Answer to Question #159355 in Linear Algebra for Abhipsita Das

Question

Let A and B be n × n matrices with A invertible. Prove that AB and BA have the

same eigenvalues. 

Answer 1

Let A and B be n × n matrices with A invertible.

using the statement: if two matrices have the same characteristic equation, then they will have the same eigenvalues.

therefore in this solution, we will show that AB and BA have the same characteristic equation.

consider,

C_AB = characteristic polynomial of a matrix AB

= |AB-$\lambda\Iota$ | since, |A| denotes the determinatnt of matrix A

= |ABAA^-1 - $\lambda$ AA^-1|

= | A(BA - $\lambda$$\Iota$ )A^-1| $\because$ A is invertible, therefore AA^-1 = $\Iota$

= |A|.|BA - $\lambda$$\Iota$|.|A^-1|

= |A|.|BA - $\lambda$$\Iota$|.$\frac{1}{|A|}$

= |BA - $\lambda$$\Iota$| = characteristic polynomial of matrix BA = C_BA

Hence,

both AB and BA have the same characteristic polynomial (equation)

by using the above statement AB and BA will have the same eigenvalues.