Solving Matrix Equations with Real Scalars

[blueberryfive]

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(A^T)=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?

 

[HallsofIvy]

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(A^T)=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 "A^*_{ij}= A_{ji}" so that, replacing j with i, "A^*_{ii}= A_{ii}"? 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.

(AB)_{ij}= \sum A_{ik}B_{kj}[/itex]. If the &quot;jth&quot; column of B is all 0s, then the &quot;jth&quot; row of A is all 0.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I can&#039;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&#039;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? </div> </div> </blockquote>

 

[blueberryfive]

Thank you.

Also,

Tr(A^TA)\geq0.

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

 

[HallsofIvy]

Thank you.

Also,

Tr(A^TA)\geq0.

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

That's a sum of squares!

 

[Fredrik]

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 (rA)_{ij}=rA_{ij}. The definition of AB where both A and B are matrices is (AB)_{ij}=\sum_k A_{ik}B_{kj}. It's not hard to use these definitions to show that the equalities you mentioned are true. Start with (A(rB))_{ij}=\sum_k A_{ik}(rB)_{kj}.

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: \operatorname{Tr}A=\sum_i A_{ii},\quad (A^T)_{ij}=A_{ji}.