GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 6

Answer 1

 $Given  Ax = \lambda x$

As A in invertible so

Option A:  $A^{-1}Ax = \lambda A^{-1}x $ (multiply by A^{-1) on both side)

 (1/ $\lambda$)x = $A^{-1}$x

$A^{-1}$x = $\lambda^{‘}$x 
Hence option A is true 

Option B:  $AAx =  \lambda A x $ (multiply by A on both side)

$A^{2}x = \lambda^{2}x$ (as Ax =$\lambda x$)

Finally $A^{2}x = \lambda^{‘}x$
hence B is true.
Option C:   $Ax + x = \lambda x + x $ (Adding x on both side)

$(A +I)x = (\lambda +1)x$

$(A +I)x = \lambda^{‘’}x$

so C is Also True.
Option D:similar to option C just add 2x on both side.
 

$Hence A,B,C,D all are true$

 

 

Comments

if A is not invertable what all are correct options?

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In think then only option A will be false as inverse of A is not possible. Other options are not depends on A inverse.

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@RK301112 Then, atleast one eigenvalue will be 0, so lambda can't not go in denominator as done in above derivation. 

Except option A,

All other option are correct.

Answer 2

Given $|A| \neq 0, \vec{V}$ is an eigenvector of 'A'.

A)
$$
\begin{aligned}
& A \vec{V}=\lambda \vec{V} \\
& \vec{V}=\lambda A^{-1} \vec{V} \\
& \Rightarrow A^{-1} \vec{V}=\frac{1}{\lambda} \vec{V} \quad
\end{aligned}
$$
                                                                 $\therefore \vec{v}$ is an eigenvector of $A^{-1}$

B) $A \vec{V}=\lambda \vec{V}$
$$
\begin{aligned}
A^2 \vec{v}=A A \vec{V} & =A \lambda \vec{V} \\
& =\lambda^2 \vec{V}
\end{aligned}
$$
                                                               $\therefore \vec{V}$ is an eigen vector of $A^2$.

C)
$$
\begin{aligned}
&(A+I) \vec{v}=A \vec{v}+\vec{v}\\
&=(\lambda+1) \vec{v}\\ \quad \therefore \vec{v} \text { is an eigen vector of }
& A+I .
\end{aligned}
$$
D)
$$
\begin{aligned}
(A+2 I) \vec{v} & =A \vec{v}+2 \vec{v} \\
& =(\lambda+2) \vec{v}
\end{aligned}
$$
                                                        $\therefore \vec{v}$ is an eigenvector of $ A+2 I \text {. } $