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2020-01-22, Wednesday

Hi, this is my post-lecture note. A short summary of what we did in class today. (Also a chance to correct my mistakes made in class)

So, what is a vector? There are many possible correct answers (in different sense). You may say one of the following

  • it is an arrow with a starting point and ending point (ha, you are thinking of a displacement vector)
  • it is an ordered set (a.k.a tuple) of numbers, like $(2,3,-1)$ a 3-dim vector, or $(2,3)$ a 2-dim vector. You represent it by an arrow from the origin to the point with that coordinate.
  • (and the most boring yet correct answer) it is an element in the vector space. (well, great, it is like asking, what is 2 + 3? and the answer is 3 + 2.) but this is correct, since, after we learned so many examples of various general vector spaces (the space of degree 2 polynomials, the solution space to differential equations, etc), you cannot simply say that vector is just a tuple of numbers.

The notion of vector space is the following.

Let $V$ be a set, we say $V$ is a vector space $V$ over a field $k$ (say, $k=\R$ or $k=\C$) if the following is true * we can add two vectors together and get a vector * we can multiply a number (in $k$) to a vector and get a vector * these two operations needs to satisfy some obvious relations, i.e., if $a, b \in k$ and $v,w \in V$, then $$ (ab)v = a(bv), \quad a(v+w) = av+aw, \quad a \neq 0, v \neq 0 \Rightarrow av \neq 0 \cdots$$.

We then give some examples of vector spaces.

Example :

  1. Fix a positive number $N$, then the set of real coefficient polynomials $f(x)$, where $deg(f) \leq N$ forms a vector space over $\R$. A basis of this vector space is $\{1, x, \cdots, x^N\}$, hence the dimension is $N+1$.
  2. The space of solution to $$f''(x) + x f(x) = 0$$ forms a 2 dimension vector space.
  3. In general, a degree $k$ homogeneous ordinary differential equation has solution space $k$-dimensional. That is why when we write down the general solution, we write $$ f(x) = c_1 f_1(x) + \cdots + c_k f_k(x)$$ where the set $\{f_i(x)\}$ are linearly independent solutions, and $c_i$ are constants.

OK, now that we see many 'weird' vector spaces, we may think twice when we say, “the length of a vector is just $\sqrt{x^2+y^2+z^2}$”, and a vector is just a tuple of numbers. But vector can be identified with a tuple of numbers, once we fix a …


A basis of a vector space $V$ is a maximal collection of linearly independent vectors $V$. The number of vectors in a basis is called the dimension.

Q: is it possible that, for a vector space $V$, I can find a basis with $3$ elements, but someone else can find a basis with $4$ elements? Why not?


Given a basis $e_1, \cdots, e_n$ of an n-dimensional vector space $V$, any element $v$ can be identified with a tuple of numbers $(v_1, …, v_n)$ by the following relation $$ v = v_1 e_1 + \cdots + v_n e_n $$ These $(v_1, \cdots, v_n)$, or $v_i$s, are coordinates of $v$ with respect of the basis $\{e_i\}$.

It is no surprise that, if you change the basis, you will also change the coordinates. Just like when you change unit, the number in front of it changes, 2 meter = 6.5 feet (NBA player Michael Jordan's height), even though they both represent the same length.

That's all. Next time, we will see what is a tensor.

math121b/01-22.txt · Last modified: 2020/01/22 21:53 (external edit)