Question

find \(A^5\) and \(A^{T}\)

\(A = B\left(B^{T}B\right)^{-1}B^{T}\)

\(B\) is a matrix of size \(m \times n\)

Answer 1

What can you deduce about A from this information alone? How many rows and columns does A have?

Answer 2

\(\large B^T B\) has n rows and n columns

Answer 3

\(A\) = (mxn) (nxn) (nxm)

so \(A\) is mxm matrix i think

Answer 4

Yeah, so far so good. That tells us a little bit. I guess it makes sense that A is a square matrix since you can't square a matrix unless it's square really.

Answer 5

oh yeah A^5 requires A to be a square matrx
so A^5 and A^T are also a mxm matrices

Answer 6

It might help to think of: \[\LARGE B^TB=C\]

Answer 7

I just mean as a whole matrix to itself. Maybe that's not useful, but apparently in the equation of A they essentially have the inverse of C, which means that B^TB represents an invertible matrix, which is also useful to know.

Answer 8

yes i believe B^TB must have an inverse eventhough they didn't specify explicitly
or is a transpose times a matrix is always invertible ?

Answer 9

\[\LARGE A=B(B^TB)^{-1}B^T\]

Well if \[\LARGE \det(B^TB)=0\] Then that means \[\LARGE (B^TB)^{-1}\] is meaningless in the equation for A, right?

Answer 10

Actually it looks like if we left multiply this equation by B^T or right multiply this equation by B we get an interesting result as well!

Answer 11

\[\LARGE AB=B(B^TB)^{-1}B^TB \]

like this ? i was never comfortable with transposes :o

Answer 12

Oh wait, do you mean to say the last two terms eat each other out ?

Answer 13

Yes exactly!

Answer 14

\[\LARGE AB=B\color{red}{(B^TB)^{-1}B^TB}\]

Answer 15

that goes away ? xD

Answer 16

Yeah, so there we have it apparently.

Answer 17

does that also make A an identity matrix ?

Answer 18

An alternate route is just to look at A^2 \[\LARGE A^2=[B(B^TB)^{-1}B^T][B(B^TB)^{-1}B^T]\]Notice how you can remove the identity matrix because of the inverse and you get:
\[\LARGE A^2=B(B^TB)^{-1}B^T = A\]

Yes, I believe this means A must be the identity matrix.

Answer 19

I think it's more easy tan that:
\(B^TB=BB^T\)
also \(BB^T\) is symetric invertibe matrix
that gives the result

Answer 20

wow! so A^5 = A = I

Answer 21

@myko You can't say this\[\LARGE B^TB = BB^T\] because B is a mxn matrix.

Answer 22

omg! yes they give different size matrices

Answer 23

In fact, to say that A is the identity matrix is not entirely true. It either has m or n columns, so which is it? =)

Answer 24

this turned out to be simpler+cuter than it appeared xD thanks a lot !!

Answer 25

Hahaha yes. XD

Answer 26

oh right, so A is an identity matrix of size m :)

Answer 27

just a quick question
\(A^2 = A \) does not mean \(A = I\) right ?

we figured out A is identity matrix from our first observation, by multiplying both sides by B... is this correct ?

Answer 28

Well I think it's a little of both. Ever since that question the other day I've been reviewing everything in linear algebra because my eyes were opened a little.

I think there are matrices that aren't the identity that are their own square. And I think there might also be matrices that you can multiply another matrix by to get the same matrix back that isn't the identity. I'm not entirely sure though since I am not sure anymore about anything I can't directly prove yet... So I'll be figuring these out now!

Answer 29

that makes sense :) atleast we can be sure that \(AB = B\) forces A be an identity

Answer 30

i think that works irrespective of A or B being invertible..

Answer 31

This is from Machine Learning right?
This is the algorithm to linear regression!

Answer 32

i thinkso.. best fit lines right ? it looks like the same matrix :
http://en.wikipedia.org/wiki/Idempotent_matrix

Answer 33

ive just started linear algebra
so im bit clueless how that works in linear regression applications :o

Answer 34

i think that works perfectly if B is a square matrix @OOOPS , then B inverse exists and A equals identity directly

Answer 35

i love this question xD good trick

Answer 36

Looks interesting too:D

Answer 37

\[\large \large AB=B\color{red}{(B^TB)^{-1}B^TB}\]

\(\large A\) becomes an Identity only when the matrix we multiply both sides is a multiple of \(B\), in other cases it could be any square matrix.

In general, Kainui's second method seems flawless :
\[\large A^2 = A = B\left(B^{T}B\right)^{-1}B^{T} = A^{T} \]

Answer 38

Well isn't that matrix you have in red in your last post the identity matrix? Afterall, it is just the product of a matrix and its inverse, it must be the identity matrix right?

Answer 39

yes, i think we can conlclude A is identity cuz AB=B.
and it became identity only because we have multiplied B both sides


A need not be an identity if we multiply any other matrix...
\[\large \large AX=B\color{red}{(B^TB)^{-1}B^TX}\]
not sure if we can make below statement :
\[\large AX = X \iff X=cB\]

Answer 40

why u study linear algebra anyway ?

Answer 41

i think ill review it as well

Answer 42

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

Answer 43

i have a book , and its good also my old notes mmm
i used to be good , i think i well do it in reguler since i have no time to see more nline lectures :o

Answer 44

i am doing lecture 17 as of now and im completely in love with gilbert strang xD
they're good for reviewing aswell...
30 lectures will take 1 month @ 1 lecture per day :)

Answer 45

haha for u xD
for me its hard sometimes to follow english lecture xD
lets see if i could match topics that u reach

Answer 46

i would give u a copy from my book :-\
if i have soft copy of itt xD
even though its the best in linear algebra