Question

If λ is an eigenvalue of A, then corresponding eigenvalue of ${{P}^{-n}}A{{P}^{n}}$ (P is a square matrix with same order as A) is
(a) ${{\lambda }^{n}}$
(b) 1
(c) ${{\lambda }^{-n}}$
(d) $\lambda $

Answer 1

Hint:For solving this problem, first let the matrix ${{P}^{-1}}AP$ be a matrix B. For obtaining the eigenvalues, we apply the formula: $B-\lambda $. By rearranging the terms and taking the determinant of both sides we obtain the eigenvalue to be the same as that of matrix A.

Complete step-by-step answer:
According to the problem statement, we are given a matrix A. The eigenvalue corresponding to the matrix A is \[\lambda \]. We are required to find the even value of the matrix ${{P}^{-n}}A{{P}^{n}}$. To evaluate this, first we tried to find the eigenvalue of the matrix \[{{P}^{-1}}A{{P}^{1}}\]. Let the matrix \[{{P}^{-1}}A{{P}^{1}}\] be another matrix B of same order.
Now, subtracting the matrix B with the product of the identity matrix with eigenvalue of matrix A. This can be expressed as:
$B-\lambda I={{P}^{-1}}AP-\lambda I$
Since, the identity matrix can be rewritten as a product of inverse and actual matrix. So, on further rearrangement, we get
$\begin{align}
  & B-\lambda I={{P}^{-1}}AP-{{P}^{-1}}\lambda IP \\
 & B-\lambda I={{P}^{-1}}\left( A-\lambda I \right)P \\
\end{align}$
Now, taking the magnitude of both sides to obtain the characteristic polynomial:
$\left| B-\lambda I \right|=\left| {{P}^{-1}} \right|\left| A-\lambda I \right|\left| P \right|$
Since, the determinant of a value can be shifted, so we get
$\begin{align}
  & \left| B-\lambda I \right|=\left| {{P}^{-1}} \right|\left| P \right|\left| A-\lambda I \right| \\
 & \left| B-\lambda I \right|=\left| {{P}^{-1}}P \right|\left| A-\lambda I \right|=\left| A-\lambda I \right|\left| I \right| \\
 & \left| B-\lambda I \right|=\left| A-\lambda I \right| \\
\end{align}$
Thus, the two matrices A and B have the same characteristic determinants and hence the same characteristic equations and the same characteristic roots. Now, the required thing is
${{B}^{n}}={{P}^{-n}}A{{P}^{n}}$
Similarly, using the above simplification, we get
$\begin{align}
  & \left| {{B}^{n}}-\lambda I \right|=\left| {{P}^{-n}} \right|\left| A-\lambda I \right|\left| {{P}^{n}} \right| \\
 & \left| {{B}^{n}}-\lambda I \right|=\left| {{P}^{-n}} \right|\left| {{P}^{n}} \right|\left| A-\lambda I \right| \\
 & \left| {{B}^{n}}-\lambda I \right|=\left| {{P}^{-n}}{{P}^{n}} \right|\left| A-\lambda I \right|=\left| A-\lambda I \right|\left| I \right| \\
 & \left| {{B}^{n}}-\lambda I \right|=\left| A-\lambda I \right| \\
\end{align}$
Hence, the eigenvalue is $\lambda $.
Therefore, option (d) is correct.

Note: Another way of solving this problem can be:
$AX=\lambda X$
Multiply both the sides with the inverse of P, we get ${{P}^{-1}}AX=\lambda {{P}^{-1}}X$. Now, operating this equation by pre-multiplying ${{P}^{-1}}P$ and some other rearrangements, we obtain
$\Rightarrow \left( {{P}^{-1}}AP \right)\left( {{P}^{-1}}X \right)=\lambda \left( {{P}^{-1}}X \right)$. So, on comparing both sides, we get the value of ${{P}^{-1}}AP$ is $\lambda $.