Exercises — Abstract vector spaces — axioms, examples beyond ℝⁿ

The zero vector is whatever element leaves every vector unchanged when added. It is different in each box.

• (a) $0=(0,0,0)$. Then $(x,y,z)+(0,0,0)=(x,y,z)$. ✓
• (b) $0=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$, the zero matrix. Adding it entry-by-entry changes nothing. ✓
• (c) $0=$ the zero function $z(x)=0$ for all $x$. Then $(f+z)(x)=f(x)+0=f(x)$. ✓

The lesson: $0$ is a role, not the digit $0$.

• (a) $\dim P_2=3$, basis $\{1,x,x^2\}$ — three independent "directions" (constant, linear, quadratic).
• (b) $\dim M_{2\times 3}=2\cdot 3=6$, basis the six elementary matrices $E_{ij}$ (a single $1$ in slot $ij$, zeros elsewhere).
• (c) $\dim =2$, basis $\{\sin x,\cos x\}$: every solution is $a\sin x+b\cos x$.

Addition and scaling are coefficient-wise (that is exactly why $P_2$ behaves like ${\mathbb{R}}^3$). $2p=4+6x-2x^2.$ $2p-q=(4-(-1))+(6-1)x+(-2-4)x^2=5+5x-6x^2.$ Coordinate vector: $\left(\begin{array}{c}5\\ 5\\ -6\end{array}\right)$.

Over $\mathbb{R}$: scalars are real, so $z=3\cdot 1+(-2)\cdot i$ gives coordinates $\left(\begin{array}{c}3\\ -2\end{array}\right)$. Over $\mathbb{C}$: scalars are complex, one basis vector $\{1\}$ suffices — $z=(3-2i)\cdot 1$, coordinate $\left(\begin{array}{c}3-2i\end{array}\right)$. The scalar field changes what counts as "one step", so ${\dim }_{\mathbb{R}}\mathbb{C}=2$ but ${\dim }_{\mathbb{C}}\mathbb{C}=1$. (See Fields and scalars.)

Run the three-part subspace test.

1. Zero: $(0,0,0)$ satisfies $0+0+0=0$. ✓ $0\in W$.
2. Closed under $+$: take $u=(x_1,y_1,z_1),v=(x_2,y_2,z_2)$ with each coordinate-sum $0$. Then $(x_1+x_2)+(y_1+y_2)+(z_1+z_2)=(x_1+y_1+z_1)+(x_2+y_2+z_2)=0+0=0.$ So $u+v\in W$. ✓
3. Closed under scaling: $a\cdot (x,y,z)=(ax,ay,az)$ and $ax+ay+az=a(x+y+z)=a\cdot 0=0$. ✓ All three pass, so $W$ is a subspace. (Geometrically it's a plane through the origin; $\dim W=2$.)

No. The fastest disqualifier is the zero test: $(0,0,0)$ gives $0+0+0=0\ne 1$, so $0\notin U$. That already kills it (Axiom 3 fails inside $U$). To make the failure vivid, closure also fails: $(1,0,0)\in U$ and $(0,1,0)\in U$, but their sum $(1,1,0)$ has coordinate-sum $2\ne 1$, so $(1,1,0)\notin U$. In the figure, the plane $U$ (coral) is shifted off the origin; the plane $W$ (mint) passes through it. Only planes through the origin can be subspaces.

Yes — this is a genuine vector space in disguise. Check the requested axioms.

• Closure: product of positives is positive; a positive raised to any real power is positive. ✓
• Axiom 3 (zero vector): we need $e$ with $x\oplus e=x$, i.e. $x\cdot e=x$, so $e=1$. The zero vector is $0=1$. ✓
• Axiom 4 (inverse): need $x\oplus x^{\prime }=1$, i.e. $x\cdot x^{\prime }=1$, so $x^{\prime }=1/x$ (positive). Thus $-v$ here means $1/v$. ✓
• Axiom 5 (scalar identity): $1\odot x=x^1=x$. ✓
• Axiom 8 (distribute over scalar sum): $(a+b)\odot x=x^{a+b}=x^a{x}^b=(a\odot x)\oplus (b\odot x)$. ✓

(The other axioms check similarly.) In fact the map $x\mapsto \ln x$ turns $\oplus \to +$ and $\odot \to$ ordinary scaling, an isomorphism to $\mathbb{R}$ — so $\dim V=1$, basis $\{e\}$ (since $\ln e=1$).

Play the two identities against each other. $0=0+0^{\prime }(\text{since }{0}^{\prime }\text{ is a zero, use it on }v=0)$ $0+0^{\prime }=0^{\prime }+0(\text{Axiom 1, commutativity})$ $0^{\prime }+0=0^{\prime }(\text{since }0\text{ is a zero, use it on }v=0^{\prime })$ Chaining: $0=0^{\prime }$. $■$ The proof used only axioms, so it holds in every vector space.

(a) Differentiate each basis vector and read off coordinates in $\{1,x,x^2\}$: $D(1)=0,D(x)=1,D(x^2)=2x.$ Columns are the coordinate vectors of these outputs: $[D]=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 2\\ 0& 0& 0\end{array}\right).$ (b) $D(p)=0$ means $p^{\prime }=0$, so $p$ is a constant: $\ker D=\{a\cdot 1\}=\mathrm{span}\{1\}$, $\dim \ker D=1$. (c) Outputs are $p^{\prime }$ for $p\in P_2$; derivatives of degree $\le 2$ give all polynomials of degree $\le 1$. So $\mathrm{im}D=P_1=\mathrm{span}\{1,x\}$, $\dim \mathrm{im}D=2$. (d) Rank–nullity: $\dim \ker D+\dim \mathrm{im}D=1+2=3=\dim P_2$. ✓

Orthogonality: $⟨\sin x,\cos x⟩={\int }_{-\pi }^{\pi }\sin x\cos xdx$. Using $\sin x\cos x=\frac{1}{2}\sin 2x$, ${\int }_{-\pi }^{\pi }\frac{1}{2}\sin 2xdx=\frac{1}{2}{\left[-\frac{\cos 2x}{2}\right]}_{-\pi }^{\pi }=\frac{1}{2}\cdot \left(-\frac{\cos 2\pi -\cos (-2\pi )}{2}\right)=0.$ So $\sin x\perp \cos x$. ✓ (Also, the integrand is an odd function, forcing $0$.) Norm-squared: using ${\sin }^{2}x=\frac{1-\cos 2x}{2}$, $⟨\sin x,\sin x⟩={\int }_{-\pi }^{\pi }\frac{1-\cos 2x}{2}dx=\frac{1}{2}[x-\frac{\sin 2x}{2}{]}_{-\pi }^{\pi }=\frac{1}{2}(2\pi -0)=\pi .$ So $\Vert \sin x\Vert =\sqrt{\pi }$. These orthogonal functions are exactly the building blocks of Fourier series.