27
Aug 18

## 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 $A.$