- #1

- 415

- 0

3)

[tex]q:R^{3}->R[/tex] is defined by [tex]q(x,y,z)=4(xy+yz+zx)[/tex]

find the minimal [tex]M\in R[/tex] so [tex]q(x,y,z)\leq M(x^{2}+y^{2}+z^{2})[/tex]

?

why in the solution the calculate the caracteristic polinomial

?

why if (t+2) is in power 2 then we have -2 in two members of

the formula q(v) ??

our polinomial doesnt separated into different lenear member

so in order to find its jordan form we need to find the minimal polinomial

etc..

but in the solution they said it straight forward why??

why did they coose in the end to put eigen vaule iside?

why its minimal?

we could put a vector which not is 0.5

?

is it true that the as the eigen vectors would diagonolise A

so is their orthonormal basis whould show Q as in sum of squares

correct?

[tex]q:R^{3}->R[/tex] is defined by [tex]q(x,y,z)=4(xy+yz+zx)[/tex]

find the minimal [tex]M\in R[/tex] so [tex]q(x,y,z)\leq M(x^{2}+y^{2}+z^{2})[/tex]

?

why in the solution the calculate the caracteristic polinomial

?

why if (t+2) is in power 2 then we have -2 in two members of

the formula q(v) ??

our polinomial doesnt separated into different lenear member

so in order to find its jordan form we need to find the minimal polinomial

etc..

but in the solution they said it straight forward why??

why did they coose in the end to put eigen vaule iside?

why its minimal?

we could put a vector which not is 0.5

?

is it true that the as the eigen vectors would diagonolise A

so is their orthonormal basis whould show Q as in sum of squares

correct?

Last edited by a moderator: