As discussed in the article Matrices of Linear Transformations, the matrix representing a linear transformation depends of the bases chosen for the vector spaces associated with the transformation. For example, let T be the linear transformation from \mathbb{R}^2 to \mathbb{R}^2. Some possible bases for \mathbb{R}^2 are B = {(1,-2), (1,1)} and C = {(-1,1), (5,3)}. By the methods explained in the article Matrices of Linear Transformation, it can be shown that the matrix of T with respect to B is K = \begin{bmatrix} -8/3 & 4/3 \\ -13/3 & 20/3 \end{bmatrix} and the matrix of T with respect to C is L = \begin{bmatrix} -2 & 0 \\ 0 & 6 \end{bmatrix}. Thus the two unequal matrices represent the same linear transformation. In the language of linear algebra, such matrices are called similar matrices.

 

An interesting relationship between similar matrices is as described in the theorem below.

 

Theorem

Let T: V → V be a linear operator on a finite dimensional vector space V. If K is the matrix of T with respect to a basis B and L is the matrix of T with respect to a basis C, the L = P-1KP where P is the transition matrix from C to B.

(As for the transition matrix, we refer the readers to the article Coordinates and Change of Basis.)

 

Diagrammatically, the theorem can be represented as follows.

In the diagram, L = P-1KP and \vec{u} is any vector in the vector space V.

 

Example 1

Find the transition matrix P such that L = P-1KP and verify that L = P-1KP.

Answer

By the theorem, P is the transition matrix from C to B. Let \vec{b}_1 = (1,-2), \: \vec{b}_2 = (1,1), \: \vec{c}_1 = (-1,1), \: \vec{c}_2 = (5,3). Consequently, B = \begin{Bmatrix} \vec{b}_1, \vec{b}_2 \end{Bmatrix} and C = \begin{Bmatrix} \vec{c}_1, \vec{c}_2 \end{Bmatrix}. It can be shown that \vec{c}_1 = - \frac{2}{3} \vec{b}_1 - \frac{1}{3} \vec{b}_2 and \vec{c}_2 = \frac{2}{3} \vec{b}_1 + \frac{13}{3} \vec{b}_2. Therefore, [\vec{c}_1]_B = \begin{bmatrix} -2/3 \\ -1/3 \end{bmatrix} and [\vec{c}_2]_B = \begin{bmatrix} 2/3 \\ 13/3 \end{bmatrix}. Then, the transition matrix from C to B is P = \begin{bmatrix} -2/3 & 2/3 \\ -1/3 & 13/3 \end{bmatrix}.

The task is now to show that L = P-1KP. It is easy to check that P^{-1} = \begin{bmatrix} -13/8 & 1/4 \\ -1/8 & 1/4 \end{bmatrix}. As a consequence,

P^{-1}KP = \begin{bmatrix} -13/8 & 1/4 \\ -1/8 & 1/4 \end{bmatrix} \begin{bmatrix} -8/3 & 4/3 \\ -13/3 & 20/3 \end{bmatrix} \begin{bmatrix} -2/3 & 2/3 \\ -1/3 & 13/3 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & 6 \end{bmatrix}.

By comparing P-1KP just obtained with L above, the relationship L = P-1KP is established.

 

The theorem motivates the following definition of similar matrices.

 

Definition

If A and B are square matrices, we say that B is similar to A if there is an invertible matrix P such that B = P-1AP.

 

Moreover, it is easy to prove that if B is similar to A then A is similar to B. Thus, B is similar to A if and only if A is similar to B. For brevity, it is customary to say that A and B are similar.

 

Example 2

Let T be the linear transformation from \mathbb{R}^2 to \mathbb{R}^2 defined by T((x1,x2)) = (2x1 – 3x2,3x1 + x2). Let 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 = (1,0), \: \vec{b}_2 = (0,1), \: \vec{c}_1 = (-1,2), \: \vec{c}_2 = (1,1).

  1. Find the matrix of T with respect to B.
  2. Find the matrix of T with respect to C by using the theorem above.
  3. Let \vec{v} = (2,8). Find T(\vec{v}) by using the results in part a, part b, and by direct computation. Compare the results.

Answer

Part a

By the methods explained in the article Matrices of Linear Transformation, it can be shown that the matrix of T with respect to B is K = \begin{bmatrix}2 & -3 \\ 3 & 1 \end{bmatrix}.

Part b

Firstly, we have to find the transition matrix from C to B, namely P. As explained in the article Coordinates and Change of Basis, the column vectors of P are the coordinate matrices of the basis vectors in the “new” basis relative to the “old” basis. Treating C as the “new” basis, \vec{c}_1 = (-1,2) = (-1) \vec{b}_1 + 2 \vec{b}_2 and consequently [\vec{c}_1]_B = \begin{bmatrix} -1 \\ 2 \end{bmatrix}. Similarly, \vec{c}_2 = (1,1) = 1 \vec{b}_1 + 1 \vec{b}_2 and [\vec{c}_2]_B = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. This yields the transition matrix P = \begin{bmatrix} -1 & 1 \\ 2 & 1 \end{bmatrix}. It is easy to check that P^{-1} = \frac{1}{3} \begin{bmatrix} -1 & 1 \\ 2 & 1 \end{bmatrix}. By applying the theorem above, the matrix of T with respect to C is L as determined below.

L = P^{-1} K P = \frac{1}{3} \begin{bmatrix} -1 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix}2 & -3 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 2 & 1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 7 & 5 \\ -17 & 2 \end{bmatrix}

Part c

By using the result in part a,

\vec{v} = (2,8) = 2(1,0) + 8(0,1) = 2 \vec{b}_1 + 8 \vec{b}_2

[\vec{v}]_B = \begin{bmatrix} 2 \\ 8 \end{bmatrix}

[T(\vec{v})]_B = K [\vec{v}]_B = \begin{bmatrix}2 & -3 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 8 \end{bmatrix} = \begin{bmatrix} -20 \\ 14 \end{bmatrix}

T(\vec{v}) = -20 \vec{b}_1 + 14 \vec{b}_2 = -20(1,0) + 14(0,1) = (-20,14)

By using the result in part b,

\vec{v} = (2,8) = 2(-1,2) + 4(1,1) = 2 \vec{c}_1 + 4 \vec{c}_2

[\vec{v}]_C = \begin{bmatrix} 2 \\ 4 \end{bmatrix}

[T(\vec{v})]_C = L [\vec{v}]_C = \frac{1}{3} \begin{bmatrix} 7 & 5 \\ -17 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} 34/3 \\ -26/3 \end{bmatrix}

T(\vec{v}) = \frac{34}{3} \vec{c}_1 - \frac{26}{3} \vec{c}_2 = \frac{34}{3} (-1,2) - \frac{26}{3} (1,1) = (-20,14)

By direct computation, substitute x1 = 2 and x2 = 8 into T((x1,x2)) = (2x1 – 3x2,3x1 + x2). Thus, T((2,8)) = (2⋅2-3⋅8,3⋅2+8) = (-20,14).

Comparing the three methods, it can be concluded that they all give the same results.

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