Question

The norm of a matrix?!

Answer 1

is the multiplication of eigen values = determinant

Answer 2

If a "unit matrix" unit vector in the direction of A is given by:

\[\frac{A}{|A|} = U\]

what power does k have to be in order for

\[\frac{U}{|U|} = U\]

if we define:\[|A| = [\det(A)]^k\]

Answer 3

oh ya ofc it is

Answer 4

The \[\frac{U}{|U|}=U\] condition is just that if you were to normalize a unit vector you will get a unit vector back.

Answer 5

U bet.

Get my pun? haha

Answer 6

lol unfortunately but I groaned @chris00 lol

Answer 7

smh

Answer 8

;) but yes, unit vectors are quite powerful

Answer 9

The thing here is these unit vectors are matrices... This is quite a strange question haha

Answer 10

Has a fun answer though I think!

Answer 11

yeah interesting.

Answer 12

heeres something

Answer 13

why this has to work

Answer 14

if u rewrite any matrix as P^-1 D P

Answer 15

dan, you cannot be an honorary professor of maths and have ur dp as some chick with a selfie a haha

Answer 16

then we proved that we can commute these matrices around and the determinant is still the same

Answer 17

wait no what did we prove again for 3 matrices multiplying

Answer 18

Yeah you don't have to diagonalize it to do this. :P

Answer 19

ya but this can show how this way, if u mulptying my 1/ det(A)^k then u will have a eigen value multiplication that is =1

Answer 20

then u know that U = U/|U| will be satisfied too

Answer 21

as u can once again rewrite U in P^-1DP form

Answer 22

looks like you have gone along way in linear algebra dan.

Answer 23

are you saying woman cannot be honorary professors of math!

Answer 24

How dare u sir

Answer 25

/mam

Answer 26

no, you are incorrectly portraying yourself

Answer 27

:o

Answer 28

Yeah if I was a mod I'd ban you right now for this dan

Answer 29

LOL pls

Answer 30

i'm trying to become a mod soley to accuse dan of fraud

Answer 31

;)

Answer 32

lolol

Answer 33

Ok no wasting times on diagonalizing dan, you're not allowed to diagonalize the matrix I forbid it for this problem since it's a waste of time

Answer 34

I'll give you a hint to make up for it

Answer 35

finee

Answer 36

what's
\[\det(\det(A)I)\]

Answer 37

det(a)^3

Answer 38

det(A)^n

Answer 39

for an n size identity matrix

Answer 40

oh i just realized u were asking a question on what k actually has to be xD

Answer 41

smh

Answer 42

lol i thought u were already saying its k for like a k by k matrix

Answer 43

No, in fact it's an nxn matrix A. K is completely to be determined still

Answer 44

Just an interesting observation I noticed right now, if the determinant of a matrix is 1, it's a "unit matrix" lol

Answer 45

This is totally made up by me by the way, like I didn't really get this from anywhere I just sorta did this out of fun