In Exercises 21 and 22, matrices are \(n \times n\) and vectors are in \({{\bf{R}}^{\bf{n}}}\) . Mark each statement True or false. Justify each answer.

22. a. The expression \({\left\| {\bf{x}} \right\|^{\bf{2}}}\) is not a quadratic form.

b. If \(A\) is symmetric and \(P\) is an orthogonal matrix, then be the change of the variable \({\rm{x}} = P{\rm{y}}\) transforms \({x^T}Ax\) into a quadratic form with no cross product term.

c. If \(A\)is a\({\bf{2 \times 2}}\)symmetric matrix, then the set of\(x\)such that\({x^T}Ax = c\) corresponds to either a circle, an ellipse or a hyperbola.

d. An indefinite quadratic form is neither positive semidefinite nor negative semidefinite.

e.If \(A\) is symmetric and the quadratic form \({x^T}Ax\) has only negative values for \({\bf{x}} \ne {\bf{0}}\) then the eigenvalues of \(A\) are all positive.

Since the identity matrix\(I\)is symmetric so,

\(\begin{aligned}{}Q({\bf{x}}) = {{\bf{x}}^T}I{\bf{x}}\\ = {{\bf{x}}^T}{\bf{x}}\\ = {\left\| {\bf{x}} \right\|^2}\end{aligned}\)

The above conclusion proves that an expression\({\left\| {\bf{x}} \right\|^2}\)is a quadratic form.

Hence the statement is false.

By the principal Axes theorem, there is an orthogonal changeof the variable \(x = Py\) which transforms \({{\bf{x}}^T}A{\bf{x}}\) into a quadratic form \({y^T}Dy\) with no cross-product term.

Hence the statement is False.

From a geometric view of Principle axes, if\(A\)is\(2 \times 2\)a symmetric matrix then the set of\({\rm{x}}\)such that\({{\bf{x}}^T}A{\bf{x}} = c\)(constant) corresponds to a circle, ellipse, a hyperbola, two intersecting lines, a single point or contains no points at all. Hence, it can take any of the forms from a circle, ellipse, a hyperbola, two intersecting lines, a single point or contains no points at all.

Hence the statement is false.

A quadratic form\(Q({\bf{x}})\)is indefinite then neither both positive and negative values. So, thequadratic form\(Q\)is indefinite form is neither positive semidefinite nor negative semidefinite.

Therefore, the statement is True.

If\(Q(x) = {x^T}Ax < 0,Q(x) = {x^T}Ax < 0\),for all\(x \ne 0\),then\(Q\)is negative definite. The\(Q\)is negative definite if and only if the eigenvalues of are all negative.

Hence the statement is false.