Example. The most classic of all examples: the set of vectors is \(\mathbb{R}^n\), and the set of scalars is \(\mathbb{R}\). \(\mathbb{R}^n\) refers to the set of \(n\)-element lists of real numbers. Consider \(\mathbb{R}^2\). Some vectors in \(\mathbb{R}^2\) are

• \((0, 1)\)
• \((1, 0)\)
• \((1, 1)\)
• \((0, -1)\)
• \((5, 6)\)
• \((-\pi, e)\)

We can add vectors via the rule: \((x_1,x_2)+(y_1, y_2) = (x_1+y_1, x_2+y_2)\). We can scale vectors via the rule: \(k(x_1,x_2) = (kx_1, kx_2)\).

This vector space very easily permits visualization, by placing the vectors in the plane.

Vector addition is accomplished gemoetrically by placing vectors “tip to tail” and going from the tip of start of the first vector to the end of the last vector.

Scalar multiplication is accomplished by shrinking/growing the vectors.

Some important properties to note about this vector space (these generalize):

Adding \((0,0)\) to vectors doesn’t change them: \((x,y)+(0,0) = (x,y)\). Multiplying vectors by \(0\) gives to \((0,0)\).

For every vector \((a,b)\), there is another vector, namely \((-a,-b)\), such that \((a,b)+(-a,-b) = (0,0)\).

Let \(x, y, z\) be vectors, let \(c, k\) be scalars. \[x+y = y+x.\]

\[(x+y)+z = x+(y+z).\]

\[c(k(x)) = (ck)x.\]

\[(c+k)x = cx + kx.\]

\[c(x+y) = cx + cy.\]

Another thing that is implicit in this all is the idea of “closure”. \(x+y\) needs to be a vector for any vectors \(x,y\), and also \(kx\) needs to be a vector for any vector \(x\) and any scalar \(k\).

Also, if you don’t know what a field is, don’t worry about it. A field is just a set with multiplication and addition that play really nicely with it. Usually this will be \(\mathbb{R}\) for us, although there are some other cool fields (e.g. finite fields, complex numbers).

Example. The vector space (over the reals) of polynomials with real coefficients \(\mathbb{R}[x]\). Addition and multiplication is defined component-wise, i.e.

\[k \sum p_j x^j = \sum kp_j x^j \]

\[\sum a_j x^j + \sum b_j x^j = \sum (a_j + b_j) x^j.\]

You can say things like \[(3x^2 + x + 2) + 5x = 3x^2 + 6x + 2.\] And \[4(x+1) = 4x+4.\]

The \(0\)-polynomial satisfies the required conditions, additive inverses are achieved by negating polynomial’s coefficients i.e. \(x+1 + (-x-1) = 0\).

Example. Functions \(f: \mathbb{N}\to \mathbb{R}\) (vector space over the reals).

Addition is defined point-wise, i.e. 

\(f + g\) is a function such that \((f+g)(x) = f(x) + g(x)\) for all \(x\in\mathbb{N}\). \(kf\) is a function with \((kf)(x) = kf(x)\).

Some examples are:

• \(f(x) = \sqrt{x}.\)
• \(f(x) = x\).
• \(f(x) = \sqrt{2} x.\)
• \(f(x) = x^2\).
• \(f(x) = 0.\)
• \(f(x) =\) the number of numbers less than \(x\) that are relatively prime to \(x\) (i.e.the number of \(y<x\) such that \(gcd(x,y) = 1\)).

Example. \(\mathbb{R}^\mathbb{N}\) (over the reals). This is the space of infinite sequences. Elements include vectors like

\[(1, 1/2, 1/3, 1/4, \ldots)\] and \[(0,0,0,\ldots).\]

Addition and multiplication is defined point-wise:

\[(a_1,\ldots) + (b_1, \ldots) = (a_1+b_1, \ldots)\] \[k(a_1,\ldots) = (ka_1, \ldots).\]

This vector space might be starting to look familiar. Could it be, this is “the same” vector space as the space of functions from the natural numbers to the real numbers? (yes!, more on this later)

Example. \(\mathbb{C}\) over \(\mathbb{C}\). Enough said.

Example. The space of continuous functions on \([0,1]\).

Example. \((\mathbb{Z}/2\mathbb{Z})^n\) the space of binary strings of length \(n\).

I’ll just comment on addition in \(\mathbb{Z}/2\mathbb{Z}\) for a second.

\(1+1 = 0\), \(1+0= 1\), \(0+0 = 0.\) Sound crazy? It’s not. You could interpret \(1,0\) as true, false, and then this is NAND (i.e. NOT AND). You could also just think of it as addition mod 2.

Anyways, there are a lot of reasons why you might want the set of binary strings of length \(n\). High quality vector space right there.

Example. Consider a map \(T : \mathbb{R}^2 \to \mathbb{R}^2\) that does \[(x, y) \mapsto (-y, x) \]

This map is linear. It does rotation by 90 degrees counter-clockwise.

Here is a visualization of this map acting on a set of points:

Example. Consider a map \(T : \mathbb{R}^2 \to \mathbb{R}^2\) that does \[(x, y) \mapsto (2x, 2y) \]

This map is linear. It scales vectors up.

Example. Consider a map \(T : \mathbb{R}^2 \to \mathbb{R}^2\) that does \[(x, y) \mapsto \left(\frac{3}{5} x + \frac{4}{5} y\right)(3/5, 4/5).\]

This map does projection onto the line \(y=\frac{4}{3}x.\)

Example. Let \(P_n\) be the space of polynomials of degree at most \(n\). The map \(T: P_n \to P_n\) \[T(p) = p'\] (the derivative) is linear.

Example. Let \(P_n\) be the space of polynomials of degree at most \(n\). The map \(T: P_n \to \mathbb{R}^3\) \[T(p) = (p(1), p(2), p(3))\] is linear.

Example. The vector space of polynomials wtih real coefficients of degree \(2\) or less, denote \(P_2\), is isomprphic to \(\mathbb{R}^3\). In particular, we can associate the polynomial \(ax^2 + bx + c \in P_2\) with, for instance, \((a,b,c) \in\mathbb{R}^3.\) This map is clearly an isomorphism.

Example. The vector space of polynomials is infinite dimensional; it clearly admits no basis: imagine it did, then take a polynomial with degree more than the highest degree of the basis elements, this yields a contradiction.

Example. The so called “standard basis” for \(\mathbb{R}^n\) is the set of vectors \(\left\{ (1,0,0,\ldots,0), (0,1,0,0,\ldots, 0), \ldots, (0,0,\ldots, 1)\right\}\) with the \(i\)-th standard basis vector \(e_i\) containing \(0\) in every component except a \(1\) in the \(i\)-th component.

Example. A basis for \(P_2\) is \(\left\{ 1, x, x^2\right\}\)

Example. An altenrative basis for \(\mathbb{R}^2\) is \(\left\{ (1,2), (2,1)\right\}\). Clearly this set spans the same space, but it is just a different way of representing points.

You can represent points like \((5,7)\) as \((5,7) = 3(1,2) + 1(2,1)\). Thus you could say that \((5,7)\)’s coordinates in the new basis rae \((3,1)\).

Example. A line through the origin in \(\mathbb{R}^2\) or \(\mathbb{R}^3\) is a subspace.

Example. Polynomials of degree at most \(2\) that have a zero at \(4\) are a subspace of the space of polynomials.

Example. Consider the linear transformation \(T: \mathbb{R}^2 \to \mathbb{R}^3\), defined by \(T(x,y) = (x,0,0).\)

\(\im(T) = [(1,0,0)]\)

Example. Consider the map \(T: P_2 \to P_2\) defined by \(ax^2 + bx + c\cdot 1 \mapsto ax+b\). \(\ker T = [1]\), \(\im T = [1, x]\). In \(P_2/\ker T\) we group together polynomials that differ by a constant, so \(P_2/\ker T = \left\{ \left\{ ax^2+bx+c\; : \;c\in \mathbb{R} \right\}\; : \;a,b\in\mathbb{R} \right\}.\) We can write this as \(P_2/\ker T = \left\{ ax^2+bx + \mathbb{R}\cdot 1\; : \;a,b\in\mathbb{R} \right\}.\) But this can easily be identified with \(\left\{ ax+b\; : \;a,b\in\mathbb{R} \right\}\). Hence \[P_2/\ker T \cong \im T.\]

Example. Consider the map \(T: \mathbb{R}^2 \to \mathbb{R}^2\) defined as \(T(x,y) = (x, y)\). \(\ker T = \left\{ 0\right\}\), and \(\mathbb{R}^2/\ker T = \left\{ \left\{ (x,y)\right\}\; : \;x,y\in \mathbb{R} \right\}\). But this is trivially identified with \(\mathbb{R}^2 = \im T\).

Example. Consider the map \(T: \mathbb{R}^3 \to \mathbb{R}^2\) defined as \(T(x,y,z) = (z, y)\). \(\ker T = [(1,0,0)]\), and \(\mathbb{R}^3/\ker T = \left\{ [(1,0,0)] + (0,y,z)\; : \;y, z\in \mathbb{R} \right\}\). But this is trivially identified with \(\mathbb{R}^2 = \im T\).

Example. Consider the map \(T: \mathbb{R}^3 \to \mathbb{R}^2\) defined as \(T(x,y,z) = (0, y)\). \(\ker T = [(1,0,0), (0,0,1)]\), and \(\mathbb{R}^3/\ker T = \left\{ [(1,0,0), (0,0,1)] + (0,y,0)\; : \;y\in \mathbb{R} \right\}\). But this is trivially identified with \([(0,1)] = \im T\).