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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?

1. $-5$
2. $-7$
3. $2$
4. $1$

edited | 4.4k views
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some one explain it in better way .
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By Cayley Hamilton theorem:- Every square matrix satisfy it's own characteristic equation.

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@Lakshman Patel RJIT could you please explain more

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@val_pro20

see the @Ashwani Kumar 2 solution.

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Nice Explanation

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

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.

by
edited
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{A−(λ+1)I}{A−(λ−1)I}=0

From here, how we get λ=−1   or  λ=1?
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Yes, that's a mistake. In hurry I did it. Thanks.
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I have edited the answer, i think this makes sense to me, is there anything i am missing?
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Is there any easy way to solve this question ? can anyone tell me please ?
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Please update the second line.-->"λ is the eigen value of A", rather it should be -"

λ is the eigen value of the matrix A I | I  A
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The eigen values of the matrix $\begin{pmatrix} A & I \\ I & A \end{pmatrix}$ are ($\lambda$+1) and ($\lambda$-1)

1. using ($\lambda$ + 1) we have eigen values: -4,-1,2,5
2. using ($\lambda$ - 1) we have eigen values: -6,-3,0,3

Please correct me if I'm wrong :)

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Are they so because

$A^2 - I =0 \implies (A -I)(A + I) = 0 \implies \text{ eigen values } \lambda -1 \space \text{and} \space \lambda +1$
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@toxicdesire

$\begin{pmatrix} A & I \\ I & A \end{pmatrix}$ is an 8 x 8 matrix.

So, we have 8 eigen values.

Given in question λ = {-5,-2,1,4}

1. using (λ + 1) we have eigen values: -4,-1,2,5
2. using (λ - 1) we have eigen values: -6,-3,0,3

Therefore, the 8 eigen values are {-4,-1,2,5,-6,-3,0,3}

In the options we have.

A) -5

B) -7

C) 2

D) 1

From these options, only valid eigen value is 2. So, option C)

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@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

$\gamma + 1 = \lambda$;

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

$\gamma - 1 = \lambda$;

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

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.

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Is this method of expantion or soln of det is aplicable for all types of Block matrices or only for this?
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Which of the following is an eigen value of  the matrix  [A  I,       where I is the 4×4 identity matrix?

I  A  ]

what does that mean ? inside this 4 * 4 matrix there is I which is again a 4 *4 matrix ?

might be very silly

but this line is confusing me
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@warrior

NO , not for every matrix

Can be used only for matrices  where the first and last and the second and third elements are the same
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A is 4x4.

I is 4X4.

And  the representation that is given would be of 8 x8.
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@gari

why then a 2*2 matrix is taken and solved
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@Ai$H in the selected solution the second method is used in this solution, first method is used see the text from wiki 0 the second methd is to be used 0 There is a typo, value of y is written wrong it is y = -5, -2, 1, 4 NOT y = -5, -4, 1, 4 [3rd last line] +26 votes Characteristics equation of given matrix is:$\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)$edited by +1 In options only$2$is mentioned so$C$is the correct choice 0$\text{The eigen value of the identity matrix $(I_{n})$ is} = 1.\$
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Eigen value of identity matrix is 1
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(Aλ)^2−I^2=0 How ?

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see the answer now, is it clear @Amal ?

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thanks

this kinds of Properties of matrixes are very difficult to conclude logically