Linear Algebra - Quadratic Forms my first time here and it's about time I actually joined this site.
So my question is about Quadratic Forms, more specifically surfaces and related problems.
We have our quadratic form just in matrix form right away:
$$
        \begin{matrix}
        1 & 3 & 2 \\
        3 & -4 & 3 \\
        2 & 3 & 1 \\
        \end{matrix}
$$
Assuming I did correctly, I found the eigenvalues to be -3, -1 and 2
And the corresponding eigenvectors to be (-1,0,1), (3,2,3) and (-1,0,-1) respectively and they are orthogonal to each other which of course creates an orthonormal basis.
Now I have to find the type of surface and the points closest to the origin and I'm not exactly sure how you go on about it.
Like how do you calculate c from this equation? Maybe I'm just insecure about how to do it.
$$
\Lambda_1(x_1')^2+\Lambda_2(x_2')^2+\Lambda_3(x_3')^2 = c
$$
According to the answer the surface is a Hyperboloid of two sheets and the closest points to the origin are +/-(3,2,3)
 A: To find quadratic form when you have the matrix, all you need to do is to perform a couple of vector/matrix multiplications, i.e 
$$Q(\vec{x}) = \vec{x}^T A \vec{x}$$
In your case, $$Q(x, y, z) = \left(\begin{matrix}x & y &z \end{matrix}\right) 
 \left(\begin{matrix}1 & 3 & 2 \\ 3 & -4 & 3 \\ 2 & 3 & 1 \end{matrix}\right) 
 \left(\begin{matrix}x \\ y \\z \end{matrix}\right) 
$$
To find characteristic polynomial, we need to find determinant of matrix $\lambda I - A$. 
$$P(\lambda)=(\lambda-1)^2(\lambda+4)-22\lambda-34$$ The roots of $P(\lambda)=0$ are $-6, -1, 5$. Thus, there is an orthogonal change of coordinates that will transform the original quadratic form into: $$-6x'^2-y'^2+5z'^2=c$$ Now you need to find the extreme points by setting two variables to zero and solving for the remaining variable. After you find these points, determine which one is the closest to the origin. Finally, you need to do reverse transformation from the new coordinate system to the old one.
A: The eigenvalues and eigenvectors that you computed are incorrect, but they do have the correct signature (pattern of signs), so we’ll work with that.
By convention, the matrix $A$ contains the coefficients of the implicit equation $\mathbf x^T A \mathbf x=1$ of the quadric surface. If the eigenvalues of $A$ are $\lambda_1$, $\lambda_2$ and $\lambda_3$, then there’s an orthonormal basis—consisting of orthogonal unit eigenvectors of $A$—in which the equation of the surface is of the form $$\lambda_1 x_1^2+\lambda_2 x_2^2 + \lambda_3 x_3^2=1. \tag1$$ The type of surface is determined by the pattern of signs of the eigenvalues. For this quadric, the pattern is $+--$, which means that the equation is of the form $${x_1^2\over a^2}-{x_2^2 \over b^2}-{x_3^2\over c^2}=1. \tag2$$ This is the equation of a hyperboloid of two sheets.
For a quadric centered at the origin, the nearest and farthest points from the origin lie on the principal axes. Comparing equations (1) and (2), we can see that the half-axis lengths are equal to $\sqrt{|1/\lambda_i|}$. For a hyperboloid of two sheets, there are nearest points to the origin—its vertices—but no farthest points. The vertices lie on the axis that corresponds to the positive eigenvalue. This eigenvalue is actually $5$, with corresponding unit eigenvector $\frac1{\sqrt{22}} (3,2,3)$, so the nearest points to the origin are $\pm\frac1{\sqrt{110}}(3,2,3)$. Note that these points are not $\pm(3,2,3)$, although they do lie along the line with that direction vector. For $(3,2,3)$ to be a vertex, the equation would have to have $110$ on the right-hand side instead of $1$, but there’s nothing in your question that would lead us to make this choice a priori.
