GATE CSE 2007 | Question: 25

Question

Let A be a $4 \times 4$ matrix with eigen values -5,-2,1,4. Which of the following is an eigen value of  the matrix$\begin{bmatrix} A & I \\ I & A \end{bmatrix}$, where $I$ is the $4 \times 4$ identity matrix?

• A. $-5$
• B. $-7$
• C. $2$
• D. $1$

Comments

@Lakshman Patel RJIT could you please explain more

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@val_pro20

see the @Ashwani Kumar 2 solution.

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Anwser of above written question starts from 5:55 minutes.

Answer 1

Ans is (C) 2

$Ax = λx$, where $λ$ is the eigen value of $A$ . Hence $(A−λI)x = 0$ or $|A−λI| = 0$
So, for our given matrix, we have

$\begin{bmatrix} A-\lambda I &I \\ I& A - \lambda I \end{bmatrix} = 0$

This is a $ 2 \times 2$ block matrix where the first and last and the second and third elements are the same. So, applying the formula for determinant of a block matrix as given here

When $A = D$ and $B = C,$ the blocks are square matrices of the same order and the following formula holds (even if $A$ and $B$ do not commute)

$\det \begin{pmatrix} A & B\\ C& D \end{pmatrix} = \det (A-B) \det(A+B)$

(second last case) https://en.wikipedia.org/wiki/Determinant#Block_matrices

we get,

$|A−\lambda I− I| \times |A-\lambda I + I|= 0 $

$\implies |A-(\lambda+1)I| \times |A - (\lambda -1)I| = 0$

Each of the eigen value of $A$ is the solution of the equation $|A- \alpha I |= 0$ ($\alpha$ being the eigen value of $A$). So, we can equate $\lambda +1$ and $\lambda -1$ to any of the eigen value of $A$, and that will get our value of $\lambda$. If we take $\alpha = 1$, we get $\lambda = 2$, and that is one of the choice. For no other choice, this equation holds. So, (c) $2$ is the answer.

Comments

@rohith1001 yeh that's not what I asked, I asked how did you get to eigen values being $\lambda \pm 1$ 

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$\lambda$ is the eigen value of the matrix A.

The characteristic equation of A is $\begin{vmatrix} A - \lambda I \end{vmatrix} = 0$

Now, let $\gamma$ be the eigen value of $\begin{pmatrix} A & I\\ I & A \end{pmatrix}$

The characteristic equation of the above matrix is $\begin{vmatrix} A-\gamma I & I \\ I & A-\gamma I \end{vmatrix} = 0$

$\begin{vmatrix} A-\gamma I - I \end{vmatrix} \begin{vmatrix} A-\gamma I + I \end{vmatrix} = 0$

The above is from (second last case) https://en.wikipedia.org/wiki/Determinant#Block_matrices

$\begin{vmatrix} A-(\gamma+1)  I \end{vmatrix} \begin{vmatrix} A-(\gamma-1)I \end{vmatrix} = 0$

Compare the above equation with the characteristic equation of A.

That gives us

$\gamma + 1 = \lambda$;

$\Rightarrow$ $\gamma = \lambda - 1$;

 

$\gamma - 1 = \lambda$;

$\Rightarrow$ $\gamma = \lambda + 1$;

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From which book to study the concepts of block matrix determinants?

Answer 2

Characteristics equation of given matrix is:

$\begin{vmatrix} A - \lambda & I\\ I& A - \lambda \end{vmatrix} =0$

$ \implies (A- \lambda)^2 - I^2 = 0$

$ \implies (A-\lambda + I)(A- \lambda -I)=0$

$ \implies \lambda = A+I, A-I$

Given eigen values of $A=-5,-2,1,4$

Then $A+I$ and $A-I$ eigen values of given matrix are: (according to property of Cayley Hamilton theorem):

$A+I=(-4,-1,2,5)$

$(A-I)=(-6,-3,0,3)$

Comments

Best Solution😊 

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wow , superb soln !!

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this will be the best answer.

Answer 3

We find the eigenvalues of given matrix by solving its characteristic equation :


|(A – x I)² – I²| = 0
|(A – (x – 1) I) * (A – (x + 1) I)| = 0
|(A – (x – 1) I)| * |(A – (x + 1) I)| = 0
So, |(A – (x – 1) I)| = 0 or |(A – (x + 1) I)| = 0

Let y be eigenvalues of A, then |(A – y I)| = 0 .

So, by comparing equations we get,
either x – 1 = y or x + 1 = y

Therefore , x = y + 1 or x = y – 1

y = -5, -4, 1, 4 (given)

So, x = −4, −1, 2, 5, −6, −3, 0, 3

 
Thus, option (C) is the answer.

Comments

@Ai$H

in the selected solution the second method is used

in this solution, first method is used 

see the text from wiki

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the second methd is to be used

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There is a typo, value of y is written wrong it is y =  -5, -2, 1, 4

NOT y = -5, -4, 1, 4 [3rd last line]

Answer 4

 

So above given matrix has 8 eigen values and we know the eigen value of identity matrix is always 1. if identity matrix size 4x4 then four eigen value={1,1,1,1}.

Comments

Can you please explain the line $\lambda_1 = eigenvalue(A) + eigenvalue(I)$ ? Is this some theorem?

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In the above picture, λ1 = A + I means eigenvalues of A + eigenvalues of I. Similarly  λ1 = A - I means eigenvalues of A – eigenvalues of I.