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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. $1$

some one explain it in better way .

By Cayley Hamilton theorem:- Every square matrix satisfy it's own characteristic equation.

@Lakshman Patel RJIT could you please explain more

@val_pro20

see the @Ashwani Kumar 2 solution.

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

{A−(λ+1)I}{A−(λ−1)I}=0

From here, how we get λ=−1   or  λ=1?
reshown by
Yes, that's a mistake. In hurry I did it. Thanks.
I have edited the answer, i think this makes sense to me, is there anything i am missing?
Is there any easy way to solve this question ? can anyone tell me please ?
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
edited

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 :)

Are they so because

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

@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)

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

edited

$\lambda$ 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

$\gamma + 1 = \lambda$;

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

$\gamma - 1 = \lambda$;

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

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

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.

by

Is this method of expantion or soln of det is aplicable for all types of Block matrices or only for this?
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
@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
A is 4x4.

I is 4X4.

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

why then a 2*2 matrix is taken and solved