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

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

plz explain

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Please explan the property you have used?

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@rahul

In other words, If sum of all the rows or columns are same, then that sum = one eigen value of that matrix .

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Is this some standard result that we need to remember or is it derivable from existing properties?

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In other words, If sum of all the rows or columns are same, then that sum = one eigen value of that matrix .

Is 15 the eigen value of matrix then?

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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.

Answer 5

Ans -1

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

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

[where $R_{1}\leftarrow R_{1}+R_{2}+R_{3}+R_{4}+R_{5}$]

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

=$15\begin{bmatrix} 1 & 1 &1 & 1 & 1\\ 14& 13 & 12 & 11 & 10\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$

[Here $R_{2}\leftarrow R_{2}+R_{3}+R_{4}+R_{5}$]

=$15\begin{bmatrix} 0 & 0 & 0 & 0 &1 \\ 1 & 1 &1 &1 & 10\\ -1&4 & -1 &-1 & 3\\ -1 & -1 & 4 &-1 &2 \\ -1 & -1 &-1 &4 & 1 \end{bmatrix}$

[Here $C_{1}\leftarrow C_{1}-C_{2}$

$C_{2}\leftarrow C_{2}-C_{3}$

$C_{3}\leftarrow C_{3}-C_{4}$

$C_{4}\leftarrow C_{4}-C_{5}$]

=$15\times 1\begin{bmatrix} 1 & 1 & 1 &1 \\ -1& 4 & -1 &-1 \\ -1 &-1 &4 &-1 \\ -1& -1 &-1 &4 \end{bmatrix}$

=$15\times 1\begin{bmatrix} 0& 0 & 0 &1 \\ -5& 5 &0 &-1 \\ 0 &-5 &5 &-1 \\ 0& 0 &-5 &4 \end{bmatrix}$

=$15\times 1\times 5\times 5\times 5\begin{bmatrix} -1& 1 &0 \\ 0 &-1 &1 \\ 0& 0 &-1 \end{bmatrix}$

=$15\times 1\times 5\times 5\times 5\begin{bmatrix} -1-\lambda & 1 &0 \\ 0 &-1-\lambda &1 \\ 0& 0 &-1-\lambda \end{bmatrix}$

=(-1-$\lambda$)(-1-$\lambda$)^2

$\lambda =-1$

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How eleminating column

$\begin{bmatrix} 0 &0 & 1\\ x& y &z \\ p& q &r \end{bmatrix}$

=$0\begin{bmatrix} y & z\\ q & r \end{bmatrix}$-$0\begin{bmatrix} x & z\\ p& r \end{bmatrix}$+$1\begin{bmatrix} x & y\\ p& q \end{bmatrix}$

=$1\begin{bmatrix} x & y\\ p& q \end{bmatrix}$

So, 1 row, and 1 column eleminated

Comments

@Srestha I think your mistake is you are applying transformations intially and then substituing the char eqn which wont work.

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ok elaborating

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@Shreshtha : Concept you are applying is 'Eigen values of lower / upper triangular matrix is nothing but their diagonal element' yes it is true.

But  determinant of two matrices are equal doesn't make same eigen values.

Answer 6

Every row sum equals $15$, so $15$ is an eigenvalue.

Comments

Nope

This is not actual procedure

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What do you mean by a actual procedure?

 If you try to expand $det(A-\lambda I )$ then we see  this matrix has four complex eigenvalues (but this is so long)

My answers shows $Ax=15x$ with $x=(1,1,1...,1)^T$

Are  you know this fact? : In a square matrix A with every row sum equals k, then k is an eigenvalue of A