in linear algebra, why do we need basis vectors when we have coordinate system?

Question

A novice question.

Can't we just use a coordinate system to express/locate anything instead of basis vectors?

I am confused, maybe because all the examples I am looking at, use 2D and 2D Cartesian coordinate systems. And x,y axes seem to be sufficient to locate or describe any point, or vector for that matter.

Say, we have basis vectors in 2D [1 0] and [0 1]. Even if we rotate them by 90 degrees, we would still keep using the original coordinate system to describe them! For instance, when transforming a vector [1 1] using the new basis vectors after rotation. The vector [1 1] will be expressed after transformation via the original Cartesian coordinates. Then why introduce basis vectors when we have a coordinate system?

Answer 1

Coordinates on one hand and basis vectors on the other represent dual points of view. Thus, coordinates $x$ and $y$ in the plane are linear maps $\mathbb R^2 \to \mathbb R$. Meanwhile, the basis vectors $e_1$ and $e_2$ are elements of $\mathbb R^2$. The fact that the $x$-coordinate of $e_1$ is $1$ corresponds to the fact that evaluating the first function on $e_1$ gives $1$, etc. Such dualities are common in mathematics and of great importance in many fields, including differential geometry.

Answer 2

Often representing a vector with a different basis enables to express it with far less coefficients. This is the idea behind many compression schemes for sound signals or images in general. Suppose you sample a sound at $N$ time instants obtaining a vector $x:=[x_0, \ldots, x_{N-1}]$. If you want to store or transmit this discretized version of the sound signals you need to store (in general) $N$ numbers.

Often, however, natural sounds can be expressed with very few sine and cosine waves, which can be captured in the DFT matrix. Suppose your signal is composed of only $2$ frequencies. This means in general that you can use only $2$ columns of the DFT matrix to express your signal (and vector):

$$x = F\hat{x}$$ where $F$ is your $N \times N$ DFT matrix and $\hat{x} \in \mathbb{R}^N$ stores the coefficients in the new basis. But this vector is now very sparse, i.e., it contains many zeros ($N-2$). This is a storage saving and often an algorithmic computational advantage.

Answer 3

Expressing/locating vectors is not the only reason to study bases. They have many applications. If you have just learned about general bases while studying linear algebra, I would advise a tiny bit of patience as the theory is developed and the applications rolled out.

Meanwhile, though, here is an extended example drawn from the theory of eigenvectors and eigenvalues, to show you one particular example of a valuable basis besides the usual standard basis. If you know this theory, you can probably jump to the end. If you don't know this theory, enjoy the ride.

Suppose that our goal is to understand a specific linear transformation, namely $$F(x,y) = (y,x+y), \quad (x,y) \in \mathbb R^2 $$ $$(x,y) \underbrace{\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}}_M = (y,x+y) $$ There it is, expressed nicely in ordinary coordinates. What more is there to know? Quite a lot, as it turns out.

Playing around for a little while, one might think of this linear tranformation as a dynamical system where the output feeds back into the input: \begin{align*} (0,1)M &= (1,1) \\ (1,1)M &= (1,2) \\ (1,2)M &= (2,3) \\ (2,3)M &= (3,5) \\ (3,5)M &= (5,8) \\ (5,8)M &= (8,13) \\ (8,13)M &= (13,21) \\ (13,21)M &= (21,34) \\ (21,34)M &= (34,55) \\ (34,55)M &= (55,89) \\ (55,89)M &= (89,144) \end{align*} Already something interesting is going on here: you can see the sequence of Fibonacci numbers unrolling in each column, with the $y$ column one step ahead of the $x$ column. And if you've ever played with Fibonacci numbers then you might anticipate what's coming next: let's compute the sequence of slopes of these vectors (to three decimal places) \begin{align*} 1/1 &= 1 \\ 2/1 &= 2 \\ 3/2 &= 1.500 \\ 5/3 &\approx 1.667 \\ 8/5 &= 1.600 \\ 13/8 &= 1.625 \\ 21/13 &\approx 1.615 \\ 34/21 &\approx 1.619 \\ 55/34 &\approx 1.617 \\ 89/55 &\approx 1.618 \\ 144/89 &\approx 1.618 \end{align*} This might lead one to wonder: There appears to be a limiting value of this slope. The line of vectors with that slope looks like it should be preserved by the matrix $M$.

Is this so?

Answer: Yes, this is one of the first examples of eigenvectors. An eigenvector for the matrix $M$ is a non-zero vector $v$ whose line is preserved by $M$, and so the effect is that $vM$ is equal to a scalar multiple of $v$, using a scalar traditionally denoted $\lambda$ and referred to as an eigenvalue: $$(*) \qquad\qquad (x,y)M = \lambda(x,y) $$ This equation can be solved for $\lambda$: $$(y,x+y) = (\lambda x, \lambda y) $$ $$(-\lambda x + y, x + (1-\lambda)y) = (0,0) $$ $$(x,y) \underbrace{\begin{pmatrix} - \lambda & 1 \\ 1 & 1-\lambda \end{pmatrix}}_{M-\lambda I} = (0,0) $$ Since $(x,y)$ is assumed to be nonzero, the matrix $M-\lambda I$ is singular, so its determinant equals zero: $$\lambda^2-\lambda-1=0 $$ $$\lambda = \frac{1}{2} \pm \frac{\sqrt{5}}{2} $$ and so there are two solutions: $$\lambda_1 = \frac{1}{2} + \frac{\sqrt{5}}{2} \qquad \lambda_2 = \frac{1}{2} - \frac{\sqrt{5}}{2} $$ When we insert each of these two values of $\lambda$ into equation $(*)$ we can solve for two eigenvectors $v_1,v_2$ (the solution is not unique, but let's pick one solution for $\lambda_1$ and one for $\lambda_2$).

Now here's the punchline: the pair of eigenvectors $v_1,v_2$ form a basis for $\mathbb R^2$, and the matrix of $M$ with respect to that matrix is a diagonal matrix whose entries are the eigenvalues $$\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} $$ This diagonal matrix is, of course, the matrix for $F$ expressed in the basis $\{v_1,v_2\}$: $$F(v_1) = \lambda_1 v_1 $$ $$F(v_2)=\lambda_2 v_2 $$ So if you were to switch over to this basis for purposes of computing the linear transformation $F$, your formula becomes particularly simple: $$F(a v_1 + b v_2) = \lambda_1 a v_1 + \lambda_2 b v_2 $$ And you can now apply this formula to do a still deeper dive into the behavior of the linear transformation $F$.

The general theorem (vaguely stated) is this: for every linear transformation $F:\mathbb R^n \to \mathbb R^n$ from $\mathbb R^n$ to itself, there is a basis of $\mathbb R^n$ in which the matrix of $F$ has a particularly nice form. It's not always a diagonal form, but its not far from it; this is known as the real Jordan form for $F$.