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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?

cmi
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1 Answers1

4

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.

cmi
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user
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  • Hiii...How are you?@gimusi – cmi Aug 06 '18 at 16:18
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    Hiiii….I don't want answer. I know the answer. I want explanations of my doubts..@gimusi – cmi Aug 06 '18 at 16:20
  • I know all options are wrong. I want explanations of my doubts..@gimusi – cmi Aug 06 '18 at 16:22
  • @cmi Sorry I made your same mistake, the matrix associated at the quadratic form is the symetric part therefore only option "a" is correct. – user Aug 06 '18 at 16:39
  • @cmi see recent https://math.stackexchange.com/questions/2873773/find-the-bilinear-form-associated-to-a-quadratic-form/2874100#2874100 for a concrete example of the matrix of second partials – Will Jagy Aug 06 '18 at 17:10
  • I still did not get you. How did you get the signature and rank? Neither did I get differentiation of a quadratic form. Can you please elaborate it?@gimusi – cmi Aug 06 '18 at 18:58
  • @cmi The matrix associated to the quadratic form is the symmetric part of A that is B. Therefore you can consider the signature of B that is (2,1,0) therefore the rank is 3. Then refer to the given link for the diferentiation or try to construct directly the Hessina matrix, you'll find that it is 2A. – user Aug 06 '18 at 19:14
  • @cmi And what about that question too? Is it clear now? – user Aug 07 '18 at 06:15
  • I am sorry @gimusi.I overlooked your comment.. – cmi Aug 14 '18 at 05:48
  • I still have the following doubts::::: 1) why sylver riterion has been needed? Can I not find signature of that quadratic form $Q$ by determining the sign of the eigen values of $B$? 2)How the second partial derivative would be determined? Can you do it explicitly? I did not get the link you mentioned.@gimusi – cmi Aug 14 '18 at 05:51
  • @cmi 1) Yes of course we can use any other method but Slvester is usually th emore effective. 2) We can consider the quadratic form on A or B and calculate the Hessian matrix (qxx,qxyqxz;qyx,qyy,qyz;qzx,qzy,qzz). Here is the link https://math.stackexchange.com/questions/2422310/differentiation-of-a-quadratic-form and for the Hessian matrix https://en.wikipedia.org/wiki/Hessian_matrix – user Aug 14 '18 at 06:18
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    @cmi Sorry but for point 2) what we obtain by Hessian matrix is $A+A^T\neq 2A$ but $A+A^T=2B$, I've made a confusion with the 2 matrices. In that case, also option A seems to be wrong. – user Aug 14 '18 at 06:42
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    Yea Hessian Matrix of this quadratic form would be $2B$. so all options are wrong? I have one more doubt. What $( + , + , 0)$ does mean?@gimusi – cmi Aug 14 '18 at 06:46
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    @cmi Usually it indicate that 2 eigenvalues are positive and 1 is equal to zero, that's true for A but not for the B matrix which determines the signature. – user Aug 14 '18 at 06:50
  • Can you please edit your answer? So that I can accept your answer?@gimusi – cmi Aug 14 '18 at 07:16
  • How do you know from th Sylvester's criteria that there are two positive eigen values ? Yea we can ensure that there will be at least one negative eigen value. But How you can get the exact number of positive eigen values?@gimusi – cmi Aug 14 '18 at 07:19
  • Yea I get it as the trace is positive there must be at least one positive eigen value.@gimusi – cmi Aug 14 '18 at 07:25
  • @cmi by Sylvester we use the fact that the determinant is the product of the eigenvalues and that the signature is an invariant. Since detB <0 the signature is - - - or ++- but since det((1,1;1,0)<0 we can exclude - - - . Of course we can deduce the latter from the trace but itbis not necessary. – user Aug 14 '18 at 07:52
  • How can you exclude (- - -)? What det(1,1 ; 1,0)<0 means?@gimusi – cmi Aug 14 '18 at 07:56
  • You can say that all the LEADING principal matrices are not positive so all eigen values are not positive. and All principal matrices are not non negative so we can say there must be at least one negative eigen value. DetB > 0 so (- - -) is excluded.@gimusi – cmi Aug 14 '18 at 08:00
  • @cmi now I can’t, later I will explain it better! – user Aug 14 '18 at 08:01
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    @cmi since $$\det\left(\begin{matrix}1&1\1&0\end{matrix}\right)=-1<0$$ we can exclude the case - - - because otherwise we could not find a minor 2-by-2 with negative determinant. – user Aug 14 '18 at 08:52
  • I got you..And I understood the whole game..@gimusi – cmi Aug 14 '18 at 12:01
  • @cmi Well done! You are welcome. Bye – user Aug 14 '18 at 12:09