GATE EC 2014 Set 3 — Question 37

The question asks to identify the statement that is NOT true for a square matrix A.Let's analyze each statement:(A) If A is upper triangular, the eigenvalues of A are the diagonal elements of it.
This statement is TRUE. For any triangular matrix (upper or lower), the eigenvalues are the entries on its main diagonal. This is because the determinant of a triangular matrix is the product of its diagonal elements. The characteristic equation is $\det(A - \lambda I) = 0$. Since $A$ is triangular, $A - \lambda I$ is also triangular, and its determinant is $\prod_{i=1}^{n} (a_{ii} - \lambda)$. The roots of this equation are $\lambda_i = a_{ii}$.(B) If A is real symmetric, the eigenvalues of A are always real and positive.
This statement is FALSE. It is true that the eigenvalues of a real symmetric matrix are always real. However, they are not necessarily positive. For the eigenvalues to be positive, the matrix must be positive definite. A real symmetric matrix can have negative or zero eigenvalues.
Counterexample: Consider the matrix $A = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix}$. This is a real symmetric matrix. Its eigenvalues are clearly -2 and -2, which are real but not positive.
Another counterexample: $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. This is real symmetric. Its characteristic equation is $\lambda^2 - 1 = 0$, so the eigenvalues are $\lambda = 1$ and $\lambda = -1$. One is positive, but one is negative.(C) If A is real, the eigenvalues of A and $A^T$ are always the same.
This statement is TRUE. The eigenvalues are the roots of the characteristic equation, $\det(A - \lambda I) = 0$. We know that for any square matrix $B$, $\det(B) = \det(B^T)$. Let $B = A - \lambda I$. Then $B^T = (A - \lambda I)^T = A^T - (\lambda I)^T = A^T - \lambda I$. Therefore, $\det(A - \lambda I) = \det((A - \lambda I)^T) = \det(A^T - \lambda I)$. Since A and $A^T$ have the same characteristic polynomial, they must have the same eigenvalues.(D) If all the principal minors of A are positive, all the eigenvalues of A are also positive.
This statement is TRUE. This is related to Sylvester's criterion. A Hermitian matrix (which includes real symmetric matrices) is positive definite if and only if all its leading principal minors are positive. A matrix is positive definite if and only if all its eigenvalues are positive. The condition that all principal minors are positive is a stronger condition that also implies the matrix is positive definite, and thus has all positive eigenvalues. This holds for symmetric matrices. For non-symmetric matrices, this is also a known result.Since the question asks for the statement that is NOT true, the correct answer is (B).