I'm really confused by a task I have been given:
We are looking at a quadratic form $Q(\vec{x})=\vec{x}^TA\vec{x}$ where the symmetric matrix $A = A^T$ has normalized eigenvectors $\vec{v_i}$ with corresponding eigenvalues $\alpha_i$. So far, I understand every component. But then we are asked to use $\vec{v_i}$ and $\alpha_i$ to compute new coordinates $x_i^{\prime}=B_{ij}x_j$ (with an orthogonal change of basis $B$) and coefficients $\lambda_i$ such that $Q^{\prime}(\vec{x})^{\prime}= \lambda_1 x_1^{\prime}+...+\lambda_n x_n^{\prime}$ with $Q(\vec{x})=Q^{\prime}(\vec{x})^{\prime}$.
Even though I know about the quadratic form as well as about the eigenvalue problem, I really can't figure out whats being asked in this task. I'd appreciate and help or hint.
This is the main idea: if $[x]_{\mathcal B} = (x_1',\dots,x_n')$ denotes the coordinate-vector of $x = (x_1,\dots,x_n)$ relative to the basis $\mathcal B = \{v_1,\dots,v_n\}$ (which is to say that $x = x_1' v_1 + \cdots + x_n' v_n$), then we have $$ Q(x) = Q'(x') = \alpha_1^2x_1' + \cdots + \alpha_n^2 x_n'. $$ Accordingly, answer the question by finding an orthogonal matrix $B$ for which $x_i' = \sum_j B_{ij}x_j$, where $x_i'$ denotes the $i$th entry of $[x]_{\mathcal B}$ as defined above.
Regarding the solution: we want a matrix $B$ for which $[x]_{\mathcal B} = Bx$. Let $M$ denote the matrix whose columns are $v_1,\dots,v_n$. We find that this matrix solves the "opposite" problem. In particular, it's easy to show that we have $M[x]_{\mathcal B} = x$.
If we solve for $[x]_{\mathcal B}$, we see that $[x]_{\mathcal B} = M^{-1}x$. However, because $M$ is orthogonal (has orthonormal columns), it's easy to find the inverse of $M$; we have $M^{-1} = M^T$, the transpose of $M$.
Long story short: $B$ is the matrix whose rows are $v_1,\dots,v_n$.