Find a 2x2 matrix A for which E(subscript -3) = span { [2, -5] }
and E(subscript 6) = span { [-1, 3] },
where E(subscript λ) is the eigenspace associated with the eigenvalue λ.
A = ?
anyone know how to do this type of question?
and E(subscript 6) = span { [-1, 3] },
where E(subscript λ) is the eigenspace associated with the eigenvalue λ.
A = ?
anyone know how to do this type of question?
-
okay, J^3, merci, ty pour l'explication !
interesting notation ...
un bonjour de la France !
et oui, i am French... nobody' s perfect.
A for which
E(subscript -3) = span { [2, -5] }
and
E(subscript 6) = span { [-1, 3] },
let B = { i , j } be the basis of E
and
u = 2i - 5j
v = -i + 3j , the eigenvectors,
then
Au = -3u
and
Av = 6v
for A =
(a c)
(b d)
this brings 4 equations with 4 unknows :
2a - 5c = -3*2 = -6 --------> eq (1)
2b - 5d =-3*(-5)= 15 --------> eq (2)
-a + 3c = 6*(-1)= -6 --------> eq (3)
-b + 3d = 6*3 = 18 --------> eq (4)
2* eq(3) gives
-2a + 6c = -12 --------> eq(5)
1° + (5) gives : c = - 18 , if my calc is alright !
etc. : the rest should not so difficult for you .
et voilà !
hope it' ll help !!
interesting notation ...
un bonjour de la France !
et oui, i am French... nobody' s perfect.
A for which
E(subscript -3) = span { [2, -5] }
and
E(subscript 6) = span { [-1, 3] },
let B = { i , j } be the basis of E
and
u = 2i - 5j
v = -i + 3j , the eigenvectors,
then
Au = -3u
and
Av = 6v
for A =
(a c)
(b d)
this brings 4 equations with 4 unknows :
2a - 5c = -3*2 = -6 --------> eq (1)
2b - 5d =-3*(-5)= 15 --------> eq (2)
-a + 3c = 6*(-1)= -6 --------> eq (3)
-b + 3d = 6*3 = 18 --------> eq (4)
2* eq(3) gives
-2a + 6c = -12 --------> eq(5)
1° + (5) gives : c = - 18 , if my calc is alright !
etc. : the rest should not so difficult for you .
et voilà !
hope it' ll help !!