Two proofs about matrices Let $A \in R$ . Show that if $A^2=0$, we can't conclude that $A=0$. Furthermore, if $A'A=0$, we can conclude that $A=0$.
 A: For the first one: Let $$A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$$
Then $A^2=0$ but $A\neq 0.$
A: Hint
Let
$$A=\begin{bmatrix}1&1\\-1&-1\end{bmatrix}$$
Then we have that $A^2=0$ but $A\neq 0$. 
A: Assuming $A'$ is the conjugate transpose of $A$, which is commonly denoted by $A^*$:
Let $A$ and $B$ be arbitrary $n\times n$ matrices. Using the definition of trace and matrix multiplication, we have:
$\DeclareMathOperator{\Tr}{Tr}$
$$\Tr(B^*A) = \sum_{j=1}^n (B^*A)_{jj} = \sum_{j=1}^n \sum_{i=1}^n (B^*)_{ji}A_{ij} = \sum_{i=1}^n \sum_{j=1}^n A_{ij}\overline{B_{ij}}$$
This implies: 
$$\Tr(A^*A) = \sum_{i=1}^n \sum_{j=1}^n A_{ij}\overline{A_{ij}} = \sum_{i=1}^n \sum_{j=1}^n |A_{ij}|^2$$
which is the sum of the squares of absolute values of all elements of the matrix $A$.
$A^*A = 0$ implies $\Tr(A^*A) = 0$ which in turn implies $A_{ij} = 0$, for every $i, j \in \{1, \ldots, n\}$. Thus $A = 0$.
A: Shorter proof for the second part, assuming that $A'$ denotes conjugate transpose:
Suppose that $A'A = 0$. Then for any vector $v$ we have $v'A'Av = 0$, which is equivalent to $(Av)'(Av) = 0$. Since the only vector of zero length is the zero vector, this forces $Av = 0$. As this holds for every vector $v$, we must have $A = 0$.
A: Another short proof for the second one:
Let $A=U\Sigma V^*$ be the singular value decomposition. $A^*A$ = 0 per assumption, therefore all singular values of $A$ are 0 $\Rightarrow \Sigma=0 \Rightarrow A=0$.
