The central proposition in this article is that for every linear transformation from a vector space to another there exists a matrix that can be viewed as a representation of the transformation and the entries of the matrix depend on the bases chosen for the vector spaces. As a motivating example, consider the linear transformation T from \mathbb{R}^3 to \mathbb{R}^2 defined by T((x1,x2,x3)) = (2x1-x2+x3,x1+3x2-2x3). The transformation can be also expressed as

\left[ T(x_1,x_2,x_3) \right]_{B_2} = \begin{bmatrix} 2 & -1 & 1 \\ 1 & 3 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}

Here \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} is the coordinate matrix of the vector (x1,x2,x3) relative to some predetermined basis B1, i.e. \left[ (x_1,x_2,x_3)\right]_{B_1}. The expression \left[ T(x_1,x_2,x_3) \right]_{B_2} denotes the coordinate matrix of the vector T((x1,x2,x3)) relative to some predetermined basis B2. In the motivating example, the bases chosen for each vector space are standard bases, that is B1 = {(1,0,0), (0,1,0), (0,0,1)} for \mathbb{R}^3 and B2 = {(1,0), (0,1)} for \mathbb{R}^2. In the example, A = \begin{bmatrix} 2 & -1 & 1 \\ 1 & 3 & -2 \end{bmatrix} is called the matrix of the linear transformation T with respect to the bases B1 and B2.

 

The questions reasonably arising are as follows.

  1. Is there always such matrix as A above for every linear transformation?
  2. If there is, how to find A?

The theorem below addresses the questions.

 

Theorem

Let V, W be vector spaces. Let B and C be the bases for V and W respectively. If T is a linear transformation from V to W then there exists a matrix A such that [T(\vec{v})]_C = A [\vec{v}]_B for every \vec{v} ∈ V. Moreover, if V is of dimension n with basis B = \begin{Bmatrix} \vec{b}_1, \vec{b}, \cdots , \vec{b}_n \end{Bmatrix} and W is of dimension m with basis C = \begin{Bmatrix} \vec{c}_1, \vec{c}, \cdots , \vec{c}_m \end{Bmatrix} then A is an m×n matrix which is determined as follows.

A = \left[ [T(\vec{b}_1)]_C \quad [T(\vec{b}_2)]_C \: \cdots [T(\vec{b}_n)]_C \right]

If \vec{v} ∈ V and [T(\vec{v})]_C = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_m \end{bmatrix} then T(\vec{v}) = c_1 \vec{c}_1 + c_2 \vec{c}_2 + \cdots + c_m \vec{c}_m.

The matrix A as determined above is called the matrix of T with respect to the bases B and C.

 

Example 1

Let Pn be the vector space whose members, or vectors, are all real polynomial functions having degree ≤ n, that is, all functions expressible in the form p(x) = a0 + a1x + … + anxn where a0, a1, …, an are real numbers. Let T be the linear transformation from P1 to P2 defined as follows.

T((a0 + a1x)) = (a0 – a1) + (2a0 – 3a1)x + (a0 + a1)x2

Given the standard bases B = {1, x} and C = {1, x, x2} for P1 and P2, respectively, find the matrix of T with respect to the bases B and C.

Answer

Let B = \begin{Bmatrix} \vec{b}_1, \vec{b}_2 \end{Bmatrix}, \quad C = \begin{Bmatrix} \vec{c}_1, \vec{c}_2, \vec{c}_3 \end{Bmatrix} where \vec{b}_1 = 1, \: \vec{b}_2 = x, \: \vec{c}_1 = 1, \: \vec{c}_2 = x, \: \vec{c}_3 = x^2. By the definition of T, it follows that

\\ T(\vec{b}_1) = T(1) = T(1 + 0x) = (1-0) + (2 \cdot 1 - 3 \cdot 0)x + (1 + 0)x^2 = 1 + 2x + x^2 \\ T(\vec{b}_2) = T(x) = T(0 + 1x) = (0-1) + (2 \cdot 0 - 3 \cdot 1)x + (0 + 1)x^2 = -1 - 3x + x^2

Since the basis for P2 is the standard basis C = {1, x, x2}, the coordinate matrices of T(\vec{b}_1) and T(\vec{b}_2) can be easily obtained as follows.

[ T(\vec{b}_1) ]_C = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} and [ T(\vec{b}_2) ]_C = \begin{bmatrix} -1 \\ -3 \\ 1 \end{bmatrix}

The theorem above implies that the matrix of T with respect to the bases B and C is A = \begin{bmatrix} 1 & -1 \\ 2 & -3 \\ 1 & 1 \end{bmatrix}.

 

Example 2

Referring to Example 1, suppose that \vec{v} = 2 + 5x.

  1. Find the coordinate matrix of T(\vec{v}) relative to the standard basis C by computing A [\vec{b}]_B.
  2. Using [T(\vec{v})]_C obtained in part a, compute T(\vec{v}).
  3. Find T(\vec{v}) by the definition of T, that is T((a0 + a1x)) = (a0 – a1) + (2a0 – 3a1)x + (a0 + a1)x2 and verify that the result is the same as the one obtained in part b.

Answer

Part a

Note that \vec{v} = 2 + 5x = 2 \vec{b}_1 + 5 \vec{b}_2. Consequently, the coordinate matrix of \vec{v} relative to B is [\vec{v}]_B = \begin{bmatrix} 2 \\ 5 \end{bmatrix} and the coordinate matrix of T(\vec{v}) relative to the standard basis C is

[T(\vec{v})]_C = A [\vec{v}]_B = \begin{bmatrix} 1 & -1 \\ 2 & -3 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 5 \end{bmatrix} = \begin{bmatrix} -3 \\ -11 \\ 7 \end{bmatrix}.

Part b

By the definition of the coordinate matrix of a vector relative to a basis,  [T(\vec{v})]_C = \begin{bmatrix} -3 \\ -11 \\ 7 \end{bmatrix} implies T(\vec{v}) = -3 \vec{c}_1 - 11 \vec{c}_2 + 7 \vec{c}_3 = -3 - 11x + 7x^2.

Part c

By the definition of T, the image of (a0 + a1x) under T is (a0 – a1) + (2a0 – 3a1)x + (a0 + a1)x2.

Thus, T(\vec{v}) = T((2 + 5x)) = (2-5) + (2⋅2 – 3⋅5)x + (2 + 5)x2 = -3 – 11x + 7x2. This agrees with the result in part b.

 

That the matrix of a linear transformation depends on the basis chosen for each vector space is illustrated in Example 3 below.

 

Example 3

Let T be the linear transformation from \mathbb{R}^2 to \mathbb{R}^2 (i.e., T is a linear operator on \mathbb{R}^2) defined by T((x,y)) = (3x+5y,3x+y). Some possible bases for \mathbb{R}^2 are the standard basis B = {(1,0), (0,1)} and C = {(-1,1), (5,3)}.

  1. Find the matrix of T with respect to the B. (Name the matrix K.)
  2. Find the matrix of T with respect to the C. (Name the matrix L.)
  3. Let \vec{v} = (1,7) \in \mathbb{R}^2. Find T(\vec{v}) from [T(\vec{v})]_B = K [\vec{v}]_B and [T(\vec{v})]_C = L [\vec{v}]_C and compare the results.

Answer

Part a

Let \vec{b}_1 = (1,0), \: \vec{b}_2 = (0,1). Then

\\ T(\vec{b}_1) = T((1,0)) = (3 \cdot 1 + 5 \cdot 0,3 \cdot 1 + 0) = (3,3) \\ T(\vec{b}_2) = T((0,1)) = (3 \cdot 0 + 5 \cdot 1,3 \cdot 0 + 1) = (5,1)

Since the basis chosen is the standard basis, it is easy to deduce that [T(\vec{b}_1)]_B = \begin{bmatrix} 3 \\ 3 \end{bmatrix} and [T(\vec{b}_2)]_B = \begin{bmatrix} 5 \\ 1 \end{bmatrix}.

By the theorem above, the required matrix is K = \begin{bmatrix} 3 & 5 \\ 3 & 1 \end{bmatrix}.

Part b

Let \vec{c}_1 = (-1,1), \: \vec{c}_2 = (5,3). Then

\\ T(\vec{c}_1) = T((-1,1)) = (3 \cdot (-1) + 5 \cdot 1,3 \cdot (-1) + 1) = (2,-2) \\ T(\vec{c}_2) = T((5,3)) = (3 \cdot 5 + 5 \cdot 3,3 \cdot 5 + 3) = (30,18)

Note that \\ T(\vec{c}_1) = (2,-2) = -2(-1,1) = -2 \vec{c}_1 + 0 \vec{c}_2. From this, we have [T(\vec{c}_1)]_C = \begin{bmatrix} -2 \\ 0 \end{bmatrix}.

Similarly, \\ T(\vec{c}_2) = (30,18) = 6(5,3)= 0 \vec{c}_1 + 6 \vec{c}_2, from which we get [T(\vec{c}_2)]_C = \begin{bmatrix} 0 \\ 6 \end{bmatrix}.

By the theorem above, the desired matrix is L = \begin{bmatrix} -2 & 0 \\ 0 & 6 \end{bmatrix}.

Part c

Finding T(\vec{v}) from [T(\vec{v})]_B = K [\vec{v}]_B :

[\vec{v}]_B = [(1,7)]_B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}

[T(\vec{v})]_B = K [\vec{v}]_B = \begin{bmatrix} 3 & 5 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 7 \end{bmatrix} = \begin{bmatrix} 38 \\ 10 \end{bmatrix}

Thus, T(\vec{v}) = 38(1,0) + 10(0,1) = (38,10).

Finding T(\vec{v}) from [T(\vec{v})]_C = L [\vec{v}]_C :

It can be easily verified that \vec{v} = (1,7) = 4(-1,1) + (5,3) = 4 \vec{c}_1 + 1 \vec{c}_2. Accordingly, [\vec{v}]_C = \begin{bmatrix} 4 \\ 1 \end{bmatrix}.

[T(\vec{v})]_C = L [\vec{v}]_C = \begin{bmatrix} -2 & 0 \\ 0 & 6 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \end{bmatrix} = \begin{bmatrix} -8 \\ 6 \end{bmatrix}

This gives T(\vec{v}) = -8 \vec{c}_1 + 6 \vec{c}_2 = -8(-1,1) + 6(5,3) = (38,10). The result is the same as obtained previously.

 

Example 4

Let T be the linear operator on P1 defined as follows.

T((a0 + a1x)) = (2a0 – a1) + (4a0 + 3a1)x

Some possible bases for P1 are the standard basis B = {1, x} and C = {(2+x), (-1+3x)}.

  1. Find the matrix of T with respect to the B. (Name the matrix K.)
  2. Find the matrix of T with respect to the C. (Name the matrix L.)
  3. Let \vec{v} = 4 + 7x. Find T(\vec{v}) from [T(\vec{v})]_B = K [\vec{v}]_B and [T(\vec{v})]_C = L [\vec{v}]_C and compare the results.

Answer

Let \vec{b}_1 = 1, \: \vec{b}_2 = x. Then

\\ T(\vec{b}_1) = T(1) = T(1 + 0x) = (2 \cdot 1 - 0) + (4 \cdot 1 + 3 \cdot 0)x = 2 + 4x \\ T(\vec{b}_2) = T(x) = T(0 + 1x) = (2 \cdot 0 - 1) + (4 \cdot 0 + 3 \cdot 1)x = -1 + 3x

Since the basis chosen is the standard basis, it is easy to deduce that [T(\vec{b}_1)]_B = \begin{bmatrix} 2 \\ 4 \end{bmatrix} and [T(\vec{b}_2)]_B = \begin{bmatrix} -1 \\ 3 \end{bmatrix}.

By the theorem above, the required matrix is K = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}.

Part b

Let \vec{c}_1 = 2 + x, \: \vec{c}_2 = -1 + 3x. Then

\\ T(\vec{c}_1) = T(2+x) = T(2+1x) = (2 \cdot 2 - 1) + (4 \cdot 2 + 3 \cdot 1)x = 3 + 11x \\ T(\vec{c}_2) = T(-1 + 3x) = (2 \cdot (-1) - 3) + (4 \cdot (-1) + 3 \cdot 3)x = -5 + 5x

It can be verified that T(\vec{c}_1) = 3 + 11x = \frac{20}{7} (2+x) + \frac{19}{7} (-1 + 3x) = \frac{20}{7} \vec{c}_1 + \frac{19}{7} \vec{c}_2. This results in [T(\vec{c}_1)]_C = \begin{bmatrix} 20/7 \\ 19/7 \end{bmatrix}.

Similarly, \\ T(\vec{c}_2) = -5 + 5x = - \frac{10}{7} (2+x) + \frac{15}{7} (-1 + 3x) = - \frac{10}{7} \vec{c}_1 + \frac{15}{7} \vec{c}_2, from which we get [T(\vec{c}_2)]_C = \begin{bmatrix} -10/7 \\ 15/7 \end{bmatrix}.

By the theorem above, the desired matrix is L = \begin{bmatrix} 20/7 & -10/7 \\ 19/7 & 15/7 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 20 & -10 \\ 19 & 15 \end{bmatrix}.

Part c

Since \vec{v} = 4 + 7x, it follows that [\vec{v}]_B = \begin{bmatrix} 4 \\ 7 \end{bmatrix} and [\vec{v}]_C = \begin{bmatrix} 19/7 \\ 10/7 \end{bmatrix}.

Using K, we have:

[T(\vec{v})]_B = K [\vec{v}]_B = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 7 \end{bmatrix} = \begin{bmatrix} 1 \\ 37 \end{bmatrix}

T(\vec{v}) = \vec{b}_1 + 37 \vec{b}_2 = 1 + 37x

Using L, we have:

[T(\vec{v})]_C = L [\vec{v}]_C = \frac{1}{7} \begin{bmatrix} 20 & -10 \\ 19 & 15 \end{bmatrix} \begin{bmatrix} 19/7 \\ 10/7 \end{bmatrix} = \begin{bmatrix} 40/7 \\ 73/7 \end{bmatrix}

T(\vec{v}) = \frac{40}{7} \vec{c}_1 + \frac{73}{7} \vec{c}_2 = \frac{40}{7} (2+x) 1 + \frac{73}{7}(-1+3x) = 1+37x

The two formulae/methods yield the same result.

 

Example 5

Let T be the linear operator on P1 defined by T(a0 + a1x) = (3a0+2a1) + (a0 – 4a1)x. Two bases for P1 are B = \begin{Bmatrix} \vec{b}_1, \vec{b}_2 \end{Bmatrix} and  C = \begin{Bmatrix} \vec{c}_1, \vec{c}_2 \end{Bmatrix} where \vec{b}_1 = (1+x), \: \vec{b}_2 = (2-3x), \: \vec{c}_1 = (-2+x), and  \vec{c}_2 = (4+3x).

  1. Find the matrix A such that [T(\vec{v})]_C = A [\vec{v}]_B for every \vec{v} ∈ P1. (In this article, such matrix is referred to as inter-bases matrix of a linear operator with respect to some fixed bases. As for how to interpret this matrix, see Example 6.)
  2. Given \vec{v} = 15 - 35x ∈ P1, find T(\vec{v}) using [T(\vec{v})]_C = A [\vec{v}]_B.
  3. Compare the result in part b to the one obtained by direct computation by substituting a0 = 15 and a1 = -35 into T(a0 + a1x) = (3a0+2a1) + (a0 – 4a1)x.

Answer

Part a

T(\vec{b}_1) = T(1+x) = (3 \cdot 1 + 2 \cdot 1) + (1 - 4 \cdot 1)x =5 - 3x = - \frac{27}{10} \vec{c}_1 - \frac{1}{10}\vec{c}_2

Thus, [T(\vec{b}_1)]_C = \begin{bmatrix} -27/10 \\ -1/10 \end{bmatrix}

T(\vec{b}_2) = T(2-3x) = (3 \cdot 2 + 2 \cdot (-3)) + (2 - 4 \cdot (-3))x =14x = \frac{28}{5} \vec{c}_1 + \frac{14}{5} \vec{c}_2.

This gives [T(\vec{b}_2)]_C = \begin{bmatrix} 28/5 \\ 14/5 \end{bmatrix}.

By the theorem above, A = \left[ [T(\vec{b}_1)]_C \quad [T(\vec{b}_2)]_C \right]. So, the required matrix is A = \begin{bmatrix} -27/10 & 28/5 \\ -1/10 & 14/5 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} -27 & 56 \\ -1 & 28 \end{bmatrix}.

Part b

Note that \vec{v} = 15 - 35x = -5(1+x) + 10(2-3x) = -5 \vec{b}_1 + 10 \vec{b}_2. So, [\vec{v}]_B = \begin{bmatrix} -5 \\ 10 \end{bmatrix}. Consequently,

[T(\vec{v})]_C = A [\vec{v}]_B = \frac{1}{10} \begin{bmatrix} -27 & 56 \\ -1 & 28 \end{bmatrix} \begin{bmatrix} -5 \\ 10 \end{bmatrix} = \begin{bmatrix} 139/2 \\ 57/2 \end{bmatrix}

T(\vec{v}) = \frac{139}{2} \vec{c}_1 + \frac{57}{2} \vec{c}_2 = \frac{139}{2} (-2+x) + \frac{57}{2} (4+3x) = -25 + 155x

Part c

T(\vec{v}) = T(15-35x) = (3⋅15 + 2⋅(-35)) + (15 – 4⋅(-35))x = -25 + 155x.

The result obtained in this part is consistent with the one obtained in part b.

 

Example 6 (Interpretation of Inter-bases Matrix of A Linear Operator)

Let T be the linear operator on P1 defined by T(a0 + a1x) = (3a0+2a1) + (a0 – 4a1)x. Two bases for P1 are B = \begin{Bmatrix} \vec{b}_1, \vec{b}_2 \end{Bmatrix} and  C = \begin{Bmatrix} \vec{c}_1, \vec{c}_2 \end{Bmatrix} where \vec{b}_1 = (1+x), \: \vec{b}_2 = (2-3x), \: \vec{c}_1 = (-2+x), and  \vec{c}_2 = (4+3x). It can be shown that [T(\vec{v})]_C = F [\vec{v}]_C for every \vec{v} ∈ P1, where F = - \frac{1}{5} \begin{bmatrix} 6 & 43 \\ 8 & -1 \end{bmatrix}.

  1. Given \vec{v} ∈ P1 with \vec{v} satisfying [\vec{v}]_B = \begin{bmatrix} -5 \\ 10 \end{bmatrix}, find [T(\vec{v})]_C.
  2. Using the inter-bases matrix A in Example 5, find [T(\vec{v})]_C and compare the results.

Answer

Part a

We first have to find the transition matrix from B to C. It can be shown that \vec{b}_1 = \frac{1}{10} \vec{c}_1 + \frac{3}{10} \vec{c}_2 and \vec{b}_2 = - \frac{9}{5} \vec{c}_1 - \frac{2}{5} \vec{c}_2. Consequently, the transition matrix from B to C is P = \begin{bmatrix} 1/10 & -9/5 \\ 3/10 & -2/5 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 1 & -18 \\ 3 & -4 \end{bmatrix}. It follows that

[\vec{v}]_C = P [\vec{v}]_B = \frac{1}{10}\begin{bmatrix} 1 & -18 \\ 3 & -4 \end{bmatrix} \begin{bmatrix} -5 \\ 10 \end{bmatrix} = \begin{bmatrix} -37/2 \\ -11/2 \end{bmatrix}

[T(\vec{v})]_C = F [\vec{v}]_C = - \frac{1}{5} \begin{bmatrix} 6 & 43 \\ 8 & -1 \end{bmatrix} \begin{bmatrix} -37/2 \\ -11/2 \end{bmatrix} = \begin{bmatrix} 139/2 \\ 57/2 \end{bmatrix}

Part b

In Example 5, A = \frac{1}{10} \begin{bmatrix} -27 & 56 \\ -1 & 28 \end{bmatrix}. As shown in part b of the example, [T(\vec{v})]_C = A [\vec{v}]_B = \frac{1}{10} \begin{bmatrix} -27 & 56 \\ -1 & 28 \end{bmatrix} \begin{bmatrix} -5 \\ 10 \end{bmatrix} = \begin{bmatrix} 139/2 \\ 57/2 \end{bmatrix}. The two methods result in identical results. The inter-bases matrix serves as a ‘shortcut’ when the matrix of a linear operator is formulated based on a basis which differs from the basis used in constructing the coordinate matrix of an input vector. 

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