A clean way to see why symmetric matrices always have real eigenvalues is through the quadratic form connection. For a real symmetric matrix $A$ and any nonzero vector $v$, the scalar $v^T A v$ is always a real number since it is built purely from real entries and real vector components. If $v$ is a (possibly complex) eigenvector with $Av = \lambda v$, then looking at $v^{*T} A v$ (using the conjugate transpose) and using the symmetry $A = A^T$ together with the fact that $A$ has real entries, the expression can be shown to equal both $\lambda \|v\|^2$ and its own complex conjugate $\bar\lambda \|v\|^2$. Since $\|v\|^2$ is a positive real number, this equality forces $\lambda = \bar\lambda$, meaning $\lambda$ has no imaginary part, i.e. it is real. This holds true no matter what the actual entries of the symmetric matrix are, so option B, "the eigenvalues are always real," is the only universally correct statement; the other options describe special cases that only sometimes hold (negative eigenvalues for negative-definite matrices, equal eigenvalues only for a scalar multiple of identity), not general properties of every real symmetric matrix.
\[\boxed{\text{Eigenvalues of a real symmetric matrix are always real}}\]For the matrix, $A = \begin{bmatrix} -4 & 0 \\ -1.6 & 4 \end{bmatrix}$, the eigenvalues ($\lambda$) and eigenvectors ($X$) respectively are:
Consider the following matrix: \[ \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix} \] The largest eigenvalue of the above matrix is \(\underline{\hspace{2cm}}\).