# Linear algebra too early.

I have started college few days ago. At the first exposition of physics, professor has been reminding us what is vector and what is definition of a vector. But he has been using linear algebra to explain us vectors and this is not covered in high school. Therefore I decided to learn it on my own.

Now I know somethings about linear algebra, but I have a few questions.

1. What is difference between Linear Space and Vector Space?
2. Subspaces have to be defined with the same laws as space to be a subspace of the space?
3. Why does the subset have to contain zero vector to be a subspace? If a set is empty multiplication should not work, because empty set times anything else should give empty set.
4. Can we define own laws for subspace?
5. Professor calls vector addition as inner product and scalar multiplication as outer product, what does mean?

1. none
2. a subspace is a subset which is also a space
3. because part of the definition of space is that it contains the zero vector; in other words, the empty set does not satisfy the definition of space
4. what do you mean? Do you mean can it have different addition and scalar multiplication, not inherited from the parent? If so, the answer is "no". That might be a space, but it would not be a subspace.
5. vector addition is not the inner product. The inner product is additional structure. This is not the place to go into a lengthy exposition. But the inner product of two vectors is a scalar. For ordinary Euclidean plane or space, the inner product $u\cdot v=|u||v|\cos\theta$, where $\theta$ is the angle between the two vectors. In coordinate terms we have $(a,b,c)\cdot(x,y,z)=ax+by+cz$, whereas the scalar product $\lambda (a,b,c)=(\lambda a,\lambda b,\lambda c)$.

Apparently "inner/outer" is translation from a different language. I am not clear what point is being made here.

The starting point for a vector space is a field $F$, which is usually the reals or the complex numbers. A field is a structure where addition, multiplication, 0, 1, inverses (1/k and -k) all work as for the reals.

We then have a set of elements, called "vectors" which form an abelian group under addition. In other words, the sum of any two vectors is a vector. The zero vector has the obvious meaning (when added to another vector it leaves it unchanged). We can think of the vectors as having coordinates which belong to $F$. So the classic example would be $\mathbb{R}^3$, the set of points in Euclidean space, with a typical vector being written $(a,b,c)$, where $a,b,c$ are real numbers.

A common convention is to write scalars (members of the field) by Greek letters, such as $\lambda,\mu,\nu$. We can multiply a vector by a scalar to get another vector, so if $\lambda$ is a scalar, and $v$ is a vector, then $\lambda v$ is a vector. The mental picture (at least where $F$ is the reals) is that we are increasing the length of the vector by a factor $\lambda$. So if $v=(x,y,z)$ then $\lambda v=(\lambda x,\lambda y,\lambda z)$.

But I am rambling, I guess I am not clear what your difficulty is. The use of outer/inner you are referring to is just a comment about the domain/range of the operations.

• Your answer to 5 is not at all what the professor meant. It's "inner" because both the argument must be in $V$ and the result is in $V$. Scalar multiplication is "outer" because one argument is in $V$ and the other is in the base field ($\mathbb{R}$, $\mathbb{C}$...). Calling it "inner" is unfortunate in this context, but it's clear this is what was meant. Besides, maybe English isn't the native language of OP, and inner was the closest translation to what the prof meant. – Najib Idrissi Oct 6 '14 at 9:23
• @NajibIdrissi Using technical words with clearly defined meaning in a non-standard sense is a recipe for disaster. You will confuse yourself and others. – almagest Oct 6 '14 at 9:35
• Then perhaps explain that instead of explaining the inner product, which will only confuse the OP more...? – Najib Idrissi Oct 6 '14 at 9:39
• @Najib idrissi the first part covers with wikipedia im my language, unfortunately, there isn't "outer" on wikipedia, and still i don' t know how to call it in english. But why result of scalar times a vector isn't in Vector Space? It look like it breaks clousure under multiplication rule. – Arlic Oct 6 '14 at 9:45
• @Arlic No, the result is in the vector space. But the argument (in $\lambda \cdot x$, the $\lambda$) is not; it's in the base field. – Najib Idrissi Oct 6 '14 at 11:02

without getting too technical here are a few thoughts which may help. if not, my apologies!

first be aware of two of the persistent problems with math terminology, which can hinder progress in the early stages of study.

1. ambiguity - same word used in different senses. one recurrent offender is normal
2. words used which are common in everyday language, but which have a much more specific meaning that may be quite different e.g. group, field, ring, root, space.

these everyday meanings may in some cases be helpful, but can also lead to confusion

a space is a set with additional structure. if a topology is specified we have a topological space. if, on the other hand, the set has a linear structure (which is algebraic in nature), it is called a linear space or in some cases a vector space. if we have both structures then the object is a topological vector space. a topological group is a set which has both a group structure and a linear structure which are (in a specific way) compatible. a vector space is an abelian group, but with further structure relating to multiplication of vector by a scalar. the scalars belong to a field associated with a vector space. if the vector space is $V$ and the field is $F$ then we say $V$ is a vector space over $F$.

science students are programmed early to think of a vector as an almost physical thing. arrows with three co-ordinates and three different kinds of 'multiplication' - two of which are rather unfamiliar and one of which (the so-called cross product) seems initially to be deliberately obscure. whilst 3-D vectors over the real numbers (the field $\mathbb{R})$ form a perfectly good vector space, one should try to not to let this specific example monopolize intuition. for example every field is a vector space over itself.

let $F$ be a field and let $X$ be any non-empty set. there is a well-defined set $F^X$ whose elements are all the functions from $X$ to $F$. even though $X$ may have no algebraic structure at all it is clear that if $f,g \in F^X$ we can add these functions together by adding their images in $F$ and the result will be another function from $X$ to $F$. likewise a function can be multiplied by a scalar in $F$ to give another function. here the functions are the vectors of a vector space whose associated field is $F$. actually this construction is always there, perhaps in the backkground, obscured by some more concrete intuition..

so, for example, if $X$ is the set $\{1,2,\dots,n\}$ and $X$ is the field of real numbers $\mathbb{R}$ then we obtain the familiar vector space $\mathbb{R}^n$

in simplistic terms you may think of a subspace as a subset of the underlying set of the space, which satisfies the same defining axioms as the space itself. for a vector space this means that (i) for any vector in the subset, all its scalar multiples are also in the subset (ii) for any two vectors in the subset, their sum is also in the subset.

zero must be in the subspace because $-1$ is in the field, and thus if $v$ is in the subspace so is $-v$ (abbreviation for $(-1)v$). since the sum of any two subspace vectors must be in the subspace so must $v-v=0$ (remember a vector space is an abelian group, and this is therefore also required of a subspace)