Euclidean norm ($\Vert x{\Vert }_2$)

The Euclidean norm of $x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n$ is

For an orthogonal matrix $Q$, the identity $Q^{T}Q=I$ gives $\Vert Qx{\Vert }_2^2=x^T{Q}^{T}Qx=x^{T}x$. Thus multiplication by $Q$ preserves the Euclidean norm. Applying this on both input and output sides proves $\Vert UA{V}^T{\Vert }_2=\Vert A{\Vert }_2$ for orthogonal matrices $U,V$ and the Euclidean operator norm. The analogous invariance of the Frobenius norm follows by summing column lengths or by a trace calculation.

For a nonzero vector $x\in {\mathbb{R}}^n$, put $r=|x|$ and $\hat{x}=x/r$. The gradient of the Euclidean norm is $\nabla r=\hat{x}$. Differentiate its components to obtain the Hessian matrixThe factor $I-\hat{x}{\hat{x}}^T$ is the orthogonal projection matrix onto the orthogonal complement of the radial direction. Thus the Hessian matrix has eigenvalue zero radially and $1/r$ on each tangential direction, explaining the convexity of the Euclidean norm. Taking the trace gives $\Delta r=(n-1)/r$ away from the origin. The Euclidean norm is not differentiable at the origin: its restriction to any line is a nonzero multiple of the absolute value.

For a real vector $v$, the square of its Euclidean norm is $\Vert v{\Vert }_2^2=v\cdot v={\sum }_j{v}_j^2$. Unlike the norm itself it is a quadratic function, so expanding sums directly exposes diagonal and cross terms. This is useful for the mean-square displacement of an isotropic planar random walk.

The Euclidean norm is self-dual:The upper bound is Cauchy-Schwarz inequality, and equality holds at $v=x/\Vert x{\Vert }_2$ when $x\ne 0$.

This is a ball for the Euclidean metric.

The closed Euclidean ball of radius $R$ about $a\in {\mathbb{R}}^n$ is $\{x:\Vert x-a{\Vert }_2\le R\}$; replacing $\le$ by $<$ gives the open ball.

The Euclidean unit ball is the closed unit ball of the Euclidean norm, centered at zero. Its radial function is identically $1$, making it a natural reference body for John ellipsoids and volume ratios.

The unit ball of a normed vector space is $\{x:\Vert x\Vert \le 1\}$ under the closed-ball convention.

The closed unit ball of a normed vector space is the set of vectors of norm at most one. In a continuous dual space it is compact in the weak-star topology by the Banach-Alaoglu theorem.

The open unit ball of a normed vector space $X$ is $B_X=\{x\in X:\Vert x\Vert <1\}$.