Show that if \({\bf{x}}\) is an eigenvector of the matrix product \(AB\) and \(B{\rm{x}} \ne 0\), then \(B{\rm{x}}\) is an eigenvector of\(BA\).

A matrix(plural matrices) is arectangular array or table ofnumbers,symbols, orexpressions, arranged in rows and columns, which is used to represent amathematicalobject or a property of such an object.

A vector \(v\)is called an eigenvector of corresponding to an eigenvalue of a matrix \(A\) if it satisfies \(A{\rm{v}} = \lambda {\rm{v}}\).

Since \({\rm{x}}\) is an eigenvector of \(AB\), hence there exist an eigenvalue \(\lambda \), such that

\(\left( {AB} \right){\rm{x}} = \lambda {\rm{x}}\).

We can rewrite this equality as \(A(B{\rm{x}}) = \lambda {\rm{x}}\).

Multiplying both sides by \(B\) from the left we will have,

\(\begin{array}{c}BA\left( {B{\rm{x}}} \right) &= B\left( {\lambda {\rm{x}}} \right)\\\left( {BA} \right)\left( {B{\rm{x}}} \right) &= \lambda \left( {B{\rm{x}}} \right)\end{array}\)

It is given that\(B{\rm{x}} \ne 0\), so by the definition \(B{\rm{x}}\) is an eigenvector of \(BA\) (\(\lambda \) is an eigenvalue).

Therefore,

\[\begin{aligned}{c}AB{\rm{x}} &= \lambda {\rm{x}}\\BA\left( {B{\rm{x}}} \right) &= B\left( {\lambda {\rm{x}}} \right)\\BA\left( {B{\rm{x}}} \right){\rm{ }} &= \lambda \left( {B{\rm{x}}} \right)\end{aligned}\]

It is proved that if \({\rm{x}}\) is an eigenvector of \(AB\) then \(B{\rm{x}}\) is an eigenvector of \(BA\).