Bilinear form with parameter [closed]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












I have a bilinear form such that the associate matrix is $ A=left(beginmatrix0&0&k\
0&k&0\
k&0&0\
endmatrixright)$.
$,$ Does exist a $k$ such that $F_k((e_1+e_2+e_3),(e_1+e_2+e_3))=4$



I don't understand how can I start!




linear-algebra matrices bilinear-form




share|cite|improve this question













closed as off-topic by Travis, Strants, José Carlos Santos, Adrian Keister, user223391 Jul 20 at 1:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Strants, José Carlos Santos, Adrian Keister, Community
If this question can be reworded to fit the rules in the help center, please edit the question.












  • How is $F_k$ defined?
    – José Carlos Santos
    Jul 16 at 16:05










  • $xz'+yy'+zx'$ right?
    – Roberto De La Fuente
    Jul 16 at 16:07














up vote
0
down vote

favorite












I have a bilinear form such that the associate matrix is $ A=left(beginmatrix0&0&k\
0&k&0\
k&0&0\
endmatrixright)$.
$,$ Does exist a $k$ such that $F_k((e_1+e_2+e_3),(e_1+e_2+e_3))=4$



I don't understand how can I start!




linear-algebra matrices bilinear-form




share|cite|improve this question













closed as off-topic by Travis, Strants, José Carlos Santos, Adrian Keister, user223391 Jul 20 at 1:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Strants, José Carlos Santos, Adrian Keister, Community
If this question can be reworded to fit the rules in the help center, please edit the question.












  • How is $F_k$ defined?
    – José Carlos Santos
    Jul 16 at 16:05










  • $xz'+yy'+zx'$ right?
    – Roberto De La Fuente
    Jul 16 at 16:07












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have a bilinear form such that the associate matrix is $ A=left(beginmatrix0&0&k\
0&k&0\
k&0&0\
endmatrixright)$.
$,$ Does exist a $k$ such that $F_k((e_1+e_2+e_3),(e_1+e_2+e_3))=4$



I don't understand how can I start!




linear-algebra matrices bilinear-form




share|cite|improve this question













I have a bilinear form such that the associate matrix is $ A=left(beginmatrix0&0&k\
0&k&0\
k&0&0\
endmatrixright)$.
$,$ Does exist a $k$ such that $F_k((e_1+e_2+e_3),(e_1+e_2+e_3))=4$



I don't understand how can I start!





linear-algebra matrices bilinear-form





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 16 at 16:10









Ivo Terek

43.9k949134




43.9k949134









asked Jul 16 at 15:59









Roberto De La Fuente

417




417




closed as off-topic by Travis, Strants, José Carlos Santos, Adrian Keister, user223391 Jul 20 at 1:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Strants, José Carlos Santos, Adrian Keister, Community
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Travis, Strants, José Carlos Santos, Adrian Keister, user223391 Jul 20 at 1:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Strants, José Carlos Santos, Adrian Keister, Community
If this question can be reworded to fit the rules in the help center, please edit the question.











  • How is $F_k$ defined?
    – José Carlos Santos
    Jul 16 at 16:05










  • $xz'+yy'+zx'$ right?
    – Roberto De La Fuente
    Jul 16 at 16:07
















  • How is $F_k$ defined?
    – José Carlos Santos
    Jul 16 at 16:05










  • $xz'+yy'+zx'$ right?
    – Roberto De La Fuente
    Jul 16 at 16:07















How is $F_k$ defined?
– José Carlos Santos
Jul 16 at 16:05




How is $F_k$ defined?
– José Carlos Santos
Jul 16 at 16:05












$xz'+yy'+zx'$ right?
– Roberto De La Fuente
Jul 16 at 16:07




$xz'+yy'+zx'$ right?
– Roberto De La Fuente
Jul 16 at 16:07










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










You have that $F_k(vecx,vecy) = vecx^top A vecy$, so $$beginalign F_k((1,1,1),(1,1,1)) &= beginpmatrix 1 & 1 & 1endpmatrix beginpmatrix 0 & 0 & k \ 0 & k & 0 \ k & 0 & 0endpmatrix beginpmatrix 1 \ 1 \ 1 endpmatrix \ &= beginpmatrix 1 & 1 & 1endpmatrix beginpmatrix k \ k \ k endpmatrix \ &= 3kendalign$$So $k = 4/3$ fits the bill.






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    You have that $F_k(vecx,vecy) = vecx^top A vecy$, so $$beginalign F_k((1,1,1),(1,1,1)) &= beginpmatrix 1 & 1 & 1endpmatrix beginpmatrix 0 & 0 & k \ 0 & k & 0 \ k & 0 & 0endpmatrix beginpmatrix 1 \ 1 \ 1 endpmatrix \ &= beginpmatrix 1 & 1 & 1endpmatrix beginpmatrix k \ k \ k endpmatrix \ &= 3kendalign$$So $k = 4/3$ fits the bill.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      You have that $F_k(vecx,vecy) = vecx^top A vecy$, so $$beginalign F_k((1,1,1),(1,1,1)) &= beginpmatrix 1 & 1 & 1endpmatrix beginpmatrix 0 & 0 & k \ 0 & k & 0 \ k & 0 & 0endpmatrix beginpmatrix 1 \ 1 \ 1 endpmatrix \ &= beginpmatrix 1 & 1 & 1endpmatrix beginpmatrix k \ k \ k endpmatrix \ &= 3kendalign$$So $k = 4/3$ fits the bill.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        You have that $F_k(vecx,vecy) = vecx^top A vecy$, so $$beginalign F_k((1,1,1),(1,1,1)) &= beginpmatrix 1 & 1 & 1endpmatrix beginpmatrix 0 & 0 & k \ 0 & k & 0 \ k & 0 & 0endpmatrix beginpmatrix 1 \ 1 \ 1 endpmatrix \ &= beginpmatrix 1 & 1 & 1endpmatrix beginpmatrix k \ k \ k endpmatrix \ &= 3kendalign$$So $k = 4/3$ fits the bill.






        share|cite|improve this answer













        You have that $F_k(vecx,vecy) = vecx^top A vecy$, so $$beginalign F_k((1,1,1),(1,1,1)) &= beginpmatrix 1 & 1 & 1endpmatrix beginpmatrix 0 & 0 & k \ 0 & k & 0 \ k & 0 & 0endpmatrix beginpmatrix 1 \ 1 \ 1 endpmatrix \ &= beginpmatrix 1 & 1 & 1endpmatrix beginpmatrix k \ k \ k endpmatrix \ &= 3kendalign$$So $k = 4/3$ fits the bill.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 16 at 16:09









        Ivo Terek

        43.9k949134




        43.9k949134












            Comments

            Post a Comment

            Popular posts from this blog

            What is the equation of a 3D cone with generalised tilt?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
            Read more

            Relationship between determinant of matrix and determinant of adjoint?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
            Read more

            Color the edges and diagonals of a regular polygon

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
            Read more