What is the interval of W(A)

Sketch of solution: Let $Q$ denote the orthogonal matrix whose columns are the eigenvectors of $A$. Note that $D = Q^TAQ$ is diagonal; moreover, we'll say that $$ D = \pmatrix{\lambda_1 & 0\\0 & \lambda_2}. $$ Show that $(Au|u) = (D(Q^Tu)|Q^Tu)$. Using this, conclude that $W(A) = W(D)$. In order to find $W(D)$, express $(Du|u)$ in terms of $\lambda_1,\lambda_2$ and the entries of $u$.

Since $A$ is real and symmetric, it can be orthogonally diagonalized by the spectral theorem: $A=PDP^T$ where $PP^T=I,D=\text{diag}(\lambda_1,\lambda_2).$

Convince yourself that for any unit vector $u$, we have $\|P^Tu\|^2=(P^Tu)^T(P^Tu)=1.$

Then convince yourself $(Au|u)=u^TAu=\sum_i \lambda_i (P^Tu)_i^2\in [\lambda_{\text{min}}\|P^Tu\|^2,\lambda_{\text{max}}\|P^Tu\|^2]=[\lambda_{\text{min}},\lambda_{\text{max}}].$