Gate ECE 2017 Eigen Value

Question

For the given matrix A, one of the Eigenvalue is real

$A=\begin{bmatrix} 1 &2 &3 &4 &5 \\ 5 &1 &2 &3 &4 \\ 4&5 &1 &2 &3 \\ 3&4 &5 &1 &2 \\ 2 &3 &4 &5 &1 \end{bmatrix}$

The real Eigen value is:

Comments

Is there any other method to solve this? Finding characteristic equation would be very time consuming...

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add all the rows u wiill get 15-x as common take So 15 is real valyue

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Can you please explain in some detail,i am not able to get

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you have already answered it here https://gateoverflow.in/117446/gate-ece-2017-eigen-value

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15 will be the answer.

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For future readers !!!

https://math.stackexchange.com/questions/347408/prove-that-if-the-sum-of-each-row-of-a-equals-s-then-s-is-an-eigenvalue-o

Answer 1

Sum of all rows .
(15-x) * |Matrix| = 0
15 is a factor .

Comments

Yes and its standard one !!

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Sorry for being silly ,but then how the answer is zero,should it be 15 then?

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how (15-$\lambda$) could take common?

Answer 2

$\begin{bmatrix} 1-\lambda & 2 &3 & 4 & 5\\ 5& 1-\lambda & 2 & 3 & 4\\ 4& 5 & 1-\lambda &2 & 3\\ 3& 4 & 5 & 1-\lambda & 2\\ 2& 3 &4 & 5 & 1-\lambda \end{bmatrix}$

=$\begin{bmatrix} 15-\lambda & 15-\lambda &15-\lambda & 15-\lambda & 15-\lambda \\ 5& 1-\lambda & 2 & 3 & 4\\ 4& 5 & 1-\lambda &2 & 3\\ 3& 4 & 5 & 1-\lambda & 2\\ 2& 3 &4 & 5 & 1-\lambda \end{bmatrix}$

Now, taking common

=$15-\lambda\begin{bmatrix} 1 & 1 &1 & 1 & 1 \\ 5& 1-\lambda & 2 & 3 & 4\\ 4& 5 & 1-\lambda &2 & 3\\ 3& 4 & 5 & 1-\lambda & 2\\ 2& 3 &4 & 5 & 1-\lambda \end{bmatrix}$

So, ans 15

Answer 3

$\begin{bmatrix} 1 & 2 &3 & 4 &5 \\ 5 &1 &2 & 3 &4 \\ 4& 5 &1 &2 &3 \\ 3& 4 & 5 & 1 &2 \\ 2&3 & 4 & 5 & 1 \end{bmatrix}$

The characteristic equation is

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

$\Rightarrow$$\begin{vmatrix} 1-\lambda & 2 &3 & 4 &5 \\ 5 &1-\lambda &2 & 3 &4 \\ 4& 5 &1-\lambda &2 &3 \\ 3& 4 & 5 & 1-\lambda &2 \\ 2&3 & 4 & 5 & 1-\lambda \end{vmatrix}$ =0

Applying $C_{1}\leftarrow C_{1}+C_{2}+C_{3}+C_{4}+C_{5}$

$\Rightarrow$$\begin{vmatrix} 15-\lambda & 2 &3 & 4 &5 \\ 15-\lambda &1-\lambda &2 & 3 &4 \\ 15-\lambda & 5 &1-\lambda &2 &3 \\ 15-\lambda& 4 & 5 & 1-\lambda &2 \\ 15-\lambda&3 & 4 & 5 & 1-\lambda \end{vmatrix}$ =0

Taking $15-\lambda$ as common

$\Rightarrow$$(15-\lambda)$$\begin{vmatrix} 1 & 2 &3 & 4 &5 \\ 1 &1-\lambda &2 & 3 &4 \\ 1& 5 &1-\lambda &2 &3 \\ 1& 4 & 5 & 1-\lambda &2 \\ 1&3 & 4 & 5 & 1-\lambda \end{vmatrix}$=0

$\Rightarrow$$(15-\lambda)=0$

$\Rightarrow$$\lambda=15$

Answer 4

$\left | A - \lambda I \right | =0$

$\Rightarrow \begin{vmatrix} 1 - \lambda & 2 & 3 & 4 & 5 \\ 5 & 1 - \lambda & 2 & 3 & 4 \\ 4 & 5 & 1 - \lambda & 2 & 3\\  3 & 4 &5 & 1 - \lambda &2 \\ 2 & 3 & 4 & 5 & 1 - \lambda \end{vmatrix} = 0$

Perform the row operation: $ R_1 \rightarrow R_1 + R_2 + R_3 + R_4 + R_5 $

$\Rightarrow \begin{vmatrix}
15 - \lambda & 15 - \lambda & 15 - \lambda & 15 - \lambda & 15 - \lambda \\
5 & 1 - \lambda & 2 & 3 & 4 \\
4 & 5 & 1 - \lambda & 2 & 3\\
3 & 4 &5  & 1 - \lambda &2 \\
2 & 3 & 4 & 5 & 1 - \lambda
\end{vmatrix} = 0 $

So, $ \lambda = 15 $ is the eigen value.