Using \(2 \times 2\) matrices, show that the Jordan Canonical Form of a matrix is NOT a continuous function of the elements of the matrix.
Show that every real quadratic form over the complexes (i.e. for every \(z \in \mathbb{C}^n\), \(f(z) = z^H A z \in \mathbb{R}\)) can be generated by a Hermitian matrix.
Show that if \(z^H A z = 0\) for all \(z \in \mathbb{C}^n\), then \(A = 0\).
Show by \(2 \times 2\) example that \(x^T A x = 0\) for all \(x \in \mathbb{R}^n\), where \(A\) is real, does not imply that \(A = 0\).
Solution
Part (a).
1. Jordan canonical form is not a continuous function of the entries of a matrix.
For \(t>0\), let \[ A_t= \begin{pmatrix} 0&1\\ t&0 \end{pmatrix}. \] Then \(A_t\to A_0\) entrywise as \(t\to0^+\), where \[ A_0= \begin{pmatrix} 0&1\\0&0 \end{pmatrix}. \] For \(t>0\), the characteristic polynomial is \[ \lambda^2-t, \] so \(A_t\) has two distinct eigenvalues \(\pm\sqrt t\) and is diagonalizable. Thus its Jordan canonical form is, up to the harmless ordering of the two blocks, \[ J(A_t)= \begin{pmatrix} \sqrt t&0\\0&-\sqrt t \end{pmatrix}, \] which tends to the zero matrix. But \(A_0\) is a nonzero nilpotent matrix with one Jordan block of size \(2\), so \[ J(A_0)= \begin{pmatrix} 0&1\\0&0 \end{pmatrix}. \] Hence \(J(A_t)\) does not tend to \(J(A_0)\).
Part (b).
2. If \(z^HAz\in\mathbb R\) for every \(z\in\mathbb C^n\), then the same quadratic form is generated by the Hermitian matrix\[ H=\frac{A+A^H}{2}. \]
For every \(z\), \[ \overline{z^HAz}=z^HA^Hz. \] By hypothesis \(z^HAz\) is real, so \[ z^HAz=z^HA^Hz. \] Therefore \[ z^HHz =\frac12\bigl(z^HAz+z^HA^Hz\bigr) =z^HAz. \] Also \[ H^H=H, \] so \(H\) is Hermitian.
Part (c).
3. If \(z^HAz=0\) for every \(z\in\mathbb C^n\), then \(A=0\).
Write \(A=(a_{ij})\) and let \(e_1,\ldots,e_n\) be the standard basis. Taking \(z=e_i\) gives \[ a_{ii}=0 \] for every \(i\). For \(i\ne j\), taking \(z=e_i+e_j\) gives \[ a_{ij}+a_{ji}=0. \] Taking instead \(z=e_i+i e_j\) gives, using the zero diagonal entries, \[ i a_{ij}-i a_{ji}=0, \] so \[ a_{ij}-a_{ji}=0. \] Combining the last two equations yields \[ a_{ij}=a_{ji}=0. \] Thus every entry of \(A\) is zero.
Part (d).
4. Over \(\mathbb R\), the identity \(x^TAx=0\) for every real \(x\) does not imply \(A=0\).
Take \[ A= \begin{pmatrix} 0&1\\-1&0 \end{pmatrix}. \] This matrix is nonzero and skew-symmetric. For every \(x\in\mathbb R^2\), the scalar \(x^TAx\) satisfies \[ x^TAx=(x^TAx)^T=x^TA^Tx=-x^TAx. \] Hence \(2x^TAx=0\), so over \(\mathbb R\), \[ x^TAx=0 \] for every \(x\), even though \(A\ne0\).