Professor Dave here, let’s change things up.
Now that we know about basis vectors, we are ready to learn how to change from one basis
to another, which can be useful in certain situations. This can help express a problem
in a way that is more easily solved, so let’s learn how to do this now.
We saw in the previous tutorial that having vectors of length one in each of the x, y,
and z coordinates, is in fact a basis. For this example consider the vector space R2,
which contains vectors in the two-dimensional x-y space. Our most simple basis here will
be a unit vector in the x direction, i, which can be expressed as (1, 0), and a unit vector
in the y direction, j, which can be expressed as (0, 1). Because we know this is a basis,
we can write any vector in R2, represented by v, as a linear combination of i and j.
Let’s say that v equals (v1, v2). We can easily write this as v1 times (1, 0) plus
v2 times (0, 1). Or equivalently, v equals v1(i) + v2(j). We have written our vector
in terms of our basis, and v1 and v2 can be considered the coordinates of v in this basis.
However, we can have some other basis for R2 and v will have some other coordinates
in that basis. By changing basis, we are essentially creating new axes in that vector space.
Let’s consider a new pair of basis vectors, u1 equals (1, 2) and u2 equals (3, 3). We
can write our vector v in terms of this basis, where v equals v1'(u1) + v2'(u2) where v1'
and v2' are our new coordinates in this basis. While we are changing basis, we’re not actually
affecting the vector, so it is still the same vector as before, when we wrote it as v1(i)
+ v2(j), meaning we can set these two expressions equal to each other. We can also write our
new basis vectors in terms of i and j. u1 equals (1, 0) + 2 times (0, 1), which is just
1(i) + 2(j). u2 equals 3 times (1, 0) + 3 times (0, 1), which is just 3(i) + 3(j). Now
see what happens when we substitute these values into our equation for the vector v.
We end up getting v1(i) + v2(j) being equal to v1'(i + 2j) + v2'(3i + 3j). Doing a little
algebra, we can quickly distribute these terms and then group in terms of i and j, so this
side becomes (v1' + 3v2')i + (2v1' + 3v2')j. We can put this back into vector form, since
both sides have an i term and a j term. And then we can write this out as a coefficient
matrix, multiplied by a vector with components v1 and v2. We now have a simple matrix equation
that relates the coefficients in our standard basis, v1 and v2, to the coefficients in our
new basis, v1' and v2'. The matrix that relates them is referred to as the “transition matrix”,
represented by the letter U, and we will note that v is equal to U times v prime. An important
thing to note is that this transition matrix has columns that are the same as the basis
vectors u1 and u2. This means that in order to easily get the transition matrix between
any basis and the standard basis, instead of going through the algebra each time, the
matrix U can be written as the new basis we are working with as its columns.
Finally, let’s say we started with a vector in the standard basis, v = (2, 3). We want
to find the new coefficients in the basis u1, u2. In other words, we want to solve for
v1' and v2'. We could jump straight to using the transition matrix, but to make things
even easier we can manipulate this equation such that v1' and v2' are isolated. In order
to isolate this v prime term, we have to get rid of U. Remember that when we learned about
inverse matrices, a matrix multiplied by its inverse will give the identity matrix, or
effectively just the number 1, just like dividing some number by itself. Let’s therefore multiply
our equation by the inverse of U, which labeled U-1. Bear in mind that order matters in matrix
multiplication, it is not commutative, so let’s make sure to make U inverse the leftmost
term on each side of the equation. We end up with U-1v = U-1Uv', and since U inverse
and U multiply to become 1, we get 1v', or simply v' on the right side. Now we can plug
in our numbers. We know our transition matrix is 1, 3, 2, 3, and the vector we want to change
basis for is (2, 3). We will need the inverse of the transition matrix, since the equation
calls for U inverse times v, so for a 2 by 2 matrix, we simply switch the diagonal terms,
flip the sign on the other terms, and then divide this by the determinant of the original
matrix. We find that the determinant of U is negative three, and this will give us an
inverse matrix of 3, -3, -2, 1, all multiplied by negative one third. Now let’s plug in
our vector, v, which was (2, 3). Going through the matrix multiplication, we get these expressions,
which we can simplify, and then each term will be multiplied by negative one third.
This leaves us with v' = (1, 1/3). These are the new coordinates of v in the basis u1,
u2. Written out, this is v = u1 – one third u2.
It’s important to note that the transition matrix made of the basis vectors as columns
relates the coordinates in that basis to the standard basis only. Relating two non-standard
bases requires an extra step using the standard basis as a stepping stone. Let’s say in
addition to our basis u1 = (1, 2), and u2 = (3, 3), we also have a basis w1 = (-1, -1),
and w2 = (3, 0). We know that the transition matrix for each basis, U and W, will relate
each to the standard basis. Relating the standard basis to the U basis: v = Uvu. And relating
the standard basis to the W basis: v = Wvw. Setting these two equal to one another, we
get Uvu = Wvw which makes getting from one basis to another as easy as taking an inverse
matrix. Going from W to U we get vu = U-1Wvw and similarly, going from U to W we get vw
= W-1Uvu. This change of basis is actually a kind of
linear transformation, which we will be covering next. But before moving on, let’s check comprehension.