Today we will return to the concept of coordinate systems given by bases of \(\mathbb{R}^n,\) and then do some review of Chapters 3 and 4.
Last Wednesday, we defined basis and discussed briefly how a basis for a subspace gives a way to impose a coordinate system on that subspace. Recall that if \(\mathbf{v}_1, \dots, \mathbf{v}_m\) is a basis for a subspace \(S \subseteq \mathbb{R}^n,\) then every vector \(\mathbf{x} \in S\) can be written uniquely as a linear combination \(c_1 \mathbf{v}_1 + \dotsm + c_m \mathbf{v}_m.\) Because of this fact, we made the following definition:
Definition (Coordinates in a basis)
Suppose that \(S\) is a subspace of \(\mathbb{R}^n\) and that \(\mathcal{B} = (\mathbf{v}_1, \dots, \mathbf{v}_m)\) is a basis for \(S.\) Given a vector \(\mathbf{v} \in S,\) the coordinates of \(\mathbf{v}\) with respect to the basis \(\mathbf{v}_1, \dots, \mathbf{v}_m\) are the unique coefficients \(x_1, \dots, x_m\) such that \[\mathbf{v} = x_1 \mathbf{v}_1 + \dotsm + x_m \mathbf{v}_m.\] To signify this, we write \[[\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} x_1 \\ \vdots \\ x_m \end{bmatrix}_{\mathcal{B}}.\]
Here's an example that we saw last week:
Example (Coordinates)
Let \(\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) and \(\mathbf{v}_2 = \begin{bmatrix} 0 \\ 2 \end{bmatrix}.\) These vectors form a basis \(\mathcal{B}\) for \(\mathbb{R}^2.\) Then, the vector \(\mathbf{v} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}\) can be written uniquely as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2,\) specifically, \(\mathbf{v} = 5\mathbf{v}_1 - 2\mathbf{v}_2.\) This means that the coordinates of \(\mathbf{v}\) with respect to the basis \(\mathcal{B}\) are \((5,-2).\) In other words, \[[\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} 5 \\ -2 \end{bmatrix}_{\mathcal{B}}.\]
Notice that the "usual" coordinates of a vector \(\mathbf{v} \in \mathbb{R}^n\) are the same as the coordinates of that vector with respect to the standard basis of \(\mathbb{R}^n.\) In other words, when we write a vector \(\mathbf{v} = (v_1, \dots, v_n),\) we are really saying that \[\mathbf{v} = v_1 \mathbf{e}_1 + \dotsm + v_n \mathbf{e}_n.\] Philosophically, there is really no difference between the standard basis and any other basis; it's all relative. Another way to think of it is, when we originally defined the space \(\mathbb{R}^n\) as the set of all vectors with \(n\) components, we were implicitly declaring some \(n\) vectors as the standard basis, and we used that basis to form our coordinate system.
Given all this, it is natural to ask how one might change coordinates from one basis to another. In other words, if we have a vector \(\mathbf{v},\) how can we convert between its coordinates in the standard basis and in some other basis?
To answer this, suppose we have a (non-standard) basis \(\mathcal{B} = (\mathbf{v}_1, \dots, \mathbf{v}_n)\) of \(\mathbb{R}^n,\) and we know the coordinates of each basis vector \(\mathbf{v}_i\) in the standard basis. If we are given a vector \(\mathbf{v}\) expressed in terms of this basis \(\mathcal{B},\) then we know that \[\mathbf{v} = [\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix}_{\mathcal{B}} = c_1 v_1 + \dotsm + c_n v_n\] where \((c_1, \dots, c_n)\) are the coordinates of \(\mathbf{v}\) with respect to \(\mathcal{B}.\) To compute the coordinates of \(\mathbf{v}\) in the standard basis, all we need to do is evaluate this linear combination \[\mathbf{v} = c_1 v_1 + \dotsm + c_n v_n.\] In fact, we can view this in a more systematic way: letting \[U = \begin{bmatrix} | & & | \\ \mathbf{v}_1 & \dotsm & \mathbf{v}_n \\ | & & | \end{bmatrix},\] we see that \[\mathbf{v} = U \cdot [\mathbf{v}]_{\mathcal{B}}.\] In other words, to convert from the basis \(\mathcal{B}\) to the standard basis, we take the coordinates with respect to \(\mathcal{B}\) and multiply by the matrix \(U\) which has the basis vectors as columns.
Example (Non-standard basis to standard basis)
You can check that the vectors \(\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix},\) \(\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\) and \(\mathbf{v}_1 = \begin{bmatrix} 5 \\ -2 \\ 4 \end{bmatrix}\) form a basis \(\mathcal{B}\) for \(\mathbb{R}^3.\) Suppose that someone is using this basis to represent vectors in \(\mathbb{R}^3,\) and says that they have a vector \(\mathbf{v}\) with coordinates \((2, 1, 1).\) To figure out what vector that is in the standard basis, we should compute \[ U \cdot [\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} 1 & 0 & 5 \\ 2 & 1 & -2 \\ 3 & 1 & 4 \end{bmatrix} \begin{bmatrix}2 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 7 \\ 3 \\ 11 \end{bmatrix}.\] Thus, when expressed in terms of the standard basis, \(\mathbf{v}\) has coordinates \((7, 3, 11).\)
We can now consider the reverse procedure: if we have a vector written normally (i.e., in terms of the standard basis), how can we find the coordinates of that vector with respect to the standard basis? In other words, given \(\mathbf{v} \in \mathbb{R}^n,\) we want to find the representation \([\mathbf{v}]_{\mathcal{B}}.\) To do this, we can use what we just learned: these two sets of coordinates should satisfy the vector equation \[\mathbf{v} = U \cdot [\mathbf{v}]_{\mathcal{B}}.\] In order to solve for \([\mathbf{v}]_{\mathcal{B}},\) we would like to "divide" by the matrix \(U,\) which we know means multiply by its inverse. Luckily, \(U\) is an invertible matrix because its columns form a basis for \(\mathbb{R}^n\) (by the unifying theorem)! So, we can indeed multiply on the left by \(U^{-1}\) and get \[U^{-1} \mathbf{v} = [\mathbf{v}]_{\mathcal{B}}.\] Therefore, to convert from the standard basis to a non-standard basis, we multiply by the inverse of the matrix corresponding to the non-standard basis.
Example (Standard basis to non-standard basis)
In the previous example, we found a vector \(\mathbf{v}\) with (standard) coordinates \((7, 3, 11).\) To convert back to the non-standard basis from above, we need to invert the matrix \(U.\) Doing so, using our row reduction method, will give \[U^{-1} = \begin{bmatrix} 1 & 0 & 5 \\ 2 & 1 & -2 \\ 3 & 1 & 4 \end{bmatrix}^{-1} = \begin{bmatrix} 6 & 5 & -5 \\ -14 & -11 & 12 \\ -1 & -1 & 1 \end{bmatrix}.\] Then, we have \[[\mathbf{v}]_{\mathcal{B}} = U^{-1} \mathbf{v} = \begin{bmatrix} 6 & 5 & -5 \\ -14 & -11 & 12 \\ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 7 \\ 3 \\ 11 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}.\] This indeed matches the coordinates given to us in the previous example.
We call the matrix \(U\) the change of basis matrix for the basis \(\mathcal{B}.\)
Changing the basis (changing the coordinate system) can be very useful because a linear transformation, or some other object, can often look much nicer with respect to a different basis (different coordinates). We'll see one example of this when we discuss diagonalization in a few weeks.
For Midterm 2 on Friday, here is a list of topics that you should be familiar with:
I recommend looking through the previous midterms given in the Week 6 Module. (The problems on this exam should be pretty similar to those.) You can also look at any past homework problems that you weren't sure about at the time.