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Entries in every row add to zero (nullspace/determinant question!)

  1. Apr 6, 2008 #1
    Edit: My friend just explained it to me, duh! If the row adds to zero then (row)dot(1,1,...,1) = 0! And it is in the nullspace. Thanks anyway! :)

    Hi all.

    I've been learning about determinants and I was reading one of the sample problems in my textbook. I thought I understood the chapter, but I don't understand their answer at all. (Note this is not homework! Just trying to learn.)

    Their question:
    "If the entries in every row of A add to zero, solve Ax=0 to prove detA=0. If those entries add to one, show that det(A-I)=0. Does this mean detA=1?"

    Their answer:
    "If the entries in every row add to zero, then (1, 1, . . . , 1) is in the nullspace: singular A has det = 0. (The columns add to the zero column so they are linearly dependent.) If every row adds to one, then rows of A − I add to zero (not necessarily detA = 1)."

    I guess I am confused as to why if the entries in each row add to zero, then (1,1,...,1) is in the nullspace? Any help would be much appreciated! Thanks!
    Last edited: Apr 6, 2008
  2. jcsd
  3. Apr 6, 2008 #2


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    What is
    [tex]\left(\begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i\end{array}\right)\left(\begin{array}{c}1 \\ 1 \\ 1\end{array}\right)[/tex]?

    What if a+ b+ c= 0, d+ e+ f= 0, g+ h+ i= 0?
  4. Apr 9, 2008 #3
    Thank you HallsofIvy, all so simple now. Just had a little trouble reading it!
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