Norms on R^n

From Vectors to Lengths

In the previous tutorial, vectors in $\mathbb{R}^n$ were defined using coordinatewise addition and scalar multiplication. The Euclidean length introduced earlier assigns a nonnegative size to each vector. There are other ways to measure that size, and each can be useful for a different purpose. The common structure behind these measurements is the definition of a norm.

A norm is a function that respects the vector operations: scaling a vector scales its length by the absolute value of the scalar, and the length of a sum is at most the sum of the lengths. It also assigns length zero only to the zero vector. These requirements capture basic properties of length without forcing one particular formula.

Definition: A norm on $\mathbb{R}^n$ is a function $\|\cdot\|:\mathbb{R}^n\to[0,\infty)$ satisfying, for all $x,y\in\mathbb{R}^n$ and all $\alpha\in\mathbb{R}$:

$\|x\|=0$ if and only if $x=0$;
$\|\alpha x\|=|\alpha|\|x\|$ (absolute homogeneity);
$\|x+y\|\leq\|x\|+\|y\|$ (the triangle inequality).

The displayed list contains three numbered conditions, with the first combining nonnegativity and the condition for zero length. Since the function takes values in $[0,\infty)$, nonnegativity is built into its codomain. The triangle inequality is named for the corresponding fact about the lengths of the sides of a triangle. Together with the vector-space laws established in “Vectors in $\mathbb{R}^n$,” these axioms ensure that the length behaves consistently under addition and scaling.

Three Standard Norms

For $x=(x_1,\ldots,x_n)$, three standard choices are the sum of the absolute coordinate values, the Euclidean length, and the largest absolute coordinate value. They are denoted by $\|x\|_1$, $\|x\|_2$, and $\|x\|_\infty$, respectively.

\|x\|_1=\sum_{i=1}^n|x_i|, \qquad \|x\|_2=\left(\sum_{i=1}^n x_i^2\right)^{1/2}, \qquad \|x\|_\infty=\max_{1\leq i\leq n}|x_i|.

The Euclidean norm $\|x\|_2$ is the length associated with the Euclidean metric from “Euclidean Space $\mathbb{R}^n$”: the distance between $x$ and $y$ is $\|x-y\|_2$. The other two formulas emphasize different features of the coordinates. The one norm totals their absolute sizes, while the infinity norm records only the largest one. The subscript $\infty$ is standard notation even though there are only $n$ coordinates and the maximum is finite.

Worked Example: Evaluating Three Norms

Let $x=(-2,1,3,-1)\in\mathbb{R}^4$. For the one norm, add the absolute values of the coordinates:

\|x\|_1=|-2|+|1|+|3|+|-1|=2+1+3+1=7.

For the Euclidean norm, square the coordinates, add, and take the square root:

\|x\|_2=\sqrt{(-2)^2+1^2+3^2+(-1)^2} =\sqrt{4+1+9+1} =\sqrt{15}.

The largest absolute coordinate is $3$, so $\|x\|_\infty=3$. Thus the same vector has lengths $7$, $\sqrt{15}$, or $3$, depending on which norm is being used.

All three formulas are norms. The one and infinity norm checks are direct, and the Euclidean check uses the Cauchy–Schwarz Inequality established in “Euclidean Space $\mathbb{R}^n$.” The following result verifies these claims; its proof also identifies where each norm axiom enters.

Theorem (The Standard Norms on $\mathbb{R}^n$): The functions $\|\cdot\|_1$, $\|\cdot\|_2$, and $\|\cdot\|_\infty$ defined above are norms on $\mathbb{R}^n$.

Proof. Each formula is nonnegative. For the one norm, $\|x\|_1=0$ means that every nonnegative term $|x_i|$ is zero, so every $x_i=0$ and $x=0$. Conversely, if $x=0$, its one norm is zero. For the infinity norm, the maximum of the nonnegative values $|x_i|$ is zero exactly when every coordinate is zero. For the Euclidean norm, a sum of squares is zero exactly when every square is zero, so $\|x\|_2=0$ exactly when $x=0$.

For any real $\alpha$, the identity $|\alpha x_i|=|\alpha||x_i|$ gives $\|\alpha x\|_1=|\alpha|\|x\|_1$. For the Euclidean norm, \[ \|\alpha x\|_2 =\left(\sum_{i=1}^n(\alpha x_i)^2\right)^{1/2} =|\alpha|\left(\sum_{i=1}^n x_i^2\right)^{1/2} =|\alpha|\|x\|_2. \] For the infinity norm, \[ \|\alpha x\|_\infty =\max_{1\leq i\leq n}|\alpha x_i| =|\alpha|\max_{1\leq i\leq n}|x_i| =|\alpha|\|x\|_\infty. \] These identities also hold when $\alpha=0$.

For the one norm, the real-number triangle inequality in each coordinate gives \[ \|x+y\|_1=\sum_{i=1}^n|x_i+y_i| \leq\sum_{i=1}^n(|x_i|+|y_i|) =\|x\|_1+\|y\|_1. \] For the infinity norm, for every $i$, \[ |x_i+y_i|\leq |x_i|+|y_i| \leq\|x\|_\infty+\|y\|_\infty. \] Taking the maximum over $i$ proves $\|x+y\|_\infty\leq\|x\|_\infty+\|y\|_\infty$.

Finally, using the dot product and the Cauchy–Schwarz Inequality, \[ \|x+y\|_2^2 =\|x\|_2^2+2x\cdot y+\|y\|_2^2 \leq\|x\|_2^2+2\|x\|_2\|y\|_2+\|y\|_2^2 =(\|x\|_2+\|y\|_2)^2. \] Both sides before taking square roots are nonnegative, and $\|x\|_2+\|y\|_2\geq0$. Taking square roots therefore gives $\|x+y\|_2\leq\|x\|_2+\|y\|_2$. Each function satisfies all the norm axioms. $\square$

Using the Norm Axioms

Once a function is known to be a norm, its axioms provide useful calculations without returning to its coordinate formula. For example, absolute homogeneity immediately determines the norm of any scalar multiple, and the triangle inequality bounds the norm of a sum. This is especially helpful when the coordinates of a vector expression are complicated.

Worked Example: Applying Homogeneity

Let $u=(1,-2,2)\in\mathbb{R}^3$. Its infinity norm is \[ \|u\|_\infty=\max\{|1|,|-2|,|2|\}=2. \] For the vector $-3u$, homogeneity gives \[ \|-3u\|_\infty=|-3|\|u\|_\infty=3\cdot2=6. \] Indeed, $-3u=(-3,6,-6)$, and the largest absolute coordinate is $6$. The coordinate calculation confirms the result from the norm axiom.

Worked Example: Bounding the Norm of a Sum

Take $a=(2,-1,0)$ and $b=(-1,3,4)$ in $\mathbb{R}^3$, and use the one norm. The individual norms are \[ \|a\|_1=|2|+|-1|+|0|=3,\qquad \|b\|_1=|-1|+|3|+|4|=8. \] The triangle inequality guarantees $\|a+b\|_1\leq 11$. In this case, \[ a+b=(1,2,4),\qquad \|a+b\|_1=|1|+|2|+|4|=7\leq3+8=11. \] The inequality need not be an equality: cancellation in a coordinate can make the norm of the sum smaller than the sum of the norms.

Comparing the Standard Norms

Different norms can assign different numbers to the same vector, but in a fixed finite-dimensional space their values are bounded in terms of one another. The constants in the following inequalities depend on $n$. The result gives a precise way to compare the three standard measurements without claiming that they are identical.

Theorem (Comparison of the Standard Norms): For every $x\in\mathbb{R}^n$, $$ \|x\|_\infty\leq\|x\|_2\leq\sqrt{n}\,\|x\|_\infty \quad\text{and}\quad \|x\|_2\leq\|x\|_1\leq\sqrt{n}\,\|x\|_2. $$

Proof. For each coordinate $i$, $x_i^2\leq\sum_{j=1}^n x_j^2$. Thus $|x_i|\leq\|x\|_2$ for every $i$, and taking the maximum gives $\|x\|_\infty\leq\|x\|_2$. Also, $|x_i|\leq\|x\|_\infty$ for every $i$, so \[ \|x\|_2^2=\sum_{i=1}^n x_i^2 \leq\sum_{i=1}^n\|x\|_\infty^2 =n\|x\|_\infty^2. \] Taking nonnegative square roots proves $\|x\|_2\leq\sqrt{n}\|x\|_\infty$.

To compare the one norm and Euclidean norm in the other direction, expand the square: \[ \|x\|_1^2 =\left(\sum_{i=1}^n|x_i|\right)^2 =\sum_{i=1}^n x_i^2+2\sum_{1\leq i<j\leq n}|x_i||x_j| \geq\sum_{i=1}^n x_i^2 =\|x\|_2^2. \] Both norms are nonnegative, so $\|x\|_2\leq\|x\|_1$. For the remaining bound, apply the Cauchy–Schwarz Inequality to the vectors $(|x_1|,\ldots,|x_n|)$ and $(1,\ldots,1)$: \[ \|x\|_1 =\sum_{i=1}^n|x_i| \leq \left(\sum_{i=1}^n|x_i|^2\right)^{1/2} \left(\sum_{i=1}^n1^2\right)^{1/2} =\sqrt{n}\,\|x\|_2. \] This proves all four inequalities. $\square$

Worked Example: Checking the Comparison Bounds

Let $v=(2,-1,2)\in\mathbb{R}^3$. Its three norms are \[ \|v\|_\infty=2,\qquad \|v\|_2=\sqrt{2^2+(-1)^2+2^2}=3,\qquad \|v\|_1=2+1+2=5. \] Here $n=3$, and the first pair of bounds reads \[ 2\leq3\leq2\sqrt{3}. \] The upper inequality holds because $3\leq2\sqrt{3}$, which follows by squaring the nonnegative sides: $9\leq12$. The second pair reads \[ 3\leq5\leq3\sqrt{3}. \] The upper inequality follows from $25\leq27$ after squaring. The example illustrates that the bounds need not be equalities.

Why the Choice of Norm Matters

A norm supplies a notion of length for vectors, but the particular norm determines what is emphasized. The one norm adds the absolute contributions from all coordinates. The infinity norm focuses on the largest coordinate. The Euclidean norm combines all coordinates through the square root of the sum of their squares. For example, in applications where the largest coordinate error is the main concern, the infinity norm may be a natural measure; when the total absolute error matters, the one norm may be more suitable.

The comparison theorem explains an important finite-dimensional feature: none of these three norms can become arbitrarily large relative to another while $n$ is fixed. It does not say that the norm values are equal, nor that the comparison constants are independent of dimension. The factors $\sqrt{n}$ appear explicitly, so changing the dimension changes the bounds. The definition also applies to norms beyond these three; a formula must still satisfy every norm axiom before it can be called a norm.

A common mistake is to check only that a proposed length is nonnegative and zero at the zero vector. Those facts do not establish the triangle inequality or homogeneity. For instance, the largest absolute coordinate is a norm because it satisfies all three conditions, not merely because it seems to measure size. In the next tutorial, norms will provide the lengths used to define open balls in $\mathbb{R}^n$.

Check Your Understanding

Use the definitions and results in this tutorial to answer each question.

State the homogeneity and triangle inequality axioms for a norm.
Compute the one, Euclidean, and infinity norms of $(1,-3,2)\in\mathbb{R}^3$.
For $x=(x_1,\ldots,x_n)$, explain why $\|x\|_\infty\leq\|x\|_2$.
Use homogeneity to express $\|5x\|_1$ in terms of $\|x\|_1$.
Why does the comparison theorem not imply that the three standard norms always have the same value?

What You'll Learn