Let \(A\) be a real \(n\times n\) matrix such that \[ \langle Ax,x\rangle\ge0 \] for every \(x\in\mathbb R^n\). Show that, for every \(u\in\mathbb R^n\), \[ Au=0 \quad\Longleftrightarrow\quad A^Tu=0. \]
Solution
Define \[ q(x)=\inner{Ax}{x}. \]
1. If \(q(u)=0\), then\[ (A+A^T)u=0. \]
1.1. For every \(x\in\RR^n\) and every \(t\in\RR\),\[ q(u+tx) = t\inner{(A+A^T)u}{x} + t^2q(x). \]
Since \(q(u)=0\), \[ \begin{aligned} q(u+tx) &= \inner{A(u+tx)}{u+tx}\\ &= t\inner{Au}{x} + t\inner{Ax}{u} + t^2\inner{Ax}{x}\\ &= t\inner{Au+A^Tu}{x} + t^2q(x). \end{aligned} \]
1.2. For every \(x\in\RR^n\),\[ \inner{(A+A^T)u}{x}=0. \]
Fix \(x\) and set \[ c=\inner{(A+A^T)u}{x}, \qquad d=q(x)\geq0. \] By the hypothesis and step 1, \[ tc+t^2d\geq0 \] for every real \(t\). If \(c\neq0\), choose \(t\) with sign opposite to \(c\) and with \[ 0<\abs{t}<\frac{\abs{c}}{d+1}. \] Then \[ tc+t^2d \leq -\abs{t}\abs{c} + \abs{t}^2d <0, \] a contradiction. Hence \(c=0\).
1.3. One has\[ (A+A^T)u=0. \]
Apply step 2 with \[ x=(A+A^T)u. \] Then \[ \norm{(A+A^T)u}^2=0, \] so \((A+A^T)u=0\).
1.4. Q.E.D.
Step 3 proves step 1.
2. If \(Au=0\), then \(A^Tu=0\).
If \(Au=0\), then \[ q(u)=\inner{Au}{u}=0. \] Step 1 therefore gives \[ Au+A^Tu=0. \] Since \(Au=0\), it follows that \(A^Tu=0\).
3. If \(A^Tu=0\), then \(Au=0\).
If \(A^Tu=0\), then \[ q(u) = \inner{Au}{u} = \inner{u}{A^Tu} =0. \] Step 1 therefore gives \[ Au+A^Tu=0. \] Since \(A^Tu=0\), it follows that \(Au=0\).
4. Therefore, for every \(u\in\RR^n\),\[ \boxed{Au=0\quad\Longleftrightarrow\quad A^Tu=0}. \]
The forward implication is step 2 and the reverse implication is step 3.
5. Q.E.D.
Step 4 is the required equivalence.