Matrix similarity

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 $\eta_i$ 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