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Question: If λ is an eigenvalue of A, then corresponding eigenvalue of \({{P}^{-n}}A{{P}^{n}}\) (P is a square...

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

Explanation

Solution

Hint:For solving this problem, first let the matrix P−1AP{{P}^{-1}}AP be a matrix B. For obtaining the eigenvalues, we apply the formula: B−λB-\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 λ\lambda . We are required to find the even value of the matrix P−nAPn{{P}^{-n}}A{{P}^{n}}. To evaluate this, first we tried to find the eigenvalue of the matrix P−1AP1{{P}^{-1}}A{{P}^{1}}. Let the matrix P−1AP1{{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−λI=P−1AP−λIB-\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
B−λI=P−1AP−P−1λIP B−λI=P−1(A−λI)P \begin{aligned} & B-\lambda I={{P}^{-1}}AP-{{P}^{-1}}\lambda IP \\\ & B-\lambda I={{P}^{-1}}\left( A-\lambda I \right)P \\\ \end{aligned}
Now, taking the magnitude of both sides to obtain the characteristic polynomial:
∣B−λI∣=∣P−1∣∣A−λI∣∣P∣\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
∣B−λI∣=∣P−1∣∣P∣∣A−λI∣ ∣B−λI∣=∣P−1P∣∣A−λI∣=∣A−λI∣∣I∣ ∣B−λI∣=∣A−λI∣ \begin{aligned} & \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{aligned}
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
Bn=P−nAPn{{B}^{n}}={{P}^{-n}}A{{P}^{n}}
Similarly, using the above simplification, we get
∣Bn−λI∣=∣P−n∣∣A−λI∣∣Pn∣ ∣Bn−λI∣=∣P−n∣∣Pn∣∣A−λI∣ ∣Bn−λI∣=∣P−nPn∣∣A−λI∣=∣A−λI∣∣I∣ ∣Bn−λI∣=∣A−λI∣ \begin{aligned} & \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{aligned}
Hence, the eigenvalue is λ\lambda .
Therefore, option (d) is correct.

Note: Another way of solving this problem can be:
AX=λXAX=\lambda X
Multiply both the sides with the inverse of P, we get P−1AX=λP−1X{{P}^{-1}}AX=\lambda {{P}^{-1}}X. Now, operating this equation by pre-multiplying P−1P{{P}^{-1}}P and some other rearrangements, we obtain
⇒(P−1AP)(P−1X)=λ(P−1X)\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−1AP{{P}^{-1}}AP is λ\lambda .