Visualizing Sylvester's law According to Sylvester's law, every $2 \times 2$ real symmetric matrix is congruent to exactly one of six standard types. List them.
I know that the symmetric matrix is congruent to the diagonal matrix, but what do they want me to list. What are the "standard types" ?
If we consider the operation of $GL_2$ on $2 \times 2$ matrices by $P \star A= PAP^t$m then Sylvester's Law asserts that the symmetric matrices form six orbits. We may view the symmetric matrices as points in $\mathbb{R}^3$, letting $(x,y,z)$ correspond to the matrix $\begin{pmatrix} x& y\\ y& z \end{pmatrix}$. Describe the decomposition of $\mathbb{R}^3$ into orbits geometrically, and make a clear drawing depicting it.
Please help, I am really having difficulty
 A: The law in question is "Sylvester's Law of Inertia", so the six types refer to the six different kinds of inertia/signature. That is, if the sign of each eigenvalue is classified as "+" (positive), "0" (zero) or "-" (negative), what are the six possible sign pairs (ignoring order)?
Edit: By Sylvester's law of inertia, the inertia of a real symmetric matrix is preserved by congruence. Therefore, the "orbit" for the sign pair (+,0), for instance, is just the set of all matrices with a positive eigenvalue and a zero eigenvalue, (i.e. the set of all singular and positive semidefinite matrices). Now, a $2\times2$ real symmetric matrix lies in this orbit if and only if its trace is positive but its determinant is zero. Hence the orbit is described by $\{(x,y,z): x+z>0,\ xz=y^2\}$ in $\mathbb{R}^3$, i.e. it is the part of the surface $xz=y^2$ lying inside the open half-space $x+z>0$. For the orbit of (+,+), Sylvester's criterion is useful. The other four cases are either easier to handle or they can be handled analogously.
A: The types refer to the sign of the diagonal entries.  So the types are $\text{diag}(1,1)$, $\text{diag}(1,0)$, $\text{diag}(1,-1)$, $\text{diag}(0,0)$, $\text{diag}(0,-1)$, $\text{diag}(-1,-1)$.  Every $2\times2$ symmetric matrix is equivalent to one of these by a mapping of the form $P^t A P$ where $P$ is invertible.
(How do I know this terminology?  I took a class from J.H. Conway 32 years ago.  He called it Sylvester's law of inertia.)
Now you understand what the question means, it should be easy to answer.
A: The eigenvalues of $$\left(\begin{matrix}x & y \\ y & z\end{matrix}\right)$$
are
$$
\lambda_{\pm}=\frac{1}{2}\left((x+z)\pm\sqrt{(2y)^2+(x-z)^2}\right).
$$
The matrix has signature $(2,0)$ or $(1,0)$ if $x+z \ge \sqrt{(2y)^2+(x-z)^2}$ (with signature $(1,0)$ with equality), or similarly $(0,2)$ or $(0,1)$ if $x+z \le -\sqrt{(2y)^2+(x-z)^2}$, or $(1,1)$ if $|x+z|<\sqrt{(2y)^2+(x-z)^2}$, or $(0,0)$ if $x=y=z=0$.  You should be able to interpret each of these conditions geometrically (the boundaries are quadratic surfaces).
A: The six matrices$$\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}-1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}, \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}, \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$$are representatives for the orbits.
The determinant of the matrix $A = \begin{pmatrix}x & y \\ y & z\end{pmatrix}$ is $xz - y^2$, and because the determinant is homogeneous, the locus $xz = y^2$ is an ellptical double cone $C$ in $\mathbb{R}^3$. To visualize the cone, one may diagonalize by the substitution $\sqrt{2}x = u + v$, $\sqrt{2}z = u - v$. The equation becomes $u^2 = v^2 + 2y^2$.
The cone splits into three orbits, the first three on the list, and it also splits into three connected parts, $\{0\}$, the "positive" cone $C^+$ with $u > 0$, and the "negative" cone $C^-$ with $u < 0$. The complement of $C$ in $\mathbb{R}^3$ splits into the remaining three orbits, and it splits into three connected parts, the interiors $D^+$ and $D^-$ of $C^+$ and $C^-$, and the exterior $\Delta$ of the cone.
It is natural to guess that the orbits of the six representatives are the sets $0$, $C^+$, $C^-$, $D^+$, $D^-$, $\Delta$, in that order, and it is obvious that the orbit of $0$ is $0$. To show this, one first checks that the representatives are in the respective sets. Then the quickest thing to do is to note that one can put a symmetric matrix $A$ into the standard form, in which $P^\text{T}AP$ is one of the six matrices listed, using a matrix $P$ that has positive determinant. The group of matrices with positive determinant is path connected. Therefore an orbit will also be path connected. The connected orbit of $e_{11}$ is contained in $C$, and since the complement of $0$ splits into two connected parts, that orbit must be contained entirely in $C^+$. Similarly, the orbit of $-e_{11}$ must be contained entirely in $C^-$. Together the orbits of $0$, $e_{11}$, and $-e_{11}$ make up $C$. So the orbit of $e_{11}$ must be all of $C^+$, etc$\dots$

In the interests of completeness, the orbits are explicitly:


*

*$1$, $1$ gives $(a^2 + c^2, ab + cd, b^2 + d^2)$ with $ad - bc \neq 0$ $($negative of $-1$, $-1)$. Then $xz - y^2 = (ad - bc)^2 > 0$, with $x$, $z > 0$.

*$1$, $-1$ gives $(a^2 - c^2, ab - cd, b^2 - d^2)$ with $ad - bc \neq 0$. Then $xz - y^2 = -(ad - bc)^2 < 0$, with no restrictions on $x$, $z$.

*$-1$, $-1$ gives $(-a^2 - c^2, -ab-c, -b^2-d^2)$ with $ad - bc \neq 0$ $($opposite of $1$, $1)$. Then $xz - y^2 = (ad-bc)^2 > 0$, with $x$, $z < 0$.

*$1$, $0$ gives $(a^2, ab, b^2)$ with $ad - bc \neq 0$ $($opposite of $-1$, $0)$. Then $xz = y^2$, with $x$, $z \ge 0$ but not both zero.

*$-1$, $0$ gives $(-a^2, -ab, -b^2)$ with $ad - bc \neq 0$ $($opposite of $1$, $0)$. Then $xz = y^2$, with $x$, $z \le 0$ but not both zero.

*$0$, $0$ gives $(0, 0, 0)$.


These are simply determinant identities.
