Question

Let $A$ be a square matrix of order $n\times n$. A constant $\lambda$ is said to be characteristic root of $A$ if there exists a $n\times 1$ matrix $X$ such that $AX=\lambda X$

If $\lambda$ is a characteristic root of $A$ and $n\in N$, then ${\lambda }^n$ is a characteristic root of

• A. $A^n$
• B. $A^{n-1}$
• C. $A^{-n}$
• D. $A-A^n+A^{-n}$

Answer 1

Answer: (A)

$A^n$

Since $X\ne 0$ is such that $(A-\lambda I)X=0,|A-\lambda I|=0\iff A-\lambda I$ is singular. If $A-\lambda I$ is non-singular the then equation $(A-\lambda I)X=0\Rightarrow X=0$
If $\lambda =0$, we get $|A|=0\Rightarrow A$ is singular.
We have $A^{2}X=A\left(AX\right)=A\left(\lambda X\right)=\lambda \left(AX\right)$
$={\lambda }^{2}X$,
$A^{3}X=A\left(A^{2}X\right)=A\left({\lambda }^{2}X\right)$
$={\lambda }^2\left(AX\right)={\lambda }^2\left(\lambda X\right)={\lambda }^{3}X$
Continuing in this way, we obtain
$A^{n}X={\lambda }^{n}X\forall n\in N$