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Prove that

  1. Each field of characteristic zero contains a copy of the rational number field.
  2. For an $n$ by $n$ matrix $A,$ if it is not invertible, then there exists an $n$ by $n$ matrix $B$ such that $AB=0$ but $B\ne0.$

For (1), I think I have to use the fact that each subfield of the complex number field contains every rational number. Am I right? For (2), I have no idea what to do first.

Thanks.

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3 Answers 3

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For the first question you can start right at the definition and build your way up from there:

a) Any field contains $1\neq 0$

b) Since the characteristic is zero it contains $1\neq 1+1\neq 1+1+1\neq \cdots$ hence a copy of the natural numbers.

c) It contains inverses wrt addition and multiplication, thus a copy of the integers and the rationals.

For your second question: There is a vector in the kernel of $A$, make this vector a column of $B$ and fill with zeros.

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Hint on 2):

You know that since $A$ is not invertible there exists $x\neq 0$ such that $Ax = 0$. Can you use this to construct $B$?

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For 2 try showing that as $A$ is non-invertible, so is has a nonzero nullity which means there is at least one nonzero vector which is mapped to 0. Now taking $B$ to be the matrix formed by taking its columns as the vector described above, this will multiply to give the nxn zero matrix.

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