# Linear Algebra: #10 Similar Matrices; Changing Bases

**Deﬁnition**

Let A and A' be n ×n matrices. If a matrix C ∈ GL(n, F) exists, such that A' = C

^{−1}AC then we say that the matrices A and A' are similar.

**Theorem 27**Let f :

**V**→

**V**be a linear mapping and let {

**u**, . . . ,

_{1}**u**}, {

_{n}**v**, . . . ,

_{1}**v**} be two bases for

_{n}**V**. Assume that A is the matrix for f with respect to the basis {

**v**, . . . ,

_{1}**v**} and furthermore A' is the matrix for f with respect to the basis {

_{n}**u**, . . . ,

_{1}**u**}.

_{n}Then we have A' = C

^{−1}AC.

*Proof*

From the definition of A', we have

for all i = 1, . . . , n. On the other hand we have

Therefore we have A' = C

^{−1}AC.

Note that we have written here

**v**(in terms of the summation from

_{k}*l*= 1, ...., n), and then we have said that the resulting matrix (which we call C

^{*}) is, in fact, C

^{−1}. To see that this is true, we begin with the definition of C itself. We have

That is, CC

^{*}= I

_{n}, and therefore C

^{*}= C

^{−1}

^{}.

Which mapping does the matrix C represent? From the equations

we see that it represents a mapping g :

**V**→

**V**such that g(

**v**) =

_{i}**u**for all i, expressed in terms of the original basis {

_{i}**v**, . . . ,

_{1}**v**}. So we see that a

_{n}*similarity transformation*, taking a square matrix A to a similar matrix A' = C

^{−1}AC is always associated with a change of basis for the vector space

**V**.

Much of the theory of linear algebra is concerned with ﬁnding a

*simple*basis (with respect to a given linear mapping of the vector space into itself), such that the matrix of the mapping with respect to this simpler basis is itself simple — for example diagonal, or at least trigonal.

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