TIFR-2015-Maths-A-3

Question

Let $A$ be a $10 \times 10$ matrix with complex entries such that all its eigenvalues are non-negative real numbers, and at least one eigenvalue is positive. Which of the following statements is always false ?

• A. There exists a matrix $B$ such that $AB-BA = B$
• B. There exists a matrix $B$ such that $AB-BA = A$
• C. There exists a matrix $B$ such that $AB+BA=A$
• D. There exists a matrix $B$ such that $AB+BA=B$

Answer 1

Statement 1: There exists a matrix B such that AB + BA = A. This statement is true and can be satisfied for certain matrices A and B. For example, if A is the zero matrix, then any matrix B will satisfy AB + BA = A.

Statement 2: There exists a matrix B such that AB - BA = B. This statement is true and can be satisfied for certain matrices A and B. For example, let A be the zero matrix and B be any matrix. Then AB - BA = 0 - 0 = 0, which is equal to B.

Statement 3: There exists a matrix B such that AB - BA = A. This statement is always false. It is not possible to find a matrix B that satisfies AB - BA = A for all matrices A with non-negative real eigenvalues and at least one positive eigenvalue. The equation AB - BA = A can be rearranged as AB = BA + A, and taking the trace of both sides yields Tr(AB) = Tr(BA + A). However, the trace is cyclic, so Tr(AB) = Tr(BA), and the equation reduces to Tr(AB) = Tr(BA + A). Since the eigenvalues of A are non-negative real numbers, the trace of A is non-negative. However, the trace of BA + A may or may not be non-negative, making this equation impossible to satisfy for all A.

Statement 4: There exists a matrix B such that AB + BA = B. This statement is true and can be satisfied for certain matrices A and B. For example, let B be the zero matrix. Then AB + BA = A0 + 0A = 0, which is equal to B.

Therefore, the statement that is always false is Statement 3: There exists a matrix B such that AB - BA = A.