# Vector norms are only on column vectors? How about row vectors?

When I studied on norms: vectors are defined by : $$x\in R^n$$.

Many books said $$x$$ is a column vector. (Also, define inner product $$x^tx$$.)

What I am curious about is : $$||x|| =||x^t||$$? (any) norms of vector and its transpose are the same.

We should see $$||x^t||$$ or calculate as like a column vector? or is it a matrix, and calculate like induced norm?

• Norms can be defined on any (real) vector space Commented Apr 6, 2021 at 9:22
• @HagenvonEitzen Yeah, then it doesn't matter if it is row vector or column vector? The result would be the same? Commented Apr 6, 2021 at 9:25
• If you have a norm $\Vert\cdot\Vert$ on the space of column vectors, you can simply define a norm on the space of row vectors by $\Vert x^t\Vert:=\Vert x\Vert$. It's quite natural to do so. But you can also define entirely different norms on both spaces. So it really depends: which norms are we talking about? The ones which satisfy your condition by definition? Then certainly yes. Any other pair of norms? Then certainly no. Commented Apr 6, 2021 at 10:28
• I think that this problem is quite simpler. «Vectors are defined by $x \in \mathbb{R}^n$», therefore a related question should be: does the definition $x \in \mathbb{R}^n$ always represent a column vector? Or a row vector? Or either of them? And if $x \in \mathbb{R}^n$ only represents a column vector, how do you write a row vector with the same notation? Commented Apr 6, 2021 at 10:41
• I believe that vectors are row vectors just because this way they are easier to handle when dealing with matrices. An $m \times n$ matrix $\mathbf{A}$ can be multiplied to an $n \times 1$ column vector $\mathbf{x} \in \mathbb{R}^n$: $\mathbf{A} \cdot \mathbf{x}$ (while it would be impossible to multiply the same matrix to a row vector $1 \times n$, $\mathbf{x}^T$). But column vectors can be switched to row vectors while maintaining exactly the same features. I think that a column vector and its corresponding row vector are just two equivalent representations of the same object Commented Apr 6, 2021 at 10:51

The elements $$x\in{\mathbb R}^n$$ are just $$n$$-tuples $$(x_1,x_2,\ldots, x_n)$$ and, a priori, neither row nor column vectors. If you use them for doing euclidean geometry, say, studying ellipses or spirals, then you usually use the norm $$|x|:=\sqrt{x_1^2+\ldots+x_n^2}\ .\tag{1}$$ When you are doing in fact linear algebra then you consider the points of the given "base space" $$X:={\mathbb R}^n$$ as column vectors and set up the standard matrix calculus. In this environment you sometimes also use norms, e.g., the above norm $$(1)$$, which then is denoted $$\|x\|_2$$. In some cases you use other norms, e.g., $$\|x\|_1:=\sum_{k=1}^n |x_k|,\qquad{\rm or}\qquad \|x\|_\infty:=\max_{1\leq k\leq n}|x_k|\ .$$ These norms refer to the "base space" $$X$$. To this space is associated its dual $$X^*$$ containing the linear functionals $$\phi: \>X\to{\mathbb R}$$. Such a functional appears as $$\phi(x)=\sum_{k=1}^n a_k x_k\ ,$$ so that "datawise" this $$\phi$$ is again encoded by an $$n$$-tuple $$(a_1,\ldots,a_n)=:a$$. In matrix calculus this $$n$$-tuple $$a$$ is then a row vector (and is sometimes written $$a^{\rm t}$$, or similar). When we have some norm $$\|\cdot\|$$ on the base space $$X$$ the $$\phi$$ has automatically its functional norm $$\|\phi\|:=\sup_{x\ne0}{|\phi(x)|\over\|x\|}\ .$$ When the norm in $$X$$ is the euclidean norm $$\|\cdot\|_2$$ then $$\|\phi\|=\|a\|_2$$ again, but when the norm in $$X$$ is $$\|\cdot\|_1$$ then the "associated" norm in $$X^*$$ is $$\|\phi\|=\|a\|_\infty$$. (The proof is an exercise.)