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  2. How to prove $(AB)^T=B^T A^T$ - Mathematics Stack Exchange

    math.stackexchange.com/questions/1440305

    Given an m × n -matrix A and an n × p -matrix B. Prove that (AB)T = BTAT. Here is my attempt: Write the matrices A and B as A = [aij] and B = [bij], meaning that their (i, j) -th entries are aij and bij, respectively. Let C = AB = [cij], where cij = ∑nk = 1aikbkj, the standard multiplication definition. We want (AB)T = CT = [cji]. That is the element in position j, i is ∑nk = 1aikbkj ...

  3. How to prove that - Mathematics Stack Exchange

    math.stackexchange.com/questions/3077312

    Then you get in total: EDIT: you could've just as easily left the expression without swapping the two and arrived at the expression: (AB)t = AtBt (A B) t = A t B t, right? No, the textbooks wants to show that (AB)t =BtAt (A B) t = B t A t. Because generally speaking (AB)t ≠AtBt (A B) t ≠ A t B t. Example:

  4. $AB=BA$ implies $AB^T=B^TA$ when $A$ is normal

    math.stackexchange.com/questions/675239

    where A and B are n × n real matrices, and A is a normal matrix, i.e., AAT = ATA - it is true for complex matrices as well, with AT replaced by A ∗. There is a non-elementary proof of this using the exponential of a matrix and properties of entire functions. Update. In the first version of the question, both A and B were supposed to be normal, but as Shlomi correctly pointed out, this is ...

  5. What is the SVD of - Mathematics Stack Exchange

    math.stackexchange.com/questions/4283968/what-is-the-svd-of-abt-bat

    What is the SVD of A = abT + baT when a and b are column vectors? It is also given that aTb = 0. I've tried finding AAT and ATA to find the e-values and e-vectors for U and Σ to no avail. This method worked for finding the SVD of A = abT, but no luck with this case. Any pointers to the right direction would be greatly appreciated! Edit: I saw this post, but I didn't quite get how they got the ...

  6. Delilah Allen, b. abt 1788 - Genealogy.com

    www.genealogy.com/forum/surnames/topics/allen/3195

    Delilah Allen, b. abt 1788 By genealogy.com user February 10, 1999 at 11:36:53 Looking for parents of Delilah Allen, b. abt 1788 in prob SC.Spouse: Jonathan Hall, b abt 1780 in Halifax Co., VA Thanks

  7. Prove $tr_B(vec(A)vec(B)^T)=AB^T$ - Mathematics Stack Exchange

    math.stackexchange.com/questions/4616342/prove-tr-bvecavecbt-abt

    This question shows research effort; it is useful and clear

  8. linear algebra - Show that $ (AB)^T = A^T B^T$ if and only if $A$...

    math.stackexchange.com/questions/2459681/show-that-abt-at-bt-if-and-only-if-a...

    Not the answer you're looking for? Browse other questions tagged linear-algebra matrices .

  9. James McDonald born abt 1786 GA - Genealogy.com

    www.genealogy.com/forum/surnames/topics/mcdonald/2061

    James McDonald born abt 1786 GA By Gava Graham January 22, 2000 at 12:08:22 Searching for James (possible middle Initial M.) McDonald born abt. 1786 in (probably) Franklin / Jackson Co. GA. died 1862 Randolph Co. AR Married three times. #1 unknown children James 1810, William 1814, Melissa 1825, # 2 - Sarah Smith married 1837 Randolph Co. AR.children Martha 1838, Sousan 1844,# 3 Sarah Lawrence ...

  10. Re: John Cave (b. abt. 1757 - d. abt. 1807) - Genealogy.com

    www.genealogy.com/forum/surnames/topics/cave/281

    In reply to: John Cave (b. abt. 1757 - d. abt. 1807) Lynna Kay Shuffield 10/13/99. John Cave was married to Sarah Wilson Brown, not Sarah Brown Wilson and according to the Brown genealogy published in the Manning papers (by Edna A. & William Harris Manning Jr ... in a history of Barnwell County SC) she was only married once to John Cave.John ...

  11. How to solve for matrix D in - Mathematics Stack Exchange

    math.stackexchange.com/questions/298192

    For example ABCTDBATC = ABT A B C T D B A T C = A B T becomes BCTDBATC =BT B C T D B A T C = B T, cause according to your assumption all matrices are invertible and of course they are squares .