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 August 29th, 2011, 07:23 PM #1 Newbie   Joined: Aug 2009 Posts: 29 Thanks: 0 Invertible matrix Let A be an nxn matrix. Prove that if A is not invertible, then there exists an nxn matrix B such that $AB=0$ but $B \neq 0$
August 29th, 2011, 08:05 PM   #2
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Re: Invertible matrix

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 Originally Posted by problem Let A be an nxn matrix. Prove that if A is not invertible, then there exists an nxn matrix B such that $AB=0$ but $B \neq 0$
A not invertible means (among other things) that the determinant of A is zero.

If A is the zero matrix, then we are done.

Otherwise, suppose that A is not the zero matrix, yet det(A) = 0.

Then A has at least one non-zero entry, say A_ij = c, for some value of c, and some 0 <= i,j <= n

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 August 29th, 2011, 08:24 PM #3 Newbie   Joined: Aug 2009 Posts: 29 Thanks: 0 Re: Invertible matrix The determinant of a matrix will be introduced after the section where this question is from. Aside from determinant, can we use any property of invertible matrix? Thanks for the help.
 August 31st, 2011, 05:30 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Invertible matrix If an n by n maA is not "invertible", then it is not "one-to-one". That is, there exist two distinct vectors, X and Y, such that AX= AY. Of course, AX- AY= A(X- Y)= 0.

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