Finite-Dimensional Normed Spaces

Coordinates Reveal the Structure

For a linear map on a finite-dimensional space, a basis reduces the problem to finitely many scalar coordinates. This makes it possible to control the map using finitely many estimates. It also connects normed-space questions to familiar facts about $\mathbb{R}^n$, such as the Heine–Borel Theorem. In this tutorial we use those connections to establish properties that are special to finite-dimensional spaces.

Recall the Equivalence of Norms in Finite Dimensions theorem from earlier in the course. In particular, if $v_1,\ldots,v_n$ is a basis of a finite-dimensional normed space $V$, then the norm of $\sum_{i=1}^n a_i v_i$ is comparable to the Euclidean norm of its coordinate vector. We will use this result without reproving it. The constants in that comparison may depend on the chosen basis and norm.

Definition: A normed vector space $V$ is finite-dimensional if it has a finite basis $v_1,\ldots,v_n$. Each $x\in V$ then has a unique representation $$ x=\sum_{i=1}^n a_i v_i. $$ The scalars $a_1,\ldots,a_n$ are the coordinates of $x$ in that basis. The zero space has dimension $0$.

For a fixed basis, define the coordinate map $\Phi:\mathbb{R}^n\to V$ by $\Phi(a_1,\ldots,a_n)=\sum_{i=1}^n a_i v_i$. This map is a linear bijection. The earlier equivalence-of-norms result implies that both $\Phi$ and its inverse are continuous. More precisely, for some constants $c,C>0$,

c\|a\|_2\leq \|\Phi(a)\|_V\leq C\|a\|_2 \qquad (a\in\mathbb{R}^n).

These inequalities explain why finite-dimensional normed spaces behave much like Euclidean spaces: a bounded set in $V$ has bounded coordinate vectors, and a convergent sequence of coordinate vectors gives a convergent sequence in $V$. They also ensure that the choice of basis does not change which sequences converge.

Closed Bounded Sets Are Compact

In $\mathbb{R}^n$, the Heine–Borel Theorem says that a set is compact if and only if it is closed and bounded. The same conclusion holds in every finite-dimensional normed space. We prove the direction that is useful most often here: closedness and boundedness imply compactness.

Theorem (Heine–Borel Property in Finite Dimensions): Every closed and bounded subset of a finite-dimensional normed space is compact.

Proof. First suppose $V$ has positive dimension $n$, choose a basis, and use its coordinate map $\Phi:\mathbb{R}^n\to V$. Let $K\subseteq V$ be closed and bounded, and put $A=\Phi^{-1}(K)$. If $x\in K$, then $\|x\|_V\leq M$ for some finite $M$. Writing $x=\Phi(a)$, the coordinate comparison gives $c\|a\|_2\leq M$. Thus $A$ is bounded in $\mathbb{R}^n$.

The set $A$ is also closed. If $a_j\in A$ and $a_j\to a$ in $\mathbb{R}^n$, continuity of $\Phi$ gives $\Phi(a_j)\to\Phi(a)$ in $V$. Every $\Phi(a_j)$ belongs to $K$, so closedness of $K$ implies $\Phi(a)\in K$, and hence $a\in A$. By the Heine–Borel Theorem, $A$ is compact in $\mathbb{R}^n$. Since $\Phi$ is continuous, its image $\Phi(A)=K$ is compact.

If $V$ has dimension $0$, then $V=\{0_V\}$. Every subset of $V$ is either empty or a singleton, and both are compact. This handles the remaining case. $\square$

A useful consequence is that, whenever $V\neq\{0_V\}$, its unit sphere is compact: it is bounded, and it is closed because the norm is continuous. Thus the Attainment on a Compact Unit Sphere theorem from earlier in the course applies to linear maps whose domain is finite-dimensional. The unit sphere of the zero space is empty, so that case should not be included in a claim about maximizing over unit vectors.

Worked Example: A Compact Set of Linear Polynomials

Consider $V=\{p:p(t)=a+bt,\ a,b\in\mathbb{R}\}$, with the supremum norm on $[0,1]$, and let $$ K=\{a+bt: |a|\leq 2,\ |b|\leq 1\}. $$ The constant polynomial $1$ and the polynomial $t$ form a basis of $V$, so $V$ is finite-dimensional. For every $p=a+bt\in K$ and every $t\in[0,1]$, $$ |p(t)|=|a+bt|\leq |a|+|b|t\leq 2+1=3. $$ Therefore $\|p\|_\infty\leq 3$, and $K$ is bounded in $V$.

The coordinate set for $K$ is the rectangle $[-2,2]\times[-1,1]$, which is closed. Equivalently, if $a_j+b_jt\in K$ converges in $V$ to $p$, then the coordinate vectors converge by continuity of the inverse coordinate map; their limit still lies in that rectangle. Thus $K$ is closed in $V$. The Heine–Borel Property in Finite Dimensions now implies that $K$ is compact. The key point is that boundedness was checked in the given norm, while compactness can be concluded using basis coordinates.

Every Linear Map from a Finite-Dimensional Space Is Bounded

A linear map is determined by what it does to a basis. Since each input is a finite linear combination of basis vectors, the triangle inequality bounds the output by a finite sum. The coordinate comparison then controls that sum by the norm of the input.

Theorem (Linear Maps with Finite-Dimensional Domain Are Bounded): Let $V$ be a finite-dimensional normed space, let $W$ be any normed space, and let $T:V\to W$ be linear. Then $T$ is bounded.

Proof. If $V=\{0_V\}$, then $T$ is the zero map and is bounded. Otherwise, choose a basis $v_1,\ldots,v_n$ of $V$. By the coordinate comparison, there is $c>0$ such that if $x=\sum_{i=1}^n a_i v_i$, then $$ \|(a_1,\ldots,a_n)\|_2\leq c^{-1}\|x\|_V. $$ In particular, $|a_i|\leq\|(a_1,\ldots,a_n)\|_2$ for each $i$. By linearity and the triangle inequality in $W$, $$ \|T(x)\|_W =\left\|\sum_{i=1}^n a_iT(v_i)\right\|_W \leq\sum_{i=1}^n |a_i|\|T(v_i)\|_W. $$ Since $|a_i|\leq c^{-1}\|x\|_V$, this gives $$ \|T(x)\|_W \leq c^{-1}\left(\sum_{i=1}^n\|T(v_i)\|_W\right)\|x\|_V. $$ The factor $c^{-1}\sum_{i=1}^n\|T(v_i)\|_W$ is finite and independent of $x$. Thus $T$ is bounded, by the Boundedness Criterion for Linear Maps. $\square$

The codomain $W$ in this theorem need not be finite-dimensional or complete. It is the finite number of coordinates in the domain that makes the argument work. In particular, every linear functional on a finite-dimensional normed space is bounded.

Worked Example: Differentiation on Linear Polynomials

Let $V=\{p(t)=a+bt\}$ carry the supremum norm on $[0,1]$, and let $W=C[0,1]$ carry its supremum norm. Define $D:V\to W$ by $D(p)=p'$, where $p'$ is regarded as a constant function. This map is linear. For $p(t)=a+bt$, we have $p(1)-p(0)=b$, so $$ \|D(p)\|_\infty=|b| =|p(1)-p(0)| \leq |p(1)|+|p(0)| \leq 2\|p\|_\infty. $$ Thus $D$ is bounded with operator norm at most $2$. The polynomial $p(t)=t-\tfrac12$ satisfies $\|p\|_\infty=\tfrac12$ and $\|D(p)\|_\infty=1$. Consequently, $$ \frac{\|D(p)\|_\infty}{\|p\|_\infty}=2, $$ so the operator norm is exactly $2$. This example also illustrates that the codomain may be an infinite-dimensional space even when every linear map from the domain is bounded.

Finite-Dimensional Subspaces Are Closed

Another consequence of coordinate control is that a finite-dimensional subspace cannot omit a limit point of its own convergent sequences. This fact is useful when constructing closed subspaces of function spaces, and it does not require the ambient normed space to be complete.

Theorem (Finite-Dimensional Subspaces Are Closed): Every finite-dimensional linear subspace $W$ of a normed space $V$ is closed in $V$.

Proof. If $W=\{0_V\}$, it is closed. Otherwise, choose a basis $w_1,\ldots,w_m$ of $W$, and suppose a sequence $(y_j)$ in $W$ converges in $V$ to some $y$. Write $$ y_j=\sum_{i=1}^m a_{j,i}w_i. $$ The sequence $(y_j)$ is Cauchy because it converges. The coordinate comparison for the norm on $W$ therefore gives a constant $c>0$ such that $$ c\|(a_{j,1}-a_{k,1},\ldots,a_{j,m}-a_{k,m})\|_2 \leq \|y_j-y_k\|_V. $$ The right-hand side tends to zero as $j,k\to\infty$. Hence the coordinate vectors form a Cauchy sequence in $\mathbb{R}^m$, so they converge to some $(a_1,\ldots,a_m)$.

Set $z=\sum_{i=1}^m a_iw_i$, which belongs to $W$. The finite-sum triangle inequality gives $$ \|y_j-z\|_V \leq\sum_{i=1}^m |a_{j,i}-a_i|\|w_i\|_V\longrightarrow 0. $$ Thus $y_j\to z$ in $V$, while by assumption $y_j\to y$. Uniqueness of norm limits gives $y=z\in W$. Every convergent sequence in $W$ therefore has its limit in $W$, so $W$ is closed. $\square$

Worked Example: The Affine Functions Form a Closed Subspace

In $C[0,1]$ with the supremum norm, let $W=\{a+bt:a,b\in\mathbb{R}\}$. The functions $1$ and $t$ are linearly independent and span $W$, so $W$ has dimension $2$. By the Finite-Dimensional Subspaces Are Closed theorem, $W$ is closed in $C[0,1]$.

The result can also be seen directly. Suppose $p_j(t)=a_j+b_jt$ converges uniformly to $f\in C[0,1]$. Then $$ a_j=p_j(0)\longrightarrow f(0), \qquad a_j+b_j=p_j(1)\longrightarrow f(1). $$ Subtracting the first convergent sequence from the second gives $b_j\to f(1)-f(0)$. For each $t\in[0,1]$, it follows that $$ f(t)=\lim_{j\to\infty}(a_j+b_jt) =f(0)+\bigl(f(1)-f(0)\bigr)t. $$ This is an affine function, so $f\in W$, as required.

What Finite Dimensionality Provides

The proofs above repeatedly use the same mechanism: choose a basis, control the finitely many coordinates, and transfer the resulting estimate back to the given norm. It yields compactness of closed bounded sets, boundedness of every linear map with finite-dimensional domain, and closedness of every finite-dimensional subspace.

These conclusions must not be assumed for arbitrary normed spaces. In particular, the compactness theorem depends on finite-dimensional coordinates and the Heine–Borel Theorem in $\mathbb{R}^n$. Likewise, the boundedness result applies to maps whose domain is finite-dimensional; it does not assert that every linear map from an infinite-dimensional space is bounded. Keeping track of which space is finite-dimensional prevents a common misuse of these results.

Takeaway: In a finite-dimensional normed space, basis coordinates provide control equivalent to Euclidean control. Consequently, closed bounded sets are compact, every linear map from the space into any normed space is bounded, and every finite-dimensional subspace of a normed space is closed.

Check Your Understanding

Use basis coordinates and the results proved above to answer the following questions.

What coordinate comparison follows from the Equivalence of Norms in Finite Dimensions theorem for a fixed basis?
Why does boundedness of a subset of a finite-dimensional normed space imply boundedness of its coordinate set?
In the proof that every linear map from a finite-dimensional domain is bounded, where is finiteness of the basis used?
Does the boundedness theorem require the codomain to be finite-dimensional? Explain.
Why does the proof that finite-dimensional subspaces are closed not require the ambient space to be complete?