GATE CSE 2017 | Set 1 | Question: 31

Question

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?

• A. Both I and II
• B. I only
• C. II only
• D. Neither I nor II

Comments

perfect

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I feel this is not an accurately framed question. 2 completely different analysis are possible for it 1} ijth elements of matrix A^2 , 2} square of ijth element. without any parenthesis or any text description this is ambiguous notation. 

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Guys I got a question regarding notation of Cofactor. Initially i thought Aij as cofactor and complicated the problem. But in lecture Aij is considered an element of the matrix. In exam, how do I not get confused with notation?

Should I startv considering Aij as element of matrix by default for every GATE matrix problem if they didn't give information that Aij is cofactor?

Answer 1

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.

Comments

Why see boring text see colorful analysis

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Truly Peak solution

 

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https://youtu.be/oykA4bXKIUY?t=164
Same question explanation by Sachin Sir

Answer 2

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

 

Comments

Sorry i  confuse between  real and rational

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hai, can u explain when hav u used the summation in your solution?

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Goated Explanation

Answer 3

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)

Comments

@Prashant_Dubey

 

This is indeed a special case where both eigenvalues are equal, specifically when \(\lambda_1 = \lambda_2 = \pm 5\). This satisfies the given condition that \(\lambda_1^2 + \lambda_2^2 = 50\). However, the characteristic polynomial \(\lambda^8 \cdot (\lambda - \lambda_1)^2\) represents just one scenario. The broader analysis considers all possible eigenvalues, not just this specific case.

Could you please clarify what additional point you are trying to make with this example?

 

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@suraj,

I could'nt able to get ur logic.

"I" can be one of the case of eigen values. 

Wrt your logic if i put 6,√14as eigen values(here also both are real values) then it also proves the equation.then according to "II", largest magnitude of eigen value can be strictly greater than 5, is also satisfying. 

Then how can we be sure to say only "I" as ans wrt your approach?

 

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@Chaithu555 the question is saying neccessarily correct means correct for all possible values not just one sepecific case for option 1 it is correct for any possioble value we can thinlk of we awlays get one of the eigen value in range [-5,5] but option second for case 5,5 this because one case where it doesnt impley hence option 2 not always neccessarliy correct

Answer 4

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.

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.

Answer 5

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)

Comments

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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rank here is 3 which is greater than 2, this is wrong i guess

Answer 6

It is given that dimension of matrix A is n x n and the rank of the matrix A is 2, from this data we can make characteristic equation:

Rank = 2: this means number pivot columns in matrix A is 2 and other are free variables. (if there are n columns in A and number of pivot columns in A is equal to 2 then number of free variable will be (n-2). 

Now, we can relate this to lecture 5 of linear algebra course where characteristic equation is given as $\lambda ^{2 }(\lambda -3)^{3}$ and A is 5 x 5 symmetric matrix(AM = GM). Now we are calculating the rank of matrix A by saying there are 2 Linear Independent eigen vectors corresponding to $(\lambda=0)$ This means, the number of free variable corresponding to $(\lambda=0)$  is equal to number of Liner independent eigen vectors corresponding to $(\lambda=0)$ since A is symmetric matrix. Rank is calculated as $(5 - fv )$ , fv is number of free variable corresponding to $(\lambda=0)$

$In$ $this$ $Question$ $Also$,

$Rank$ = $(n - free Variable )$,  from this we can infer that $FreeVariable = (n - Rank)$ = $(n - 2)$ free variable is corresponding to  $(\lambda=0)$, Now we can easily make characteristic equation from this information as:

 $\lambda ^{n-2 }(\lambda -\lambda1)^{1}(\lambda -\lambda2)^{1}$ from this equation we can say that Rank of original matrix A should be 2.

We have the characteristic equation, (We can say that there are only 2 non zero eigen values and (n-2) eigen values are all zero)

Now come to the second given information:

Sum of squares of all elements of matrix is equal to 50. ( Now to connect this information with the question we have to find (A.A-Transpose, and replace A-transpose with A, you will get A square) 

Therefore, eigen values must be between [-5 to 5]. 

Comments

matrix 1 0 0

           0 7 0

            0 0 0 satisfies the condition

but why nobody is talking about it?