Test by Bikram | Mock GATE | Test 4 | Question: 20

Question

Consider the  $5\times 5$ matrix below :

$\begin{bmatrix} 1&0 &0 &0 &1 \\ 0& 1 & 1 & 1 & 0\\ 0& 1 &1 &1 &0 \\ 0& 1 &1 &1 &0 \\ 1 & 0 & 0 & 0 & 1 \end{bmatrix}$

The product of the non-zero eigenvalues of the matrix is _____.

Comments

Trace = 5. (Sum of diagonal elements)

There will be 5 (not necessarily distinct) eigenvalues, because this is a $5\times 5$ matrix.

Let the eigen values be $a, b,c,d,e$

$a+b+c+d+e=5$

 

The rank of this matrix is 2.

---

How?

$R_5 \leftarrow R_5-R_1$. Last row becomes 0. So, rank isn't 5.

For any $4\times 4$ matrix, we'll still not get non-zero determinant.

Same for $3\times 3$ matrix.

But for this $2\times 2$ matrix

$\begin{bmatrix} 0 & 1\\ 1& 0 \end{bmatrix}$

Determinant is non-zero.

So, rank =2.

---

 

Since rank = 2, we only have two non-zero eigenvalues. So, the equation reduces to

$a+b=5$

Now, eigenvalues can be $(1,4)$ or $(2,3)$

 

=> The answer could be 4 or 6.

Answer 1

6 will be answer...

Comments

@Bikram @akash.dinkar12 @Arjun

I am little doubtful about this solution. From first look it looks correct.

But I have below doubt

All Five equations

x1 + x5 = lambda * x1

x2 + x3 + x4 = lambda * x2

x2 + x3 + x4 = lambda * x3

x2 + x3 + x4 = lambda * x4

x1 + x5 = lambda * x5 

must hold true for lambda =2 and also for lambda = 3

but according to this explanation  

x1 + x5 = lambda * x1 &

x1 + x5 = lambda * x5 

hold true for lambda =2

and 

x2 + x3 + x4 = lambda * x2

x2 + x3 + x4 = lambda * x3

x2 + x3 + x4 = lambda * x4

holds true for lambda = 3

To solve system of five equation , we should substitute one equation in another to find relation between all five variable. 

Now if we use lambda =2 and try to find eigen vector , we can see that x1 and x5 are related (same)

x2 , x3 , x4 are related (same). but there is no correlation between this two group. hence we fail to find eigen vector

I am 70% sure about my explanation and not 100% . Please correct me If my explanation is wrong