I have put it as answer too -
https://gateoverflow.in/118312/gate-cse-2017-set-1-question-31?show=407380#a407380
let $A=\left[\begin{array}{lll}\uparrow & \uparrow & \uparrow \\ a_{1} & a_{2} & a_{3} \\ \downarrow & \downarrow & \downarrow\end{array}\right]$ what will be $A^{\top} A=$ ?
$$ A^TA =
\left[\begin{array}{c}
\longleftarrow a_{1}^{\top} \longrightarrow \\
\longleftarrow a_{2}^{\top} \longrightarrow \\
\longleftarrow a_{3}^{\top} \longrightarrow
\end{array}\right]\left[\begin{array}{lll}\uparrow & \uparrow & \uparrow \\ a_{1} & a_{2} & a_{3} \\ \downarrow & \downarrow & \downarrow\end{array}\right]
$$
$\therefore$ Diagonal elements of $A^{\top} A=\left\|a_{1}\right\|^{2},\|a_2\|^{2},\|a_3|^{2}$ etc. Here $a_{1}, a_{2}, a_{3}$ are vector (columns of $A$ )
Now, what is trace $\left(A^{\top} A\right)=$ ?
$$
\begin{array}{r}
\operatorname{trace}\left(A^{\top} A\right)=\left\|a_{1}\right\|^{2}+\left\|a_{2}\right\|^{2}+\left\|a_{3}^{2}\right\|
\end{array}
$$
It is in fact sum of the square of all elements of $A$ (Think about it, it is very crucial)
Given that, trace ($A^TA$ ) is 50 .
Also, given that $\operatorname{rank}(A)=2$.
let $A$ is $10 \times 10$ then number of pivot variables = $2$, number of free variables = $8$.
Free variables $=8$ means Number of linearly independent eigenvectors for $\lambda=0$ is 8 .
Now it means in characteristic polynomial, we have at least $\lambda=0$ power 8 .
This means characteristic polynomial could be one of the following –
Option a) $\lambda^{8} \cdot\left(\lambda-\lambda_{1}\right)\left(\lambda-\lambda_{2}\right)$
Option b) $\lambda^{9}\left(\lambda-\lambda_{1}\right)$
Option C) $\lambda^{10}$
Since real Symmetric matrix is given So, $$\text{ Algebraic multiplicity(AM)} = \text{Geometric multiplicity(GM)}$$.
Hence we have characteristic polynomial as $\lambda^{8} \cdot\left(\lambda-\lambda_{1}\right)\left(\lambda-\lambda_{2}\right)$.
Eigen values of $A=\overbrace{0,0,\dots 0}^\text{8 times}, \lambda_{1}, \lambda_{2}$
Eigen values of $A^2=\overbrace{0,0,\dots 0}^\text{8 times}, \lambda_{1}^{2}, \lambda_{2}^{2}$
For symmetric matrix $A^{\top} A=A^{2}$
$\begin{aligned} \operatorname{trace}\left(A^{\top} A\right)=\operatorname{trace}\left(A^{2}\right)=& 50 \\ \end{aligned}$
$0+0 \cdots 0+\lambda_{1}^{2}+\lambda_{2}^{2}=50$
$\Rightarrow \quad \lambda_{1}^{2}+\lambda_{2}^{2}=50$
If both lambda’s are outside the range of $[-5, 5]$ then sum of $\lambda^{2}+\lambda_{2}^{2}$ can not be 50.