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 \xi_i^\prime of x in another basis v_1,...,v_n

(2) x=\sum\xi_i^\prime 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 \xi^\prime as a column vector.

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

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

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

Definition 1. The matrix in (3) is called a transition matrix from u_1,...,u_n to v_1,...,v_n.

Changing bases to analyze matrices

Suppose we want to analyze A. Fix a basis u_1,...,u_n and take any x\in R^n. We can decompose x as in (1). Then we have a vector of coefficients \xi. x can be considered an original and \xi - its reflection in a mirror or a clone in a parallel world. Instead of applying A to x we can apply it to its reflection \xi, to obtain A\xi. To get back to the original world, we can use A\xi as a vector of coefficients of a new vector \sum(A\xi )_iu_i and call this vector an image of x under a new mapping A_U:

(4) A_Ux\overset{def}{=}\sum(A\xi)_iu_i.

The transition x\rightarrow\xi is unique and the transition A\xi\rightarrow A_Ux is also unique, so definition (4) is correct.

Exercise 3. Show that

(5) A_U=UAU^{-1}.

Proof. By Exercise 1, \xi=U^{-1}x. (4) can be written as A_Ux=UA\xi. Combining these two equations we get A_Ux=UAU^{-1}x. Since this is true for all x, the statement follows.

The point of the transformation in (5) is that A_U may be in some ways simpler or better than A. Note that when we use the orthonormal basis of unit vectors, U=I and A_U=A.

Definition 2. The matrix A_U is called similar to A.

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