An exercise of non-degenerate Hermitian form.

Question

I am studying Hermitian forms and I encountered the exercise below. For me it seems to be wrong, because $\dim (V)$ should be the dimension of the subspace $h$ is positive plus the dimension $h$ is negative plus the dimension of the kernel, right? (This occurs because of a version of Sylvester theorem for hermitian forms)

But I could not prove that the exercise is wrong or right.

Exercise: If a non-degenerate hermitian form $h: V \times V \to \mathbb{C}$ on a $2k$-dimensional complex vector space $V$ satisfies $h(v, v) = 0$, $\forall v \in E$, $E\subset V$, $\dim(E) = k$, then $h$ has signature $(k, k).$

If anyone could help me! Thank you very much.

Answer 1

two notes
(i) Since the form is non-degenerate but there is $h\big(\mathbf v,\mathbf v\big)=0$ for some $\mathbf v\neq \mathbf 0$ it is clear that the form cannot be negative definite or positive definite.

(ii) Also note all vectors $\in E$ are necessarily orthogonal to each other under the form i.e. for $\mathbf v, \mathbf v' \in E$
$h\big(\mathbf v' +\mathbf v, \mathbf v' +\mathbf v\big) = 0\longrightarrow re\Big(h\big( \mathbf v', \mathbf v\big)\Big) = 0$
so $h\big( \mathbf v', \mathbf v\big)$ is purely imaginary $\longrightarrow h\big( \mathbf v', \mathbf v\big) = 0$

To see this implication, suppose $h\big( \mathbf v', \mathbf v\big) = \lambda$ for some purely imaginary $\lambda$ and re-run the above argument on $h\big( \lambda\mathbf v' +\mathbf v, \lambda\mathbf v' + \mathbf v\big)$

main argument
suppose the signature is $\big(r, 2k-r\big)$

Each $\mathbf v\in E$ may be written as a linear combination of an ortho-'normal' set $\{\mathbf w_1, ...,\mathbf w_k, \mathbf w'_1, ..., \mathbf w_{2k-r}'\}$ where this is an ortho-'normal' basis associated with the signature, which generates $V$
'normal' is in quotes since $h\big(\mathbf w_i,\mathbf w_i\big)=1$ and $h\big(\mathbf w_i',\mathbf w_i'\big)=-1$

Begin by constructing a basis for $E$ using these generators
$\mathbf v_i := \big(\mathbf w_i + \mathbf w_i'\big)$
for $1\leq i\leq m= \min(r, 2k-r)$
if $m=r =k$ then we are done.

WLOG suppose for a contradiction that $r\gt k$. Then our basis creation algorithm for $V$ halted prematurely and we have
$\{\mathbf v_1, \mathbf v_2, ...., \mathbf v_m\}$
since $\dim E = k\gt m$ there is some vector in $E$ not in the span of the above set. Call this $\mathbf v_{m+1}$ and mimic the techniques used in the proof of Sylvester's Law of Inertia.

$\mathbf v_{m+1} = \big(\sum_{j\gt m}\alpha_j \mathbf w_j\big) +\big(\sum_{i\leq m}\alpha_i \mathbf w_i\big) + \big(\sum_{i\leq m}\alpha_i' \mathbf w_i'\big) $

$h\big(\mathbf v_{m+1}, \mathbf v_{m+1}\big)=0\longrightarrow \big(\sum_{j\gt m}\vert\alpha_j\vert^2\big) +\big(\sum_{i\leq m}\vert\alpha_i\vert^2\big) = \big(\sum_{i\leq m}\vert \alpha_i'\vert^2\big)\gt 0$

now if $\alpha_i = \alpha_i'$ for all $i \in \{1,2,...,m\}$ then $\alpha_j=0$ for all $j$ and $\mathbf v_{m+1} = \sum_{i\leq m }\alpha_i\mathbf v_i$ which cannot happen since $\mathbf v_{m+1}$ is linearly independent of those vectors by construction. Thus we know there must be some $i$ where $\alpha_i \neq \alpha_i'$.

and by (ii) we know
$0=h\big(\mathbf v_{m+1}, \mathbf v_i\big) = \alpha_i - \alpha_i'\neq 0$
which is a contradiction

Answer 2

Let $A:V \to V$ be such that $h(v,w) = \langle Av,w \rangle$ relative to some inner product $\langle \cdot, \cdot \rangle$. Let $\mathcal B = \{v_1,\dots,v_{2k}\}$ be an orthonormal basis such that $\{v_1, \dots, v_k\}$ is a basis for $E$. We find that the (Hermitian) matrix of $A$ relative to $\mathcal B$ is $$ M = \pmatrix{0 & B\\B^* & C}. $$ Because the form is non-degenerate, $M$ must be invertible. It follows that $B$ is invertible. In the case that $C$ is invertible, $M$ is congruent to the matrix $$ \pmatrix{I & -BC^{-1}\\0 & I} M \pmatrix{I & -BC^{-1}\\0 & I}^* = \pmatrix{-BC^{-1}B^* & 0\\0 & C}. $$ Since $C$ and $-BC^{-1}B^*$ have opposite signatures, it is clear that the signature of $M$ is indeed $(k,k)$, as desired.

In the general case, note that $M$ is congruent to $$ \pmatrix{I & 0\\0 & B^{-1}}^*M\pmatrix{I & 0\\0 & B^{-1}} = \pmatrix{0 & I\\ I & B^{-*}CB^{-1}}. $$ Now, let $U$ be a unitary matrix such that $D = U^*BU$ is diagonal, with non-zero eigenvalues before the zero eigenvalues. We have $$ \pmatrix{U & 0\\0 & U}^*\pmatrix{0 & I\\ I & B^{-*}CB^{-1}}\pmatrix{U & 0\\0 & U} = \pmatrix{0 & I \\I & D}. $$ Write $D$ as a diagonal sum $D = D_1 \oplus 0$, where $D_1$ has only non-zero diagonal entries. We can rewrite $$ \pmatrix{0 & I \\I & D} = \pmatrix{0 & 0 & I & 0\\ 0 & 0 & 0 & I\\ I & 0 & D_0 & 0\\ 0 & I & 0 & 0} \sim \pmatrix{0 & I & 0 & 0\\ I & D_0 & 0 & 0\\ 0 & 0 & 0 & I\\ 0 & 0 & I & 0}, $$ where $\sim$ denotes a permutation similarity. So, $M$ is the direct sum of two matrices that have equal positive and negative signature, which means that $M$ itself has equal positive and negative signature, as was desired.

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In fact, it turns out that we can generalize the first argument and shorten the proof using the Moore-Penrose pseudoinverse. In particular, we have $$ \pmatrix{I & -BC^{\dagger}\\0 & I} M \pmatrix{I & -BC^{\dagger}\\0 & I}^* = \pmatrix{-BC^{\dagger}B^* & 0\\0 & C}, $$ where we exploit the "weak-inverse" property $C^\dagger C C^\dagger = C^\dagger$.

Answer 3

Another, simple approach: after selecting a basis we may identify $h$ with $\mathbf y_1^* A \mathbf z$
for Hermitian $A$ and prove $A$ has $k$ eigenvalues $\gt 0$ and $k$ eigenvalues $ \lt 0$.

Since $\dim E =k$ and for all vectors $\mathbf v, \mathbf v' \in E$
$h\big(\mathbf v, \mathbf v'\big)=0$

with our choice of coordinates this tells us
$\mathbf 0 = W^* A W$
where $W$ is a 2k x k matrix with rank k and $\mathbf 0$ is a k x k matrix. Running QR factorization on W
$W = QR$ where $Q$ is tall and skinny and $R$ is invertible.

$\mathbf 0 = W^* A W = R^*Q^*AQR\longrightarrow \mathbf 0 = Q^*AQ$
by Cauchy Eigenvalue Interlacing $A$ has $k$ eigenvalues $\geq 0$ and $k$ eigenvalues $\leq 0$. Since the Form is non-degenerate the inequalities are strict and the signature is $\big(k,k\big)$ as desired.