If $X^TX=0$ show that $X=0$ [closed]

Question

Let $X\in \mathbb R^{m\times n}$. If $X^T X=0$, show that $X=0$.

Any thoughts on how to prove this?

Answer 1

pick any vector $a \in R^n.$ let $b = Xa.$ then $$b^Tb = (Xa)^T(Xa) = a^TX^TXa = 0 $$ therefore $b =0.$ that is $$Xa = 0 \text{ for any $a$ in } R^n \implies X = 0 $$

Answer 2

$X^TX$ is $n\times n$. Consider the element $(j,j)$ on its diagonal: $$ 0=(X^TX)_{j,j}=\sum_{i=1}^m(X^T)_{j,i}X_{i,j}=\sum_{i=1}^mX_{i,j}X_{i,j}=\sum_{i=1}^mX_{i,j}^2. $$ This proves that $X_{i,j}=0$ for all $i=1,\ldots,m$ so that the $j$-th column of $X$ is $0$. Now let $j$ run from $1$ to $n$.

Alternatively, to show that the $j$-th column of $X$ is $0$, consider the unit vector $(0,\ldots,1,\ldots,0)^T$ ($1$ in the $j$-th position and $0$ elsewhere) and observe that $$ X^TXe_j=0\implies 0=e_j^TX^TXe_j=(Xe_j)^TXe_j=\left|Xe_j\right|^2\implies Xe_j=0. $$