Let \(A\in M_n\). Show that if \(z^HAz=0\) for all \(z\in\mathbb{C}^n\) then \(A=0\).
Give a \(2\times 2\) example of \(A\in M_2(\mathbb{R})\), where \(A\neq 0\), but \(x^TAx=0\) for all \(x\in\mathbb{R}^2\).
Show that \(A\in M_n\) is normal iff \(\|Ax\|_2=\|A^Hx\|_2\) for all \(x\in\mathbb{C}^n\).
Show that \(A\in M_n\) is normal iff \((Ax,Ay)=(A^Hx,A^Hy)\) for all \(x,y\in\mathbb{C}^n\).
(Notation: \(M_n\) denotes the set of \(n\times n\) complex matrices; \(M_2(\mathbb{R})\) denotes the set of \(2\times 2\) real matrices.)
Solution
1. If \(z^HAz=0\) for every \(z\in\mathbb C^n\), then every diagonal entry of \(A\) iszero.
Write \(A=(a_{ij})\) and let \(e_i\) be the \(i\)th standard basis vector. Taking \(z=e_i\) gives \[ 0=e_i^HAe_i=a_{ii}. \] Thus \(a_{ii}=0\) for every \(i\).
2. Every off-diagonal entry of \(A\) is also zero; hence \(A=0\).
Fix \(i\neq j\). Taking \(z=e_i+e_j\) and using 1 gives \[ 0=(e_i+e_j)^HA(e_i+e_j)=a_{ij}+a_{ji}. \] Taking \(z=e_i+i e_j\) gives \[ 0=(e_i+i e_j)^HA(e_i+i e_j)=i a_{ij}-i a_{ji}. \] Hence \(a_{ij}+a_{ji}=0\) and \(a_{ij}-a_{ji}=0\), so \(a_{ij}=a_{ji}=0\). Since this holds for every \(i\neq j\), all entries of \(A\) vanish.
3. For part (b), the matrix\[ A=\begin{pmatrix}0&1\\-1&0\end{pmatrix} \]is nonzero and satisfies \(x^TAx=0\) for every \(x\in\mathbb R^2\).
For \(x=(u,v)^T\), \[ x^TAx=(u,v)\begin{pmatrix}v\\-u\end{pmatrix}=uv-vu=0. \] The matrix is visibly nonzero.
4. If \(A\) is normal, then \(\|Ax\|_2=\|A^Hx\|_2\) for every \(x\in\mathbb C^n\).
Normality means \(A^HA=AA^H\). Therefore \[ \|Ax\|_2^2=x^HA^HAx=x^HAA^Hx=\|A^Hx\|_2^2. \] Both norms are nonnegative, so they are equal.
5. Conversely, if \(\|Ax\|_2=\|A^Hx\|_2\) for every \(x\), then \(A\) is normal.
For every \(x\), \[ 0=\|Ax\|_2^2-\|A^Hx\|_2^2 =x^H(A^HA-AA^H)x. \] Applying part (a), proved in <1>1–<1>2, to the matrix \(A^HA-AA^H\) gives \[ A^HA-AA^H=0. \] Thus \(A\) is normal.
6. If \(A\) is normal, then \((Ax,Ay)=(A^Hx,A^Hy)\) for every \(x,y\in\mathbb C^n\).
Using the standard Hermitian inner product \((u,v)=u^Hv\) and \(A^HA=AA^H\), \[ (Ax,Ay)=x^HA^HAy=x^HAA^Hy=(A^Hx,A^Hy). \]
7. Conversely, if \((Ax,Ay)=(A^Hx,A^Hy)\) for all \(x,y\), then \(A\) is normal.
Setting \(y=x\) gives \[ \|Ax\|_2^2=\|A^Hx\|_2^2 \] for every \(x\). By 5, \(A\) is normal.