Aug 18

Matrix similarity

Matrix similarity

Switching bases

The basis consisting of unit vectors is simple in that the coefficients in the representation x=\sum x_ie_i are exactly the components of the vector x. With other types of bases it is not like that: the dependence of coefficients in

(1) x=\sum\xi_iu_i

on x for a general basis u_1,...,u_n is not so simple.

Exercise 1. Put the basis vectors side by side, U=(u_1,...,u_n) and write the vector of coefficients \xi as a column vector. Then (1) becomes U\xi =x, so that \xi =U^{-1}x.

Proof. By the basis definition, \sum\xi_iu_i runs over R^n and therefore \text{Img}(U)=R^n. This implies \det U\neq 0. The rest is obvious.

The explicit formula from Exercise 1 shows, in particular, that the vector of coefficients is uniquely determined by and depends linearly on x. The coefficients \eta_i of x in another basis v_1,...,v_n

(2) x=\sum\eta_i v_i

may be different from those in (1). For future applications, we need to know how the coefficients in one basis are related to those in another. Put the basis vectors side by side, V=(v_1,...,v_n), and write \eta as a column vector.

Exercise 2. Let u_1,...,u_n and v_1,...,v_n be two bases in R^n. Then

(3) \eta=V^{-1}U\xi.

Proof. With our notation (1) and (2) become x=U\xi and x=V\eta. Thus, U\xi=V\eta and (3) follows.

Definition 1. The matrix C=V^{-1}U in (3) is called a transition matrix from u_1,...,u_n to v_1,...,v_n.

Matrix representation of a linear transformation

This topic in case of an orthonormal basis was considered earlier. Some people find the next general construction simpler.

Let u_{1},...,u_{n} be a basis and decompose x as in (1). Let \mathcal{A} be a linear transformation. From

(5) \mathcal{A}x=\sum \xi _{i}\mathcal{A}u_{i}

we see that the vectors \mathcal{A}u_{i},\ i=1,...,n, uniquely determine \mathcal{A}. Decompose them further as \mathcal{A}u_{i}=\sum_{k}a_{ki}u_{k}. Let us introduce a column-vector A^{(i)} by A^{(i)}=(a_{1i},...,a_{ni})^{T}. Then the last equation takes the form \mathcal{A}u_{i}=UA^{(i)}, i=1,...,n. With the matrix A=(A^{(1)},...,A^{(n)}) we can write one equation instead of n:

(6) \mathcal{A}U=UA.

Combining (5), (6) and Exercise 1 we get \mathcal{A}x=\mathcal{A}U\xi=UAU^{-1}x. Since x is arbitrary, the linear transformation \mathcal{A} in the basis U can be identified with the matrix UAU^{-1}:

(7) \mathcal{A}=UAU^{-1}.

Definition 2. The matrix A that is defined by (6) or (7) is called a representation of the linear transformation \mathcal{A} in the basis u_{1},...,u_{n}.

Note two special cases: 1) if the basis u_1,...,u_n is orthonormal, then U is an orthogonal matrix and 2) when we use the orthonormal basis of unit vectors, U=I and \mathcal{A}=A.

Changing bases to analyze matrices

We want to see how the representations of a linear transformation \mathcal{A} in two bases are related to each other. For a basis u_{1},...,u_{n} summarized in the matrix U we have (7). To reflect dependence of A on the basis, let us denote it A_{U}. Then from (7) \mathcal{A}=UA_{U}U^{-1}. Similarly, for another basis v_{1},...,v_{n} we have \mathcal{A}=VA_{V}V^{-1}. The last two equations lead to UA_{U}U^{-1}=VA_{V}V^{-1}. Hence, A_{U}=U^{-1}VA_{V}V^{-1}U. Here C=V^{-1}U is the transition matrix, so this can be written as

(8) A_{U}=C^{-1}A_{V}C.

Definition 3. If there is a nonsingular matrix C such that (8) is true, then the matrix A_{U} is called similar to A_{V}.

Depending on the choice of the bases, one matrix may be simpler than the other.

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