Question

Polpak here's the question!

Answer 1

the inner product defined in 4 was..

Answer 2

<A|B> = trace(A^T B)
trace of a transpose times b

Answer 3

Just tell me what I need to do. Like what do I need to find here? Ughh i hate lin algebra notation..

Answer 4

Ok, firstly, to be orthonormal they must be orthogonal.

So check that the inner products of each are all 0.

Answer 5

So <A,B> must be 0, <A,C> must be 0, <A,D> must be 0
<B,C> must be 0, <B,D> must be 0
<C,D> must be 0.

Answer 6

DAAAAAMMMNNN SO MUCH WORK??? Is there an easier way? lol I will need to multiply matrices 6 times...

Answer 7

there's an easier way

Answer 8

if the determinants are all zero, they are all lin ind and detA times detB = detAB

Answer 9

oops

Answer 10

I'm not sure how the determinate of the individual matrix can show that it and another are linearly independent.

Answer 11

meant not all zero

Answer 12

if the det of a matrix is NOT zero, it's lin ind

Answer 13

wait what about the with respect to the inner product which is:
<A|B> = tr(A^TB)

Answer 14

Yes, but we need to show that the set of these matrices is linearly independent. Not that each one on its own is linearly independent.

Answer 15

there's a property that says detAdetB = detAB

Answer 16

oh yea..

Answer 17

Anyway, you're being a baby bahrom. They all have 1 or -1, multiplying them is trivial ;p

Answer 18

lol polpak i've been doing linear algebra for the past 3 days.. i deserve being a baby for a while..

Answer 19

\((c_1A)(c_2B) =( c_1c_2)AB\)

Answer 20

bahrom, do yourself a favor: memorize the properties in your book...they all tie together

Answer 21

math i have no space left.. anyway we just did determinants yesterday this was assigned three days ago.. so what do I do? Just show that every single determinant is not equal to zero, since if the determinant is 0, then the matrices are lin dependent and the basis is NOT orthonormal?

Answer 22

I mean if i were to use determinants, would I just show that every single determinant is not 0 and that's it?

Answer 23

well if it's a square matrix, if you can at least remember that if the det is not 0 it has lin ind rows/vectors

Answer 24

it will help you later on

Answer 25

wait guys but how does the trace(A^TB) tie in here?

Answer 26

\[trac((\frac{1}{2\sqrt{2}})\left[\begin{matrix}0 & 1\\1 &0\end{matrix}\right]\left[\begin{matrix}1 & -1\\1 &1\end{matrix} \right])=trace((\frac{1}{2\sqrt{2}})\left[\begin{matrix}1&1\\1&-1\end{matrix}\right]) = 0\]

Answer 27

So certainly those two are orthogonal.

Answer 28

You know how to do the 'whole row' method of matrix multiplication right?

Answer 29

cause you _will_ need to multiply them to solve this.

Answer 30

what do u mean? i mean i know how to multiply matrices..

Answer 31

There are different ways to think about matrix multiplication.

Answer 32

okay whatever i will multiply this crap out to determine if they are orthogonal.. then what do i do to show they are orthonormal?

Answer 33

You just have to show that they have unit length. (whatever that means for matrixes)

Answer 34

Oh, I remember.

Answer 35

This is bringing back bad memories of a class _long_ dead... :P

Answer 36

<U,U> = (magnitude of U)^2

Answer 37

So you have to take their inner product with themselves and show that it's 1.

Answer 38

So that's 4 more multiplications to do.

Answer 39

You should explain how you do multiplication, cause I may be able to show you a faster way.

Answer 40

okay great so first step:

Show that:
<A,B> must be 0, <A,C> must be 0, <A,D> must be 0
<B,C> must be 0, <B,D> must be 0
<C,D> must be 0.

2nd step:
<A|A> = 1
<B|B> = 1
<C|C> = 1
<D|D> = 1

That's 10 matrix multiplications.. just kill me..

Answer 41

Anyway i just do row * column

Answer 42

Ouch. Yeah. that takes more thinking (IMO)

Answer 43

Give me two 2 by 2 matrices. for example and I'll show you how to multiply them 'faster' (for me anyway)

Answer 44

a1 a2
a3 a4
and
b1 b2
b3 b4

Answer 45

oh fine.. be complicated ;p

Answer 46

no i just want a general formula so that i know it..

Answer 47

\[AB = \left[\begin{matrix}a1<b1,b2> + a2<b3,b4>\\ a3<b1,b2> + a4<b3,b4>\end{matrix}\right]\]

Answer 48

But it's easier to explain with actual examples.

Answer 49

okay then:
1 2
3 4
and
5 6
7 8

Answer 50

Except I forgot the 2.

Answer 51

<5,6> + <14,16> = <19,24>

Answer 52

That's the first row of AB.

Answer 53

the second row is:
<15,18> +<28, 32> = <43,50>

Answer 54

It's easier for me to do on paper than typing obviously. But even though it's essentially the same operation it feels like less because I have fewer things to remember.

Answer 55

okay..

Answer 56

6+16 = 22, not 24

Answer 57

bleh.. I'm not making a good case here, but look at the problem you have for example.

Answer 58

\[\left[\begin{matrix}1&0\\0 &-1\end{matrix}\right]\left[\begin{matrix}1&-1\\1 &1\end{matrix}\right]=\left[\begin{matrix}1&-1\\-1 &-1\end{matrix}\right]\]

Answer 59

The first row of the resulting matrix is just 1 of the first row of matrix B plus 0 of the second row of matrix B.

Answer 60

The second row of the result is just 0 of the first row of B plus -1 of the second row of matrix B

Answer 61

k

Answer 62

u know what polpak i appreciate all the input but instead of tryin to memorize even more info im just gonna use the traditional way..

Answer 63

np. When you get some free time, try understanding what I said and give it a try. It might be faster. But for now sure, use the traditional method.

Answer 64

i have a final tomorrow ima need as much memory as i can..

Answer 65

i will look at this on friday.. once im done with my final..

Answer 66

*as possible..

Answer 67

I know that for me, it was faster because using the traditional way I'd always get screwed up on which column times which row I was working on...

Answer 68

I'm sure u will do well...:-)

Answer 69

thank you guys! U guys saved my booty like 11 times so far with this class.. nobody got to help me with ODEs though.. but i did good in those..

Answer 70

I unfortunately recall very little from diff-eqs. I need to re-learn that material.

Answer 71

it's much better than linear algebra!

Answer 72

much harder but much better!

Answer 73

I dunno, I find linear stuff pretty cool.

Answer 74

Linear algebra gets a bad rep because the way it is taught hasn't kept up with developments (If you get a chance you really should take a look at vector spaces (proper ones with actual vectors in 'em, not polynomials) minus all the nonsense).

Answer 75

i keep makin arith mistakes on this stuff it freaks me out all the time.. Last time i lost like 20 minutes redoing the same problem OVER AND OVER AGAIn like 3 times and i got a sign wrong..

Answer 76

I agree, used to be you could show off by calculating the determinant of a 5 by 5 matrix, not really a useful skill any more...

Answer 77

almost there.. did 3 matrix multips.. 7 to go..