Positive Definite or Negative Definite Matrix Say I have a matrix A and I'm trying to determine for which values of $a$ makes the matrix positive definite and which values make it negative definite. $$A = \begin{bmatrix}
2+a &2  & \sqrt 2\\ 
 2&3-a  & \sqrt 2\\ 
 \sqrt2&  \sqrt 2& 1+a
\end{bmatrix}.$$
Should I try to put my A in RREF and work it out from there? I'm not sure how to do that without altering the matrix itself and am worried that it may, in turn, screw up my final answer!
Another thought I had, according to an online definition, is to take the determinate of A and if it is greater than $0$ than it is positive definite. If that is the case, would I just straight up take the determinate as I would any other 3x3 matrices?
 A: Let us form the characteristic polynomial of $A$:
$$P(y)=det(A - yI)= \begin{vmatrix}
2+a-y &2  & \sqrt 2\\ 
 2&3-a-y  & \sqrt 2\\ 
 \sqrt2&  \sqrt 2& 1+a-y
\end{vmatrix}.$$
(the reason for taking $y$ instead of the traditional $\lambda$ will appear hereafter).
Making operations on rows and columns ($R_1  \leftarrow R_1-R_2$ and $C_2\leftarrow C_2-C_1$) gives:
$$P(y)=\begin{vmatrix}
a-y & 2y-1  & 0\\ 
 2&1-a-y  & \sqrt 2\\ 
 \sqrt2&  0& 1+a-y
\end{vmatrix}.$$
We have now enough zeros to expand the determinant wrt its first row.
$$P(y)=(a-y)(1-a-y)(1+a-y)-(2y-1)(2(1+a-y)-2)$$
Factoring $(a-y)$, we get the following characteristic polynomial:
$$P(y)=-(y-a)(y^2-6y+(3-a^2))$$
with solutions $$y=a, y=3+\sqrt{a^2+6}, y=3-\sqrt{a^2+6}$$
For the rest, here is a graphical discussion. I represent the dependence of the roots $y$ with respect to parameter $a$ on an "a,y" plane. The first one is a straight line. The second and the third ones are represented by the two branches of a hyperbola, the second branch intersecting (little computation...) the a-axis in $a=\pm \sqrt{3}$. Thus it suffices to read on the graphic that:


*

*There is a domain where all curves are above the a-axis : $a \in [0,\sqrt{3}]$ i.e., the domain for which matrix $A$ is definite positive.

*There is no domain where the curves are all under the a-axis, thus no values for which matrix $A$ is negative definite. 

