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in Linear Algebra
Sep 28, 2014

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34 votes

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

- If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative.
- If the trace of the matrix is positive, all its eigenvalues are positive.
- If the determinant of the matrix is positive, all its eigenvalues are positive.
- If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive.

45 votes

Best answer

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.

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.

9

12 votes

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.**

0 votes

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.

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.

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.

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:

- both a and b are positive
- both a and b are negative.

Hence option C is not correct.

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

- both trace and determinant are positive.
- 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.

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.