Assume that A is a square matrix of size n such that A2 5A

Assume that A is a square matrix of size n such that A^2 + 5A = I where I is the identity matrix of size n. Prove that A is non-singular.

Solution

We know that, if AB is non-singular, then both A and B are non-singular

Now A² + 5A = I_n . Since I_n is non-singular, A² + 5A is also non- singular. Now A² + 5A = A (a + 5I_n). Therefore, both A and A + 5I_n are non- singular.

NOTE:

Suppose that AB is non-singular. Then, the only solution to ABx = 0 is x = 0. If B were singular, then there would be some y 0 such that By = 0, in which case, we would have ABy = A(By) = A0 = 0, which gives us a nontrivial solution to ABx = 0, meaning that AB would be singular. Therefore, B is nonsingular. Further, If A were singular, then we would have some y 0 such that Ay = 0. Since B is nonsingular, there is a unique solution x = z to Bx = y. Furthermore, we have z 0, as B0 = 0 y. For this z we get ABz = Ay = 0, so this is a nontrivial solution to ABx = 0, meaning again that AB is singular. Hence, A cannot be singular, either.