# For matrix A, A*A = 0 implies A = 0

1. Jan 15, 2014

### jahlex

1. The problem statement, all variables and given/known data

Given that $A$ is a square matrix and $A^\dagger$ is its Hermitian conjugate, and that $A^\dagger A = 0$, show that $A = 0$.

3. The attempt at a solution

Let $\{|i\rangle\}$ be some orthonormal basis. Find the matrix elements of $A$ by taking $0 = \langle j| A^\dagger A |j \rangle = \displaystyle\sum_n \langle j|A^\dagger |n \rangle \langle n|A|j \rangle = \displaystyle\sum_n | \langle n|A|j \rangle |^2 \ge | \langle i|A|j \rangle |^2 = |A_{ij}|^2$, so $A_{ij} = 0$ and $A = 0$.

3. Question

Is there a way to prove the result that does not rely on finding matrix elements?

2. Jan 15, 2014

### Office_Shredder

Staff Emeritus
Yes, the proof is very similar but does not require picking a specific vector to multiply by. Suppose that Av is not zero. Then by observing
$$\left<v| A^{\dagger} A| v \right> = \left<Av| Av \right>$$
You should be able to prove that this number is both zero and non-zero, giving a contradiction.