Finding rank and signature of a quadratic form.

Question

Let $$ A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & -2 \\ 0 & 0 & 1 \end{bmatrix} $$ and define for $x,y,z \in R$ , $ Q \begin{bmatrix} x,y,z \end{bmatrix} = \begin{bmatrix} x,y,z \end{bmatrix} A \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}$

Which of the following true ?

a) The matrix of second order partial derivatives of the quadratic form $Q$ is $2A$.

b) The rank of the quadratic form $Q$ is $2$.

c)The signature of the quadratic form $Q$ is (+ + 0).

In option a) I can not understand the term "second order partial derivatives of the quadratic form $Q$". What does that mean? How to define partial derivative of a quadratic form?

In option b) I can not understand what is rank of a quadratic form? Is it same with the rank of the Matrix $A$?

In option c) signature should be the difference between the number of positive roots and the number of negative roots. So it should be $2$ here. But how come it becomes (+ + 0)?

Can somebody please help me to understand the problem?

Answer 1

We have that

$$A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & -2 \\ 0 & 0 & 1 \end{bmatrix}=\frac12(A+A^T)+\frac12(A-A^T)=B+C=\\\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix}+ \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix}$$

and

$$Q \begin{bmatrix} x,y,z \end{bmatrix} = \begin{bmatrix} x,y,z \end{bmatrix} A \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}= \begin{bmatrix} x,y,z \end{bmatrix} B \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}$$

and by Sylvester criterion we have

• $\det(1)=1>0$
• $\det\left(\begin{matrix}1&1\\1&0\end{matrix}\right)=-1<0$
• $\det(B)=-2<0$

then the signature is $(n_+,n_-,n_0)=(2,1,0)$ and $\operatorname{rank}(B)=3$.

a) can not be correct as Hessian Matrix of $Q$ would be $2B$.

No option is correct.