Using diagonalisation of a symmetric on quadratic form Let $f (x, y) = 7x_2 + 4y_2 + 4xy$. Use the diagonalisation of a symmetric matrix to write this quadratic form in the form $λ_1u_2^2 +λ_2v_2^2$, with u and v linear combinations of x and y.
Here's what I have tried:
step 1. get the symmetric matrix
$$f (x, y) = 7x_2 + 4y_2 + 4xy=\left(\begin{matrix}7&2\\2&4\\\end{matrix}\right)$$
step 2. find the eigenvalues
$$det(A-\lambda I) \\
\left(\begin{matrix}7-\lambda&2\\2&4-\lambda\\\end{matrix}\right)=(7-\lambda)(4-\lambda)-4=(\lambda-3)(\lambda-8)$$
step 3. find the eigenvectors
case for $\lambda =3$
$$\left(\begin{matrix}7-3&2\\2&4-3\\\end{matrix}\right)=\left(\begin{matrix}4&2\\2&1\\\end{matrix}\right) 
$$
Row reduced
$$\left(\begin{matrix}4&2\\2&1\\\end{matrix}\right)=\left(\begin{matrix}1&\frac{1}{2}\\0&0\\\end{matrix}\right) \implies \left(\begin{matrix}x\\y\\\end{matrix}\right)=y_1\left(\begin{matrix}-\frac{1}{2}\\1\\\end{matrix}\right)$$
Doing the same for $\lambda =8$ (i'll skip the calculation), I get $\left(\begin{matrix}x\\y\\\end{matrix}\right)=y_2\left(\begin{matrix}2\\1\\\end{matrix}\right)$
Finding the orthogonal eigenvectors
$$u_1 = \frac{y_1}{||y_1||} = \frac{2}{\sqrt{5}}\left(\begin{matrix}-\frac{1}{2}\\1\\\end{matrix}\right) \\
u_2 = \frac{y_1}{||y_1||} = \frac{1}{\sqrt{5}}\left(\begin{matrix}2\\1\\\end{matrix}\right) \\
P=(u_1,u_2)=\left(\begin{matrix}-\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\\\end{matrix}\right) \\
D = P^TAP = \left(\begin{matrix}-\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\\\end{matrix}\right)\left(\begin{matrix}7&2\\2&4\\\end{matrix}\right)\left(\begin{matrix}-\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\\\end{matrix}\right) = \left(\begin{matrix}3&0\\0&8\\\end{matrix}\right) \\
x^TAx = \lambda^TD\lambda =  \left(\begin{matrix}\lambda_1&\lambda_2\\\end{matrix}\right)\left(\begin{matrix}3&0\\0&8\\\end{matrix}\right)\left(\begin{matrix}\lambda_1\\\lambda_2\\\end{matrix}\right) = 3\lambda_1^2+8\lambda_2^2$$
However, my result does not look like it's in the same form as the question:
$λ_1u_2^2 +λ_2v_2^2 \ne  3\lambda_1^2+8\lambda_2^2$, how do I proceed from here?
 A: You start with the quadratic form
$$
 7x^2 + 4y^2 + 4xy=
\left(\begin{matrix}x&y\end{matrix}\right)A
\left(\begin{matrix}x\\y\\\end{matrix}\right)=
\left(\begin{matrix}x&y\end{matrix}\right)
\left(\begin{matrix}7&2\\2&4\\\end{matrix}\right)
\left(\begin{matrix}x\\y\\\end{matrix}\right)
$$
Write the eigenvalues decomposition of $A$ as
$$
A=PDP^T=\left[\frac{1}{\sqrt{5}}\left(\begin{matrix}-1&2\\2&1\end{matrix}\right)\right]
\left(\begin{matrix}3&0\\0&8\\\end{matrix}\right)
\left[\frac{1}{\sqrt{5}}\left(\begin{matrix}-1&2\\2&1\end{matrix}\right)\right]
$$
the quadratic form becomes
$$
\left[\left(\begin{matrix}x&y\end{matrix}\right)
\frac{1}{\sqrt{5}}\left(\begin{matrix}-1&2\\2&1\end{matrix}\right)\right]
\left(\begin{matrix}3&0\\0&8\\\end{matrix}\right)
\left[\frac{1}{\sqrt{5}}\left(\begin{matrix}-1&2\\2&1\end{matrix}\right)
\left(\begin{matrix}x\\y\\\end{matrix}\right)\right]$$
can you see what the two new variables $u,v$ are?
Note that your notation is a bit inaccurate and confusing.
A: This is a longer comment rather than an full answer:
If you want to "diagonalize" a quadratic form, completing the square is simpler than your approach via the eigendecomposition. Note that
$$ f (x, y) = 7x^2 + 4y^2 + 4xy = 6 x^2 + (2 y +x)^2 = 6 u^2 + v^2$$
with
$$u = x, \qquad v = 2 y +x$$
is a diagonalization in terms of a linear combination of $x$ and $y$.
