Discussion
Let \(g : M_{2}(\mathbb{R}) \to \mathbb{R}\) be given by \(g(A) = \text{Trace}(A^{2})\). Let 0 be the \(2 \times 2\) zero matrix. The space \(M_{2}(\mathbb{R})\) may be identified with \(\mathbb{R}^{4}\) in the usual manner. Which one of the following is correct?
Answer: C
0 is a saddle point of g.
For \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), \(\text{Trace}(A^{2}) = a^{2} + 2bc + d^{2}\). This quadratic form has gradient zero at the origin, so 0 is a critical point.
Along \(b = c = t\) (others zero) \(g = 2t^{2} \gt 0\), and along \(b = t, c = -t\) it is \(-2t^{2} \lt 0\). The form takes both signs in every neighbourhood of 0, so 0 is neither a local minimum nor a local maximum: it is a saddle point.
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