Let \(V\) be the complex vector space of complex \(2\times2\) matrices \(X\). Find all quadratic forms \(Q\) on \(V\) such that \(Q(X)=Q(AXA^{-1})\) for any complex invertible \(2\times2\) matrix \(A\).
Solution
For \(\alpha,\beta\in\CC\), define \[ Q_{\alpha,\beta}(X) \coloneqq \alpha\,\operatorname{tr}(X^2) +\beta\,\operatorname{tr}(X)^2. \]
1. Every \(Q_{\alpha,\beta}\) is a conjugation-invariant quadraticform.
Both \(\operatorname{tr}(X^2)\) and \(\operatorname{tr}(X)^2\) are homogeneous quadratic polynomials in the entries of \(X\). If \(A\) is invertible, then \[ (AXA^{-1})^2=AX^2A^{-1}, \] so invariance of trace under conjugation gives \[ \operatorname{tr}((AXA^{-1})^2) =\operatorname{tr}(X^2), \qquad \operatorname{tr}(AXA^{-1}) =\operatorname{tr}(X). \] Thus \(Q_{\alpha,\beta}(AXA^{-1})=Q_{\alpha,\beta}(X)\).
2. The restriction of any conjugation-invariant quadratic form\(Q\) to diagonal matrices is of the form\[ Q\!\left( \begin{pmatrix} a&0\\ 0&b \end{pmatrix} \right) = \alpha(a^2+b^2) +\beta(a+b)^2 \]for unique \(\alpha,\beta\in\CC\).
Because \(Q\) is quadratic, its restriction to diagonal matrices has the form \[ c\,a^2+d\,ab+e\,b^2. \] Conjugation by \[ P= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \] exchanges the two diagonal entries. Invariance therefore gives \[ c\,a^2+d\,ab+e\,b^2 = c\,b^2+d\,ab+e\,a^2 \] for all \(a,b\), so \(c=e\). Hence \[ Q(\operatorname{diag}(a,b)) =c(a^2+b^2)+d\,ab. \] Taking \[ \beta=\frac d2, \qquad \alpha=c-\frac d2 \] gives the displayed expression. Uniqueness follows by comparing the coefficients of \(a^2\) and \(ab\).
3. Matrices with two distinct eigenvalues are dense in\(M_2(\CC)\).
Let \[ X= \begin{pmatrix} a&b\\ c&d \end{pmatrix}, \qquad D= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}. \] The discriminant of the characteristic polynomial of \(X+tD\) is \[ \begin{aligned} \Delta(t) &=\operatorname{tr}(X+tD)^2-4\det(X+tD)\\ &=(a+d)^2 -4\bigl(ad-bc+t(d-a)-t^2\bigr). \end{aligned} \] This is a polynomial in \(t\) with leading coefficient \(4\), so it has only finitely many zeros. Hence there are arbitrarily small nonzero \(t\) for which \(\Delta(t)\ne0\). For such \(t\), the matrix \(X+tD\) has two distinct eigenvalues and is therefore diagonalizable. Since \(X+tD\to X\) as \(t\to0\), the claim follows.
4. Every conjugation-invariant quadratic form \(Q\) equals one of theforms in step
1.
Choose \(\alpha,\beta\) from step 2 and set \[ R\coloneqq Q-Q_{\alpha,\beta}. \] Then \(R\) vanishes on every diagonal matrix. If \(X\) is diagonalizable, there is an invertible \(A\) such that \(AXA^{-1}\) is diagonal. Since both \(Q\) and \(Q_{\alpha,\beta}\) are conjugation invariant, \[ R(X)=R(AXA^{-1})=0. \] Thus \(R\) vanishes on all matrices with distinct eigenvalues. By step 3 this set is dense, and \(R\) is a polynomial, hence continuous. Therefore \(R\) vanishes on all of \(M_2(\CC)\).
5. Consequently the complete family is\[ \boxed{ Q(X) = \alpha\,\operatorname{tr}(X^2) +\beta\,\operatorname{tr}(X)^2, \qquad \alpha,\beta\in\CC. } \]
Step 1 shows that every displayed form is admissible, and step 4 shows that there are no others.
6. Q.E.D.
Step 5 gives the requested classification.