Answer
$4y_3^2+8y_4^2$
Work Step by Step
Find eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
3-\lambda & 1 & 3 & 1\\
1 & 3-\lambda & 1 & 3\\
3 & 1 & 3-\lambda & 1\\
1 & 3 & 1 & 3-\lambda
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \\ v_4\end{bmatrix}\\
\begin{bmatrix}
3-\lambda & 1 & 3 & 1\\
1 & 3-\lambda & 1 & 3\\
3 & 1 & 3-\lambda & 1\\
1 & 3 & 1 & 3-\lambda
\end{bmatrix}=0$
$\rightarrow \lambda^2(\lambda-4)(\lambda-8)=0$
$\lambda_1=\lambda_2=0,\lambda_3=4,\lambda_8$
From that we can find eigenvectors for $\lambda=0$: $\{(0,1,0,-1);(1,0,-1,0)\}$
The orthonormal basis: $\{(0,\frac{1}{\sqrt 2},0,-\frac{1}{\sqrt 2}),(\frac{1}{\sqrt 2},0,-\frac{1}{\sqrt 2},0)\}$
For $\lambda=4$ the eigenvectors are: $\{(-1,1,-1,1)\}$
For $\lambda=8$ the eigenvectors are: $\{(1,1,1,1)\}$
Hence, the set of principal axes for the given quadratic form is: $\{(0,\frac{1}{\sqrt 2},0,-\frac{1}{\sqrt 2}),(\frac{1}{\sqrt 2},0,-\frac{1}{\sqrt 2},0),(-\frac{1}{2},\frac{1}{2},-\frac{1}{2},\frac{1}{2}),(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})\}$
The sum of squares now can be $4y_3^2+8y_4^2$
