Consider the function $F: M_{2}(\mathbb{R}) \to \mathbb{R^{2}}$ defined by $F(X)=\left[trX ,detX\right]^{T}$. Show that the critical points form a straight line and critical values form a parabola
Try:
$$DF_{X}(U)=\lim_{t \to 0}\frac{F(X+tU)-F(X)}{t}$$
Then by letting $$ X=\begin{bmatrix}x_1&x_2\\ x_3&x_4 \end{bmatrix}$$ $$U=\begin{bmatrix}u_1&u_2\\ u_3&u_4 \end{bmatrix}$$
I get $$DF_{X}(U)=\begin{bmatrix}u_1+u_4\\x_1u_4+x_4u_1-x_2u_3-x_3u_2 \end{bmatrix}$$
Equating these to $0$ I get two equations. But I am unable to conclude from there.
Thanks for the help!!
A point $X\in M^2(\mathbb{R})$ is a critical point for $F$ if the differential $DF_X:M^2(\mathbb{R})\rightarrow\mathbb{R}^2$ haven't maximum rank. In your case, you have to impose $x_1u_4+x_4u_1-x_2u_3-x_3u_2$ to be linearly dipendent to $u_1+u_4$, for every $(u_1,u_2,u_3,u_4)$. This happens if and only if $x_2=x_3=0$ and $x_1=x_4$. The critical points are then the line in $M^2(\mathbb{R})$ spanned by $\begin{pmatrix}1&0\\0&1\end{pmatrix}$. The critical values are $$F\begin{pmatrix}t&0\\0&t\end{pmatrix}=(2t,t^2)$$ that is a parabola in $\mathbb{R}^2$.