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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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6 Answers

Best answer
49 votes
49 votes
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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84 votes
84 votes

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

 

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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)
22 votes
22 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.
Answer:

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