Question

Find a matrix who's kernel is spanned by the two vectors U = ( 1,3,2) and V = ( -2, 0, 4).

Answer 1

Not sure what to do with this one

Answer 2

cross the vectors maybe?

Answer 3

I thought of doing that but I can't justify it

Answer 4

the kernal is the set of vectors which map to 0

Answer 5

u and v are a plain that has a normal

Answer 6

plane lol

Answer 7

Right so if i take the cross product of both of them then what am I really doing? Just fixing them in that plane on the origin?

Answer 8

hmm, lets try to write up a better definition of a kernel and see what we come up with. im trying to work from memory which can often times be faulty

the kernel is the set of all elements in A that map to 0 in B

Answer 9

does that sound right?

Answer 10

Yeah

Answer 11

how do we set that up generically?

Answer 12

Ax-b= 0 ?

Answer 13

for some linear combination of vectors in A: Ax = 0 defines the kernel of A

Answer 14

a1U + a2V = 0 for some x=a1,a2

Answer 15

That makes sense yeah

Answer 16

So using the cross product I get those value of a1, a2,.?

Answer 17

you need a singular matrix to have something in kernel

Answer 18

still considering my errors :)

the kernel of a vector space that is a plane, is the normal thru the origin.

Answer 19

So my vectors span a plane in R^3

Answer 20

and since they arent multiples they should be linearly dependent

Answer 21

Since the given vectors are basis of kernel, you need to find a vector thats perpendicular to both the solution vector

smply take cross product as amistre said

Answer 22

we were trying to justify the process :)

Answer 23

\[\large \pmatrix{a_1&a_2&a_3\\a_1&a_2&a_3\\a_1&a_2&a_3\\} \pmatrix{-2\\0\\4} = \pmatrix{0\\0\\0}\]

from the matrix multiplication, is it clear that the dot product of row vector and the given solution vector has to be 0 ?

Answer 24

Havent performed the operation but yes it seems like it would be

Answer 25

it is true in general,
the row space is perpendicular to the kernel

Answer 26

Any vector in kernel has to be perpendicular to the rows of a matrix

Answer 27

you can reason that by simply interpreting the matrix multiplication as dot product of row and a column

Answer 28

Ohh okay that make sense

Answer 29

Thank you

Answer 30

Lets say i had 3 vectors would the cross product still apply?

Answer 31

If the kernel was spanned by 3 vectors

Answer 32

The kernel would then be perpendicular to all 3? so A = U x V x W

Answer 33

i think crossproduct is defined only in R^3

Answer 34

if your kernel has a basis of 3, then you need a matrix of atleast 4x4

Answer 35

you need to find a vector such that the dot product with each of the vector in basis of kernel gives 0

Answer 36

you wont ever get to work such a painful problem unless your prof is a psycho :P

Answer 37

Ha ha well we wont just yet but apparently we would be allowed to use any calc or software if we do

Answer 38

thats nice :) matlab solves all your problems then

Answer 39

do you know any good references to books or online resources so that i can understand the concepts better? Im using a book but not many examples

Answer 40

heard of gilbert strang ?

Answer 41

Yeah MIT professor i believe

Answer 42

yes he is the god of linear algebra

Answer 43

Gotta check him out ha ha thank you

Answer 44

http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/

Answer 45

watch his 6th lecture,
it may not be so much useful to you as you know most things about subspaces already but still you may find some new things...

Answer 46

Awesome an OCW by him, free always is good!

Answer 47

Alright i will do that now thank you very much ganeshie8

Answer 48

np :)