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+43 votes

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?

- $-5$
- $-7$
- $2$
- $1$

+38 votes

Best answer

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.

0

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

λ is the eigen value of the matrix A I | I A

+2

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

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

Please correct me if I'm wrong :)

0

Are they so because

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

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

+1

I didn't get your question.

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

So, we have 8 eigen values.

Given in question* λ = *{-5,-2,1,4}

- using (
*λ*+ 1) we have eigen values: -4,-1,2,5 - 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)

0

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

+2

$\lambda$ 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

$\gamma + 1 = \lambda$;

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

$\gamma - 1 = \lambda$;

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

+68 votes

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

|(A – x I)^{2} – I^{2}| = 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.

0

Is this method of expantion or soln of det is aplicable for all types of Block matrices or only for this?

0

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

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

+1

@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

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

+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):

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

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

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

$ \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):

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

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

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