Answer
True
Work Step by Step
Assume that $A$ is an $n \times n$ real, symmetric matrix with eigenvalues $λ_1, λ_2,..., λ_n$ corresponding to a complete orthonormal set of eigenvectors for $A$ and $x$ is a vector of $R^n$ and the quadratic form $X^TAX$
Matrix $A$ can be diagonalized with an orthogonal matrix $S$ such that $S^TAS=D=diag(\lambda_1,\lambda_2,...\lambda_n)$
and $S^T$ has the eigenvectors of $A$:
$X^TAX$$=X^TSDS^TX\\
=[S^TX]^TD[S^TX]\\
=B^TDB$
where $B=\begin{bmatrix}
y_1\\
y_2\\
.\\
.\\
y_n
\end{bmatrix}$
Hence, $X^TAX=B^TDB=\begin{bmatrix}
y_1, y_2 ,....,y_n
\end{bmatrix}\begin{bmatrix}
\lambda_1 & 0 &...& 0\\
. & . & .\\
0 & 0 & ... & \lambda_n
\end{bmatrix}\begin{bmatrix}
y_1\\
y_2\\
.\\
.\\
y_n
\end{bmatrix}\\
=\lambda_1y_1^2+\lambda_2 y_2^2+...\lambda_ny_n^2$
Therefore, the statement is true.
