91Ó°ÊÓ

Given a quadratic form , where A is a symmetric n x n matrix. Also there exists a nonzero vector in Rⁿ such that .

Let $\mathrm{λ}$be an eigenvalue of matrix A and its corresponding eigenvector is $\stackrel{→}{v}$.

Now, it is written as $A\stackrel{→}{v}=\mathrm{λ}\stackrel{→}{v}$.

According to definition it is written as:

$$\begin{gathered} q\left(\stackrel{→}{v}\right)=\stackrel{→}{v}·A\stackrel{→}{v} \\ =\stackrel{→}{v}·\mathrm{λ}\stackrel{→}{v} \\ =\mathrm{λ}\stackrel{→}{v}·\stackrel{→}{v} \end{gathered}$$

Since $\stackrel{→}{v}$is a unit vector, so $\stackrel{→}{v}·\stackrel{→}{v}=1$.

Thus $q\left(\stackrel{→}{v}\right)=\mathrm{λ}$.

If $q\left(\stackrel{→}{v}\right)=0$then $\mathrm{λ}=0$.

Thus, 0 is an eigenvalue of A.

To determine whether the matrix is invertible or not, find whether the determinant is zero or non-zero.

Determinant is product of the eigenvalues. Since one eigenvalue is zero, therefore the determinant of matrix is zero.

Determinant is zero implies that the matrix fails to be invertible.

Hence, A is not invertible.

Therefore, the statement is true.