I have a quadratic form $5x^2 – 2xy + 5y^2 = 4$. The corresponding matrix ( for vectors $[x y ]^T$ ) is $$A=\begin{bmatrix} 5 & -1 \\ -1 & 5 \end{bmatrix}$$ Now I need to construct an orthonormal basis for $\Bbb{R^2}$ via the eigenvectors of $A$. The eigenvalues are $6$ and $4$, the corresponding eigenvectors are now $[-1 ,1]^T$ and $[1,1]^T$. $A$ is symmetric, and thus its eigenvectors are orthogonal. I think that the set of these eigenvectors should form a basis for $\Bbb{R^2}$. Is this correct?
A question about quadratic forms and eigenbasis
eigenvalues-eigenvectorslinear algebra
Best Answer
Yes, they are automatically a basis if the matrix is symmetric. Othogonality implies linear independence (and the inverse of the basis transform matrix is its (scaled) transpose).