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A matrix is said to be a negative definite matrix when it is symmetric and all its eigenvalues are non-positive.

If \({d_i}\) is the \({i^{{\rm{th }}}}\) diagonal entry of \(A\), then \(e_i^TA{e_i} = {d_i}\),where \({e_i}\) is the \({i^{{\rm{th }}}}\)standard basis vector.

For example, consider a matrix\(A = \left[ \begin{array}{l}1{\rm{ 3}}\\{\rm{3 5}}\end{array} \right]\)

To prove that the given matrix is negative definite, find the symmetric of the matrix.

Because the symmetric of the matrix is said to be positive definite matrix.

Find the symmetric of the matrix A:

\(\begin{array}{c}{A^T} = \left[ {\begin{array}{*{20}{c}}1&3\\3&5\end{array}} \right]\\ = A\end{array}\)

Here,\({A^T} = A\)which means it鈥檚 symmetric.

Since \(A\) is negative definite, therefore \(e_i^TA{e_i} < 0\). That is, \({d_i} < 0\).

The given statement is TRUE.