hello
in this question we're asked to show that any
eigenvalue for matrix A is also an
eigenvalue of the matrix A transpose and
so we'll see that the matrices A and A transpose actually
share the same eigenvalues although the
eigenvectors might be completely
different the eigenvalues are the same
okay so we're actually given a hint
which is to use the characteristic
equation in properties of the
determinants it's a good idea and we'll do that
remember that the eigenvalues are the
all solutions to the characteristic
equation so lambda is an eigenvalue A if
and only if it satisfies this equation
and now I can use the properties of
determinants and transposes
since the determinant of a matrix is equal to the
determinant of its transpose I can just
take the transpose of this thing in here
and it's still equal to zero which is
the determinant of a minus lambda I and
now I can bring the transpose inside
these brackets this becomes a transpose
and the transpose of lambda I is still
lambda I and this is the characteristic
equation for A transpose which means
that a solution to this is an eigenvalue
of a transpose
thank you
