Answer
$2y_1^2+2y_2^2-y_3^2$
Work Step by Step
Find eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & 1 & -1\\
1 & 1-\lambda & 1\\
-1 & 1 & 1-\lambda
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix}
1-\lambda & 1 & -1\\
1 & 1-\lambda & 1\\
-1 & 1 & 1-\lambda
\end{bmatrix}=0$
$\rightarrow (\lambda-2)^2(1+\lambda)=0$
$\lambda_1=\lambda_2=2,\lambda_3=-1$
From that we can find eigenvectors for $\lambda=2$: $\{(1,1);(1,-1)\}$
The orthonormal basis: $\{(\frac{1}{\sqrt 2},0,-\frac{1}{\sqrt 2});(\frac{1}{\sqrt 6},\frac{2}{\sqrt 6},\frac{1}{\sqrt 6})\}$
For $\lambda=-1$: $\{(-1,1,-1)\}$
Hence, the set of principal axes for the given quadratic form is: $\{(-\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3})\}$
The sum of squares now can be $2y_1^2+2y_2^2-y_3^2$
