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I have the following equation: $a^HUa=0$ where '$a$' can be any arbitrary vector and $U$ is a matrix ($H$ means Hermitian). Can we conclude that $U=0$? Any reference?

Thanks.

Cameron Buie
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user99866
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1 Answers1

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We can conclude that $U$ needs to be a square matrix for the involved expression to make sense. Let $U$ be an $n\times n$ matrix.

$$ \text{ the $i$th component of vector $Ua$: } (Ua)_i = \sum_{k=1}^n U_{ik}a_{k}. $$

and similarly $$ \begin{align*} a^HUa &= \sum_{j = 1}^n a^H_{j} (Ua)_j \\ &= \sum_{j} \sum_{k} a_j^*U_{jk}a_k \end{align*} $$

Let $a_j = \delta_{r,j}$, which is $1$ only index $r$ and zero elsewhere. Then, $$ a^HUa = 0 = U_{rr} $$ So you can conclude that the diagonal of $U$ is zero.

Letting $a_j = (1, 0, \dots, 1, \dots, 0)$ (a $1$ in the $s$ and $t$th places).

Then you have $U_{st} + U_{ts} = 0$. So the matrix is anti-symmetric.

Lets try the same thing with $i$ and $1$. Then we have $$ (-i)U_{st} + iU_{ts} = 0 $$

Thus the off-diagonal elements are anti-symmetric and equal, hence zero. QED

Daniel Donnelly
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