# Introduction to similar matrix and similar matrix transformation

How to perform similar matrices transformation?  Two square matrices A and B of the same order nxn are said to be similar if there exists an invertible matrix P such that B=P-1AP. It can also be written as A=PBP-1

The process of transforming a matrix A into another matrix B that is similar to it is called similar matrix transformation.

The similar matrices have same characteristic equation that’s why their eigen values are always same. If x is an eigen vector of A and Y is the eigen vector of B, then the relation between their eigen vectors is:

Y=P-1X

## Prove that similar matrices have same eigen values

If B=P-1AP

then

B-λI=P-1AP-λI

As

P-1P=I so put it in the above expression in place of I

B-λI=P-1AP-λP-1P

B-λI=P-1(A-λI)P

Taking det on both sides, we get

det(B-λI)=det[P-1(A-λI)P]

by the product rule of determinant, we get

det(B-λI)=det(P-1)det(A-λI)det(P)

or

det(B-λI)=det(P-1).det(P).det(A-λI)

or

det(B-λI)=det(P-1P).det(A-λI)

since

P-1P=I , so

det(B-λI)=det(I).det(A-λI)

and det(I)=1

so

det(B-λI)=det(A-λI)

## Prove that similar matrices have same eigen vectors that are related as Y=P-1X

Ax=λx

P-1Ax=P-1λx

or

P-1Ax=λP-1x

P-1Ax=λP-1Ix

P-1AIx=λP-1x

P-1A(PP-1)x=λP-1x

or

(P-1AP)P-1x=λP-1x

As

P-1AP=B

so

BP-1x=λP-1x

call P-1x=Y

so

BY=λY

where Y is the eigen vector of B corresponding to eigen value λ.

### Application of similar matrix transformation

This transformation helps to find the higher powers of A. For example if A=A=PDP-1

Then

A2=(PDP-1)(PDP-1)

or

A2=PD(P-1P)DP-1

A2=PD(I)DP-1

A2=PD.DP-1

A2=PD2P-1

Similarly

A3=A2.A

A3=(PD2P-1).A

A3=(PD2P-1).PDP-1

A3=PD2(P-1P)DP-1

A3=PD2(I)DP-1

A3=PD3P-1

Similarly

AK=PDK P-1

## Example of similar matrices transformation

How to perform similar matrices transformation?

Eigen Values of similar Matrices

Example of similar matrices transformation