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Linear Proofs

by blueberryfive
Tags: linear, matrix
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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?
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HallsofIvy
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Sep5-11, 08:17 AM
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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
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Sep5-11, 08:28 AM
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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
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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].


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