Let A and B be n×n complex matrices. A and B are said to be similar if there exists an n×n invertible matrix P such that

P^-1AP = B
or, equivalently
AP = PB

It is not just a single matrix P that exists; it is almost a vector space. The invertibility requirement precludes the inclusion of a matrix of zeroes Vhich is required for a vector space. For the moment let the invertibility requirement be ignored. Let V be the set of all P such that AP=PB. If P and Q belong to V then (P+Q) belongs to V. Since AP=PB and AQ=QB then

A(P+Q) = AP + AQ = PB + QB = (P+Q)B

Likewise if P belongs to V then for any scalar k, kP belongs to V since

A(kP) = k(AP) = k(PB) = (kP)B

Also the n×n matrix of zeroes, 0, belongs to V since A0=0=0B. And lastly, if P belongs to V then there exists a Q such that (P+Q)=0; namely (-1)P. So V is a vector space.

Without the invertibility requirement any two n×n matrices A and B would be similar because A0=0B. Let V_A,B the vector space of n×n matrices P such that AP=PB and let W_A,B be the set of invertible n×n matrices P such that AP=PB. Then

W_A,B = V_A,B - {0}

Consider now the process of attempting to find for A and B a matrix P such that AP=PB. This is the same as seeking a solution to the linear homogeneous set of equations

AP − PB = 0

Note that if A and B are similar and X is an eigenvector of B and λ is its eigenvalue then

APX = PBX = P(λX) = λPX

and hence λ is also an eigenvalue of A and PX is an eigenvector. Thus in order for two matrices to be similar they must have the same eigenvalues.

(To be continued.)