Register to reply

Linear Proofs

by blueberryfive
Tags: linear, matrix
Share this thread:
blueberryfive
#1
Sep4-11, 10:33 PM
P: 36
Hello,

A(rB) = r(AB) =(rA)B where r is a real scalar and A and B are appropriately sized matrices.

How to even start? A(rbij)=A(rB), but then you can't reassociate...

Also, a formal proof for Tr(AT)=Tr(A)?

It doesn't seem like enough to say the diagonal entries are unaffected by transposition..

Lastly, let A be an mxn matrix with a column consisting entirely of zeros. Show that if B is an nxp matrix, then AB has a row of zeros.

I can't figure out how to make a proof of this. I know how to say what such and such entry of AB is, but I don't know how to designate an entire column. How do you formally say it will be equal to zero, then...just because the dot product of a zero vector with anything is 0?
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
HallsofIvy
#2
Sep5-11, 08:17 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568
Quote Quote by blueberryfive View Post
Hello,

A(rB) = r(AB) =(rA)B where r is a real scalar and A and B are appropriately sized matrices.

How to even start? A(rbij)=A(rB), but then you can't reassociate...
You don't have to. Every entry in "rB" has a factor of r so every entry in A(rB) has a factor of r so A(rb)= r(AB)= (rA)B

Also, a formal proof for Tr(AT)=Tr(A)?

It doesn't seem like enough to say the diagonal entries are unaffected by transposition..
Why not? Would it be better to say "[itex]A^*_{ij}= A_{ji}[/itex]" so that, replacing j with i, "[itex]A^*_{ii}= A_{ii}[/itex]"? That may look more "formal" but it is really just saying that "the diagonal entries are unaffected by transposition".

Lastly, let A be an mxn matrix with a column consisting entirely of zeros. Show that if B is an nxp matrix, then AB has a row of zeros.
[tex](AB)_{ij}= \sum A_{ik}B_{kj}[/itex]. If the "jth" column of B is all 0s, then the "jth" row of A is all 0.

I can't figure out how to make a proof of this. I know how to say what such and such entry of AB is, but I don't know how to designate an entire column. How do you formally say it will be equal to zero, then...just because the dot product of a zero vector with anything is 0?
blueberryfive
#3
Sep5-11, 08:26 AM
P: 36
Thank you.

Also,

Tr(ATA)[itex]\geq[/itex]0.

I can't even see how to begin...

HallsofIvy
#4
Sep5-11, 08:28 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568
Linear Proofs

Quote Quote by blueberryfive View Post
Thank you.

Also,

Tr(ATA)[itex]\geq[/itex]0.

I can't even see how to begin...
That's a sum of squares!
Fredrik
#5
Sep5-11, 08:57 AM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 9,416
Quote Quote by blueberryfive View Post
Hello,

A(rB) = r(AB) =(rA)B where r is a real scalar and A and B are appropriately sized matrices.

How to even start? A(rbij)=A(rB), but then you can't reassociate...
The definition of rA where r is a real number and A is a matrix is [itex](rA)_{ij}=rA_{ij}[/itex]. The definition of AB where both A and B are matrices is [itex](AB)_{ij}=\sum_k A_{ik}B_{kj}[/itex]. It's not hard to use these definitions to show that the equalities you mentioned are true. Start with [itex](A(rB))_{ij}=\sum_k A_{ik}(rB)_{kj}[/itex].

All your other questions are also quite easy to answer if you just use these definitions, and the definition of the trace and the transpose: [itex]\operatorname{Tr}A=\sum_i A_{ii},\quad (A^T)_{ij}=A_{ji}[/itex].


Register to reply

Related Discussions
Linear algebra proofs (linear equations/inverses) Calculus & Beyond Homework 6
Linear Algebra proofs Calculus & Beyond Homework 3
Linear Mappings & Proofs Calculus & Beyond Homework 2
2 Linear Algebra Proofs about Linear Independence Calculus & Beyond Homework 2
Linear Algebra Proofs Calculus & Beyond Homework 4