in Linear Algebra edited by
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Let $A$ be $n\times n$ real valued square symmetric matrix of rank $2$ with $\sum_{i=1}^{n}\sum_{j=1}^{n}A^{2}_{ij} = 50.$ Consider the following statements.

  1. One eigenvalue must be in $\left [ -5,5 \right ]$
  2. The eigenvalue with the largest magnitude must be strictly greater than $5$

Which of the above statements about eigenvalues of $A$ is/are necessarily CORRECT?

  1. Both I and II
  2. I only
  3. II only
  4. Neither I nor II
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@set2018

it means there exist a symmetric matrix whose rank is 2 and sum of the square of all elements = 50

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Double summation

http://www.tcd.ie/Economics/staff/thijssej/ec1030/summation.pdf

 

here matrix represented as vector format

So, we could derive it like

$(a11+a12+a13)+(a21+a22+a23)$

Much easier to solve now. comment by 

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6 Answers

49 votes
49 votes
Best answer
Eigen values of $\begin{bmatrix} 0 &5 \\ 5&0 \end{bmatrix}$ are $\pm 5$. Therefore second statement is false.

Since, the rank of matrix $A$ is $2,$ therefore atleast one eigen value would be zero for $n \ge 3$.

For $n= 2,$  It can be proven that  $\lambda_1^2 + \lambda_2^2 \le \sum_{i=1}^{n}\sum_{j=1}^{n}A_{ij}^2$.

$\lambda_{1}^{2} + \lambda_{1}^{2} \le 50$

Both $\lambda_{1}$ and  $\lambda_{2}$ would be real because $A$ is a real symmetric matrix. Which implies that atleast one eigen value would be in $[-5,5].$

Hence, correct answer is $(B)$

Now, to prove $\lambda_{1}^{2} + \lambda_{2}^{2} \le \sum_{i=1}^{n}\sum_{j=1}^{n}A_{ij}^2$ for $2\times2$ matrix, let us consider the matrix is $\begin{bmatrix} a &c \\ b&d \end{bmatrix}$ and $\lambda$ is the eigen value of this matrix.

${\begin{vmatrix}a-\lambda&c\\ b&d-\lambda\end{vmatrix}} = 0$

$\lambda^{2} - (a+d)\lambda + ad -bc =0$

Let $\lambda_{1}$ and $\lambda_{2}$ are roots of this equation.

$\lambda_{1}^{2} + \lambda_{2}^{2} = (\lambda_{1} + \lambda_2)^{2} - 2 \lambda_{1}  \lambda_{2}$

$= (a+d)^{2} -2 (ad-bc)$

$=\sum_{i=1}^{2}\sum_{j=1}^{2}A_{ij}^2  - (b-c)^2$
For real  valued matrix,
$\le \sum_{i=1}^{2}\sum_{j=1}^{2}A_{ij}^2 $ (For real  symmetric matrix, $b=c$ and $\le$ would be replaced by equal sign)
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4 Comments

clearify me something. The question says about an $n \times n$ matrix of rank $2$ but all the proves work with $n=2$. Why suddenly making this assumption is correct?
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Made Easy copying answers from here. Pathetic!
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edited by
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.
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83 votes
83 votes

Another approach...........................

 

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4 Comments

How can you say that statement 1 is true just by taking an example?
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@Gyanu very nice approach

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best answer

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root 14 is not an real number in 2 example
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22 votes
22 votes
let n=2 and matrix

5     0

0     5

here clearly (A11)^2 + (A22)^2 =50

option 1 is correct as eigen value is 5

option 2 is incorrect as no eigen value is strictly greater than 5

another example

3    0    0

0   4     0

0   0     5

i am taking only diagonal matrix just for ease of solution

so answer is option (B)

4 Comments

what is the meaning of  A2ij= 50.

and  (A11)^2 + (A22)^2 =50 kindly explain as soon as  possible

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here matrix represented as vector format

So, we could derive it like

$\sum_{i=1}^{n} \sum_{j=1}^{n}A^{2}_{ij}=\sum_{i=1}^{n} \sum_{j=1}^{n}A_{ij}.A_{ij}$

$\sum_{i=1}^{3} \sum_{j=1}^{2}A_{ji}=\left ( a_{11}+a_{12}+a_{13} \right )+\left ( a_{21}+a_{22}+a_{23} \right )$

http://www.tcd.ie/Economics/staff/thijssej/ec1030/summation.pdf

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Best and easy solution.
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21 votes
21 votes
Rank of $A_{n*n}=2$   ,    means n-2 eigen values are zero .

Let $\lambda _1$,$\lambda_2$,0,0,.....be eigen values .

We know that $\sum_{I=1}^{n}\sum_{j=1}^{n}$$A_{ij}^2$= Trace of ($AA^T$)

                          =Trace of $A^2$ (A is symmetric)

                          =$\lambda _1^2$+$\lambda_2^2$+0+0+0....+0...…..(1)

As given ${\sum_{I=1}^{n}}\sum_{j=1}^{n}$ $A_{ij}^2$=50.......(2)

From 1 and 2 ,   $\lambda_1^2 +\lambda_2^2$ =50

So we can say that at least one eigen value lies between [-5,5], so statement 1 is true and 2nd is false becuase eigen values can be $\lambda_1$=-5 or +5 , $\lambda_2$=-5 or +5.
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4 Comments

@Prateek Raghuvanshi Please correct me if I'm wrong.

If $\lambda$ is an eigen value of A, then $\lambda$$^{k}$ is the root of A$^{k}$.

This is correct but A$^{k}$ can have some other extra roots right? So,

$\sum _{i=1}^{n}$ $\sum _{j=1}^{n}$ A$_{ij}^{2}$ =$\underbrace{\lambda _{1}^{2} + \lambda _{2}^{2}+ 0^{2} +...........}$ + some more extra roots

                               terms from squaring the n-roots of A

$\geq$ $\lambda _{1}^{2} + \lambda _{2}^{2} + 0^{2} +....$

 

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Rank of $A_{n*n}$=2   ,  means n-2 eigen values are zero .

But this is not always correct, right? 

For example, https://math.stackexchange.com/questions/3710069/is-the-rank-of-a-matrix-equal-to-the-number-of-non-zero-eigenvalues

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@reboot for an A(nxn) matrix if x eigen values are 0 then atmost there can be atmost n-x non zero eigen values.

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