GATE CSE 2014 | Set 3 | Question: 4

Question

Which one of the following statements is TRUE about every $n \times n$ matrix with only real eigenvalues?

• A. If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative.
• B. If the trace of the matrix is positive, all its eigenvalues are positive.
• C. If the determinant of the matrix is positive, all its eigenvalues are positive.
• D. If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive.

Comments

Totally based on properties of trace and eigen values

Answer 1

Trace is the sum of all diagonal elements of a square matrix.
Determinant of a matrix = Product of eigen values.

$(A)$ Is the right answer. To have the determinant negative, at least one eigen value has to be negative (but reverse may not be true).
We can take simple example with upper or lower triangular matrices. For options $(b) , (c)$ and $(d)$ reverse is always true.

Comments

Counter ex for (B) is, Trace = 1 but eigen values are 2,-1 .

Counter ex for (C) is, Det = 2 but eigen values are -2,-1 .

Counter ex for (D) is,Trace * Det = 2 but eigen values =1,-2.

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Good counter example

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Ya it's absolutely brilliant example.

@LakshmanPatel RJIT

Answer 2

Taking a simple example we can get a clear idea.
Let matrix A be

10	0	0
0	-2	0
0	0	-4

 Here trace(A)=4 and determinant(A)=80.

As matrix is diagonal matrix the principal diagonal elements are the eigenvalues for matrix A.

Even though a trace of a matrix is positive we have two negative eigenvalues(i.e. -2,-4). So option B is false.

Even though determinant of a matrix is positive we have two negative eigenvalues(i.e. -2,-4). So option C is false.

the product of trace and determinant is 320 which is positive still we have two negative eigenvalues(i.e. -2,-4). So option D is false.

Now if the determinant is negative then at least one eigenvalue of the matrix is negative because the product of eigenvalues of a given matrix is equal to the determinant of a given matrix.
So OPTION A is True.

Comments

Thank you bro😊

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Thank you

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I think knowing properties is better than solving by taking example.  :)

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Where are the Properties?

Answer 3

Let’s consider a 2x2 matrix P consisting of elements 

a   0

0   b

We know that trace of a matrix  = sum of its eigen values 

and determinant of a matrix = product of its eigen values

hence, tr(P) = a+b and det(P) = ab holds true.

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Now, option A says that

if the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative,

which means if trace is positive there are two ways :

1. either both a and b are positive or

2 one is positive and other is negative.​​​​​​​

we are given that determinant is negative which means the product ab is negative. This happens only when one of either a and b is negative. 

Hence option A is correct.

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option B says that

If the trace of the matrix is positive, all its eigenvalues are positive.

As discussed above that it is not necessary that a and b are always positive. It’s because any one of a or b (which is greater) can be positive and other can be negative (the smaller value) and still the sum would result to positive. (Just like 9+(-2)). Also a good mention of real eigenvalues in the question supports this very fact.

Hence option B is not correct.

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option C says that

If the determinant of the matrix is positive, all its eigenvalues are positive.

This is simple. if det(P) = ab is positive there can be two cases:

1. both a and b are positive 
2. both a and b are negative.

Hence option C is not correct.

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option D says that

If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive.

If trace (P) * det (P) = ab * (a+b) is positive it means

1. both trace and determinant are positive.
2. both trace and determinant are negative.

First case is trivial one, let’s look onto second case

if trace (P) = a+b is negative then one of them is sure to be negative . Also if det (P) = ab is negative then one of a or b is negative which contradicts the above statement that  all eigenvalues are positive.

Hence option D is not correct.

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The crux is in the fact that the sum of the eigenvalues is the trace of the matrix and product of the eigenvalues is determinant of the matrix. Rest is simple thinking.

Hope it helps.

Answer 4

TRACE = Addition of all eigen values -------1

DETERMINANT = product of all eigen values -----2

Explanation : 

                  -> By 2 We know that in the multiplication one of the value is negative then whole product becomes negative 

                  -> Apply the same logic here one of the eigen value is negative then product of all eigen values becomes negative hence determinant becomes negative

 

HENCE OPTION A) IS CORRECT

 

Answer 5

Answer: A

$$\text{Which one of the following statements is TRUE about every } n \times n \text{ matrix}$$

 

Since the question mentions "for every matrix", best way is eliminating options. 

 

Let n=2 for simplicity

 

Option B: FALSE ❌

If the trace of the matrix is positive, all its eigenvalues are positive.

$$\text{Let } \lambda_1=5 \text{ and } \lambda_2=-3 \text{ then trace = 5 - 3 = 2 (positive) but } \lambda_2 \text{ is negative.}$$

 

Option C: FALSE ❌

If the determinant of the matrix is positive, all its eigenvalues are positive.

$$\text{Let } \lambda_1=-5 \text{ and } \lambda_2=-3 \text{ then determinant = 5 x 3 = 15 (positive) but both eigenvalues are negative.}$$

 

Option D: FALSE ❌

If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive.

If the product of the trace and determinant of the matrix is positive,  both trace and determinant can simultaneously be negative meaning atleast one eigenvalue is negative.

Answer 6

option A.

 Trace is positive but determinant is negative. then atleast one of its eigen value is negative - TRUE

determinant becomes negative if there is an odd number of -ve eigen values.

option B.

trace is positive, so all eigen values are positive - FALSE

suppose 2X2  matrix has eigen value 4,-2; still we get trace as +2. 

option C

If the determinant of the matrix is positive, all its eigenvalues are positive. - FALSE

suppose 2X2  matrix has eigen value -2,-3; still we get deteminant as +6.

 

option D

If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive. FALSE

  case 1: trace and determinant both are positive.     

                  3X3 matrix with eigen values 4,-2,-1 ->  trace =+1 determinant =+8

   

case 2: trace and determinant both are negative.     

                  3X3 matrix with eigen values -4,1,2 ->  trace =-1 determinant =-8