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 $A$, then :

• A. $A-\lambda I=0$
• B. $A-\lambda I$ is singular
• C. $A-\lambda I$ is non-singular
• D. none of these

Answer 1

The correct option is B $A-\lambda I$ is singular

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
$A"X={\lambda }^{\prime \prime }X\forall n\in N$

Also,

$|P^{-1}AP-\lambda I|=|P^{-1}(A-\lambda I)P|$
$=|P^{-1}||A-\lambda I||P|=|A-\lambda I|$