that forces the orthogonal direct sum.

By [L2], $⟨i,1⟩=i$, whereas $⟨1,i⟩=-i$. These are unequal.

They are complex conjugates, exactly as [L1] requires. Thus conjugate symmetry, not symmetry, is the correct law.

A singleton is orthogonal because it has no distinct pair to check. But [L2] gives $\parallel (2,0)\parallel =2$, and its span is only a line in ${\mathbb{R}}^2$.

Thus it is neither normalised nor a basis, exposing both missing requirements in [L1]. The empty orthogonal list supplies the same spanning warning in every nonzero space.

Direct multiplication gives $P^2=P$, so $P$ is idempotent.

But $P^{\mathsf{T}}=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)\mathrm{\ne }P$, so [L2] shows that $P$ is not self-adjoint. Hence [L1] shows it is not an orthogonal projection.

Concretely, $\mathrm{im}P=\mathrm{span}(1,0)$ and $\ker P=\mathrm{span}(-1,1)$, whose generators have dot product $-1$, so this is an oblique projection.

By [L1], $(iI{)}^{\ast }=\bar{i}{I}^{\ast }=-iI$, which is not $iI$.

Directly, [L2] gives $⟨iz,w⟩=i⟨z,w⟩=⟨z,-iw⟩$, confirming that the adjoint scalar must be $-i$.