Unit 4, lesson 19

Change of Basis for Vector Coordinates

How a basis matrix rebuilds a vector from its coordinates, how to solve for coordinates in an ordered basis, and how the change-of-basis matrix and its inverse convert coordinates while the vector stays fixed.

  • Video 19
  • Change of Basis

The arrow from the origin to (3,2)(3, 2) has standard coordinates (3,2)(3, 2). Measured with the basis vectors (1,0)(1, 0) and (1,1)(1, 1) instead, the same arrow has coordinates (1,2)(1, 2), because one copy of the first plus two copies of the second lands exactly at (3,2)(3, 2). Here and below, a tuple stands for a column vector. The two lists are different, yet the arrow has not moved. This lesson shows how to compute coordinates in any ordered basis of Rn\mathbb{R}^n, how one matrix converts coordinates from one basis to another, and how to keep a change of description apart from a motion of the vector.

Key terms

Basis matrix

Definition Let B=(b1,…,bn)\mathcal{B} = (\mathbf{b}_1, \dots, \mathbf{b}_n) be an ordered basis of Rn\mathbb{R}^n. Its basis matrix is the n×nn \times n matrix

B=[b1b2⋯bn],B = \begin{bmatrix} \mathbf{b}_1 & \mathbf{b}_2 & \cdots & \mathbf{b}_n \end{bmatrix},

whose columns are the basis vectors written in standard coordinates. The coordinate vector [v]B=(c1,…,cn)[\mathbf{v}]_{\mathcal{B}} = (c_1, \dots, c_n) of a vector v\mathbf{v} lists the weights with v=c1b1+⋯+cnbn\mathbf{v} = c_1\mathbf{b}_1 + \cdots + c_n\mathbf{b}_n, so

v=B[v]B.\mathbf{v} = B[\mathbf{v}]_{\mathcal{B}}.

The weights exist because the basis spans Rn\mathbb{R}^n, and they are unique because the basis is independent. If Bc=BdB\mathbf{c} = B\mathbf{d}, then B(c−d)=0B(\mathbf{c} - \mathbf{d}) = \mathbf{0}, and independent columns force c=d\mathbf{c} = \mathbf{d}. In particular BB is invertible. The order of the basis vectors matters, because the jjth coordinate always multiplies bj\mathbf{b}_j.

In plain words A coordinate list is a recipe, and the basis matrix carries it out. Multiplying BB by the list adds up the right multiples of the basis vectors and returns the actual vector.

Example Take B=(b1,b2)\mathcal{B} = (\mathbf{b}_1, \mathbf{b}_2) with b1=(1,0)\mathbf{b}_1 = (1, 0) and b2=(1,1)\mathbf{b}_2 = (1, 1). Then

B=[1101],B[12]=[32],B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \qquad B\begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix},

so the vector v=(3,2)\mathbf{v} = (3, 2) has [v]B=(1,2)[\mathbf{v}]_{\mathcal{B}} = (1, 2). Listing the same two vectors in the order (b2,b1)(\mathbf{b}_2, \mathbf{b}_1) gives the coordinates (2,1)(2, 1) instead. By contrast, the pair (1,1)(1, 1) and (2,2)(2, 2) is not a basis of R2\mathbb{R}^2. Both vectors lie on one line, so a vector off that line such as (1,0)(1, 0) has no coefficient list at all, and the vector (2,2)(2, 2) has several, for example (2,0)(2, 0) and (0,1)(0, 1).

Solving for basis coordinates

Definition To find [v]B[\mathbf{v}]_{\mathcal{B}}, solve the square linear system Bc=vB\mathbf{c} = \mathbf{v} for the unknown column c\mathbf{c}. Since BB is invertible, the system has exactly one solution,

[v]B=B−1v.[\mathbf{v}]_{\mathcal{B}} = B^{-1}\mathbf{v}.

The inverse records the solution of this system for every right side at once, but elimination or substitution on the system itself gives the same answer. A computed coordinate list is checked by rebuilding v\mathbf{v} as B[v]BB[\mathbf{v}]_{\mathcal{B}}.

In plain words Finding coordinates means asking which combination of the basis vectors produces the given vector, and that question is a linear system.

Example With B\mathcal{B} as above and a general v=(x,y)\mathbf{v} = (x, y), write [v]B=(α,β)[\mathbf{v}]_{\mathcal{B}} = (\alpha, \beta). Then αb1+βb2=(α+β,β)\alpha\mathbf{b}_1 + \beta\mathbf{b}_2 = (\alpha + \beta, \beta) must equal (x,y)(x, y), so β=y\beta = y and α=x−y\alpha = x - y. Hence

B−1=[1−101].B^{-1} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}.

For w=(−1,3)\mathbf{w} = (-1, 3) this gives [w]B=(−1−3,3)=(−4,3)[\mathbf{w}]_{\mathcal{B}} = (-1 - 3, 3) = (-4, 3). Rebuilding confirms it, since −4(1,0)+3(1,1)=(−1,3)-4(1, 0) + 3(1, 1) = (-1, 3). A negative coordinate is not an error. It means a step against the direction of that basis vector.

Change-of-basis matrix

Definition Let B\mathcal{B} and C\mathcal{C} be two ordered bases of Rn\mathbb{R}^n with basis matrices BB and CC. The change-of-basis matrix from B\mathcal{B} to C\mathcal{C} is the n×nn \times n matrix

PC←B=C−1B.P_{\mathcal{C} \leftarrow \mathcal{B}} = C^{-1}B.

It satisfies [v]C=PC←B[v]B[\mathbf{v}]_{\mathcal{C}} = P_{\mathcal{C} \leftarrow \mathcal{B}}[\mathbf{v}]_{\mathcal{B}} for every v\mathbf{v} in Rn\mathbb{R}^n. The reason is a two-step route. Multiplying by BB rebuilds v\mathbf{v} from its B\mathcal{B}-coordinates, and multiplying by C−1C^{-1} then reads off its C\mathcal{C}-coordinates. The jjth column of PC←BP_{\mathcal{C} \leftarrow \mathcal{B}} is [bj]C[\mathbf{b}_j]_{\mathcal{C}}, the old basis vector bj\mathbf{b}_j described in the new basis. The subscript is read as ”C\mathcal{C} from B\mathcal{B}”. The output basis stands on the left, just as the output column stands on the left of the equation.

In plain words To translate a coordinate list from one basis to another, rebuild the vector and then describe it again. The product C−1BC^{-1}B does both steps, read from right to left.

Example Keep B\mathcal{B} from above and let C=(c1,c2)\mathcal{C} = (\mathbf{c}_1, \mathbf{c}_2) with c1=(1,1)\mathbf{c}_1 = (1, 1) and c2=(1,−1)\mathbf{c}_2 = (1, -1). Then

C−1=12[111−1],PC←B=C−1B=[121120].\begin{gathered} C^{-1} = \frac{1}{2}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, \\[0.4em] P_{\mathcal{C} \leftarrow \mathcal{B}} = C^{-1}B = \begin{bmatrix} \tfrac{1}{2} & 1 \\[0.2em] \tfrac{1}{2} & 0 \end{bmatrix}. \end{gathered}

Applied to [v]B=(1,2)[\mathbf{v}]_{\mathcal{B}} = (1, 2), this matrix gives (12+2,12)=(52,12)(\tfrac{1}{2} + 2, \tfrac{1}{2}) = (\tfrac{5}{2}, \tfrac{1}{2}). A direct check agrees, since 52(1,1)+12(1,−1)=(3,2)\tfrac{5}{2}(1, 1) + \tfrac{1}{2}(1, -1) = (3, 2). The columns tell the same story. The first says b1=12c1+12c2\mathbf{b}_1 = \tfrac{1}{2}\mathbf{c}_1 + \tfrac{1}{2}\mathbf{c}_2, and the second says b2=c1\mathbf{b}_2 = \mathbf{c}_1.

Inverse change-of-basis matrix

Definition Every change-of-basis matrix is invertible, and its inverse converts in the opposite direction,

(PC←B)−1=(C−1B)−1=B−1C=PB←C.\begin{aligned} \left(P_{\mathcal{C} \leftarrow \mathcal{B}}\right)^{-1} &= (C^{-1}B)^{-1} \\[0.4em] &= B^{-1}C = P_{\mathcal{B} \leftarrow \mathcal{C}}. \end{aligned}

Conversions chain in the same way. For a third ordered basis D\mathcal{D} with basis matrix DD,

PD←C PC←B=D−1C C−1B=D−1B=PD←B.\begin{aligned} P_{\mathcal{D} \leftarrow \mathcal{C}}\,P_{\mathcal{C} \leftarrow \mathcal{B}} &= D^{-1}C\,C^{-1}B \\[0.4em] &= D^{-1}B = P_{\mathcal{D} \leftarrow \mathcal{B}}. \end{aligned}

The order of the factors matters. The product BC−1BC^{-1}, with the factors of C−1BC^{-1}B swapped, is in general neither conversion.

In plain words Converting from B\mathcal{B} to C\mathcal{C} and then back again changes nothing, so the two conversion matrices undo each other. Reversing a conversion swaps the roles of the two bases.

Example For the bases above,

PB←C=B−1C=[021−1].P_{\mathcal{B} \leftarrow \mathcal{C}} = B^{-1}C = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix}.

It sends [v]C=(52,12)[\mathbf{v}]_{\mathcal{C}} = (\tfrac{5}{2}, \tfrac{1}{2}) to (0+1,52−12)=(1,2)(0 + 1, \tfrac{5}{2} - \tfrac{1}{2}) = (1, 2), which is [v]B[\mathbf{v}]_{\mathcal{B}} again. The product of the two conversions is the identity,

[021−1][121120]=[1001].\begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} \tfrac{1}{2} & 1 \\[0.2em] \tfrac{1}{2} & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

Coordinate change versus moving a vector

Definition A change of coordinates is passive. The vector v\mathbf{v} stays where it is, and only its description changes, from [v]B[\mathbf{v}]_{\mathcal{B}} to [v]C=PC←B[v]B[\mathbf{v}]_{\mathcal{C}} = P_{\mathcal{C} \leftarrow \mathcal{B}}[\mathbf{v}]_{\mathcal{B}}. Applying a linear transformation is active. It replaces v\mathbf{v} by the vector AvA\mathbf{v}, described in the same coordinates as before, so the arrow itself can move. The same invertible matrix can play either role, so the role has to be stated, and a coordinate list means something only together with the name of its basis.

In plain words Changing coordinates changes the grid you measure with, not the arrow you measure. Moving the vector changes the arrow and keeps the grid.

Example Read the matrix B−1B^{-1} found above in both ways, with v=(3,2)\mathbf{v} = (3, 2). As a coordinate change, B−1v=(1,2)B^{-1}\mathbf{v} = (1, 2) is the list [v]B[\mathbf{v}]_{\mathcal{B}}, and it still describes the arrow ending at (3,2)(3, 2), because 1b1+2b2=(3,2)1\mathbf{b}_1 + 2\mathbf{b}_2 = (3, 2). As a linear transformation of the plane, the map x↦B−1x\mathbf{x} \mapsto B^{-1}\mathbf{x} moves that arrow to a different arrow, the one ending at the standard point (1,2)(1, 2). The numbers are the same, but the first statement is about one fixed vector and the second is about two different vectors.

Check yourself

  1. Find the coordinates of v=(1,5)\mathbf{v} = (1, 5) in the ordered basis B=((2,1),(−1,1))\mathcal{B} = ((2, 1), (-1, 1)), and check the answer by rebuilding the vector.
Show the answer

Write [v]B=(α,β)[\mathbf{v}]_{\mathcal{B}} = (\alpha, \beta). Then 2α−β=12\alpha - \beta = 1 and α+β=5\alpha + \beta = 5. Adding the equations gives 3α=63\alpha = 6, so α=2\alpha = 2 and β=3\beta = 3. The coordinates are (2,3)(2, 3), and indeed 2(2,1)+3(−1,1)=(4−3,2+3)=(1,5)2(2, 1) + 3(-1, 1) = (4 - 3, 2 + 3) = (1, 5).

  1. Let B=((1,0),(1,1))\mathcal{B} = ((1, 0), (1, 1)) and C=((1,2),(0,1))\mathcal{C} = ((1, 2), (0, 1)). Find PC←BP_{\mathcal{C} \leftarrow \mathcal{B}}, then find [v]C[\mathbf{v}]_{\mathcal{C}} for the vector with [v]B=(3,−1)[\mathbf{v}]_{\mathcal{B}} = (3, -1).
Show the answer

The basis matrices and the conversion are

B=[1101],C=[1021],C−1B=[10−21][1101]=[11−2−1].\begin{gathered} B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \qquad C = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}, \\[0.4em] C^{-1}B = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix}. \end{gathered}

So PC←BP_{\mathcal{C} \leftarrow \mathcal{B}} sends (3,−1)(3, -1) to (3−1,−6+1)=(2,−5)(3 - 1, -6 + 1) = (2, -5). To check, rebuild from B\mathcal{B} to get v=3(1,0)−(1,1)=(2,−1)\mathbf{v} = 3(1, 0) - (1, 1) = (2, -1), and rebuild from C\mathcal{C} to get 2(1,2)−5(0,1)=(2,−1)2(1, 2) - 5(0, 1) = (2, -1), the same vector.

  1. With the bases and the vector v\mathbf{v} of question 2, a classmate converts [v]B[\mathbf{v}]_{\mathcal{B}} to C\mathcal{C}-coordinates by multiplying by B−1CB^{-1}C. What does that matrix actually do, and how does rebuilding v\mathbf{v} expose the mistake?
Show the answer

The product B−1CB^{-1}C is PB←CP_{\mathcal{B} \leftarrow \mathcal{C}}, which takes C\mathcal{C}-coordinates to B\mathcal{B}-coordinates. It is the inverse of the matrix from question 2,

B−1C=[−1−121].B^{-1}C = \begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}.

Applied to (3,−1)(3, -1) it gives (−2,5)(-2, 5). Rebuilding that list in C\mathcal{C} gives −2(1,2)+5(0,1)=(−2,1)-2(1, 2) + 5(0, 1) = (-2, 1), which is not v=(2,−1)\mathbf{v} = (2, -1), so the answer cannot be the C\mathcal{C}-coordinates of v\mathbf{v}. The correct conversion is C−1BC^{-1}B.

How this lesson connects