Question

Linear Algebra - picture attached

Answer 1

what i know is that since its invertible then there are M independant rows as well as N independent columns so a column must contain m, components of a vector

Answer 2

invertibility is a term valid only for square matrices
so consider m=n

Answer 3

correct

Answer 4

to make it less confusing, lets rephrase the problem :

If A is a mxm matrix, show that \(A^{-1} \implies Col(A) = \mathbb{R}^m\)

Answer 5

okay. that looks good

Answer 6

So basically you want to prove that if the inverse exists, the dimension of column space is \(m\)

Answer 7

yep! ive interpreted it like that but i am not sure how to start it

Answer 8

you're back!

Answer 9

By definition, `dimension` is the number of vectors in `basis`

`basis` contains a full set of linearly independent vectors of a vector space.

So we need to show that the `basis` of column space contains \(m\) linearly independent vectors :

Since \(A\) is invertibel :
\[Ax = 0 \iff x = A^{-1}0 = 0\]

that precicely means all the \(m\) column vectors are linearly dependent. QED

Answer 10

so can we use that to answer b then?

Answer 11

so for b that means Col(A) cannot equal R^m without A being invertible because that means it won't be a square matrix (or detA=0)

Answer 12

\[Ax = 0 \iff x = A^{-1}0 = 0\]

yes we can use this to jusstify bpth parts

Answer 13

\(Ax = 0 \iff x = 0\) is the definition of linearly independent column vectors

Answer 14

and if we let A not be invertible then we cannot end up with that previous statement and so Col(A) cannot equal R^m

Answer 15

one sec, im eating... typing with one hand

Answer 16

haha no worries

Answer 17

For part \(b\) :

\(\large Col(A) = R^m \implies \) the column vectors of \(A\) span \(m\) dimensional space and form a basis and thus are `linearly independent`. So there will be \(m\) pivots when you row reduce and consequently the inverse exists.

Answer 18

go thru these related proofs when free :

http://personalpages.manchester.ac.uk/student/joshua.dawes/notes/invertible-matrix.pdf

Answer 19

ah nice!! thanks for that. could you help with this prove. this involves orthnormal basis'

Answer 20

i should be fine with that proof we did now

Answer 21

need to review my notes, wil reply in 30 minutes

Answer 22

thanks man! youre such a great help!

Answer 23

@rational

Answer 24

\[AB = I \]

Answer 25

that means \(B\) is the inverse of \(A\), right ?

Answer 26

So we need to show :
\[\large AA^{t} = I\]

Answer 27

YEP!

Answer 28

oh isn't there a proof in the link you gave me? Except it doesn't involve orthonormal basis

Answer 29

nupi i was mistaken

Answer 30

Say \(\large x_1, x_2, \cdots x_n\) are the orthonormal vectors of \(A\)

Since the vectors are perpendicular, \(\large x_i . x_j = 0 \) if \(\large i \ne j\)
Since the vectors are unit vectors, \(\large x_i . x_j = 1 \) if \(\large i = j\)

Answer 31

fine ?

Answer 32

yep! i got that far but thats it

Answer 33

next consider \(\large A A^{t}\)

Answer 34

\(\large [ AA^{t}] = [x_i . x_j] \)

Answer 35

when \(i=j\), the element equals 1
when \(i \ne j\), the element equals 0

giving you Identity matrix

Answer 36

where i=j is along the diagonals?

Answer 37

yes, \(\large AA^{t} = [x_i . x_j] = I \)

Answer 38

\[\large AA^{t} = I \]

\[\large \implies A^{t} = A^{-1}\]

Answer 39

hmm. that seems quite short as a proof. i'm not sure if i'm making it more complex than it looks but is what you've written is suffice as a proof?

Answer 40

yes, you just need to state the definition of orthogonal and normal :

1) orthogonal \(\implies \) dot product of two distinct vectors = 0 \(\implies\) all elements other than diagonal are 0
2) normalized \(\implies\) magnitude of each vector = 1\(\implies\) dot product of two same vectors = 1 \(\implies \) all diagonal elements = 1

Answer 41

the arguments have to follow above logical order ^^

Answer 42

as you can see the proof is just about stating the definition and showing that \(AA^t = I\)

Answer 43

ah. i will keep reviewing what you have written again and again since its in my head! thanks so much for your help

Answer 44

np :) remember that `dot product of a \(i\)th row of \(A\) and \(j\)th column of B gives you the \(ij\)th element in the matrix \(AB\) :


\[\large (AB)_{ij} = A_{ik} \bullet B_{kj}\]

Answer 45

and the ijth element is the diagonals of the matrix?

Answer 46

i=j is the diagonal element

Answer 47

oh yes. my bad!

Answer 48

see
http://image.mathcaptain.com/cms/images/101/multiplication-matrix.png

Answer 49

oh your saying that when you dot A.B you just get all elements in AB ah

Answer 50

exactly !

to get the "row 1, column 2" element in AB :

take "1st" row of A
take "2nd" column of B
do dot product

Answer 51

and this dot product equals 0 when \(i \ne j\) because the vectors are orthogonal

Answer 52

AHH! that explained it alot better! i have got it now! yay