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I'm having trouble proving this equivalence:

Given $A$ a $3\times 3$ complex symmetric matrix. Let $q \in \mathbb{C}^3$ be such that $Aq\neq 0$ and $\langle Aq, q\rangle = 0$. Then, the following are equivalent:

  1. $\exists\; v\in \mathbb{C}^3$ linearly independent of $q$ such that $$\langle Aq, v\rangle = \langle Av, v \rangle = 0;$$

  2. $\det A=0$.

Here $\langle (v_1,v_2,v_3)^T, (u_1,u_2,u_3)^T\rangle = v_1u_1+v_2u_2+v_3u_3$.

Comment: $ 2 \implies 1$ is direct. Proving $ 1 \implies 2$ in $\mathbb{R}^3$ is easy since we have that $\langle v, v\rangle = 0 \iff v=0$ but this isn't true for $\mathbb{C}^3$ (take $(1,i,0)$). Any ideas? I'm guessing the symmetry of $A$ is crucial in the proof.

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Firstly, let's prove that $A$ is a symmetric operator: for all $x,y\in\mathbb{C}^3$ we have $$\langle Ax, y\rangle = (Ax)^Ty = x^TA^Ty = x^TAy = \langle x, Ay \rangle$$ as $A^T = A$ (here $^T$ means transpose).

Now let's consider first condition. From $\langle Aq, v\rangle = \langle Av,v\rangle = 0$ we get that for all $\alpha, \beta \in \mathbb{C}$ $$ \langle A(\alpha q + \beta v), v\rangle = 0. \tag1 $$ Then, using already proved symmetry of $A$ and symmetry of given "dot product" we get that $$\langle Av, q\rangle = \langle q, Av\rangle = \langle Aq,v\rangle = 0.$$ Also we know that $\langle Aq,q\rangle = 0$, so by the same way for all $\alpha, \beta \in \mathbb{C}$ $$ \langle A(\alpha q + \beta v), q\rangle = 0. \tag2 $$

Let's denote linear span $\mathcal{L}\{q,v\}$ of vectors $q$ and $v$ as $L$. $(1)$ means that subspace $AL\subset \mathbb{C}^3$ is orthogonal to the vector $v$, $(2)$ means that it is orthogonal to the vector $q$, so $\dim(AL\cap L) = 0$. Now, if $v$ and $q$ are linearly independent, then $\dim L = 2$ and $\dim AL \leq 3 - \dim L = 1$, i.e. $\dim AL < \dim L$ which implies $\det A = 0$ as $L$ is a subspace of $\mathbb{C}^3$.

If, vice versa, we know that $\det A = 0$ then $\dim\ker A > 0$ and there exist such vector $v\neq 0$ that $Av = 0$ (obviously $v$ linearly independent of $q$ as $Aq \neq 0$). Thus $\langle Aq, v\rangle = \langle q, Av\rangle = \langle q, 0 \rangle = 0$ and $\langle Av, v \rangle = \langle 0, v \rangle = 0$.

QED.

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  • $\begingroup$ Great!!! THANK YOUUUUUU $\endgroup$
    – mfcp2904
    Jul 11 '17 at 23:56
  • $\begingroup$ WAIT... Why do you claim $dim(AL\cap L)=0$? Remember that $(1,i,0)$ is "orthogonal" to itself (using the $\langle, \rangle$ defined above) $\endgroup$
    – mfcp2904
    Jul 12 '17 at 0:07
  • $\begingroup$ @mfcp2904 you may see it by manual solving of system $\langle q,q\rangle = 0$, $\langle v,v\rangle = 0$, $\langle q,v\rangle = 0$ - according to Wolfram, all the solutions are linearly dependent, so there is no non-zero vectors in $L$ orthogonal to it (when $\dim L = 2$). I've been thinking about proving this in some more elegant way, but it seems to be not trivial. I guess one may try to prove that $qv^T = qv^T$ if $q$ and $v$ are solutions of this system - this will imply their linear dependence. $\endgroup$ Jul 12 '17 at 12:02
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Suppose the contrary that $A$ is nonsingular. By Takagi factorisation, $A=C^TC$ for some nonsingular matrix $C$. If we redefine $q\leftarrow Cq$ and $v\leftarrow Cv$, the premise of the question and condition 1 would imply that the two redefined vectors are linearly independent vectors in $\mathbb C^3$ but $q^Tq=q^Tv=v^Tv=0$. You may try to prove that this is impossible. This shouldn't be hard.

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