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Linear algebra-Basis of a linear map

  • Thread starter manuel325
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  • #1
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Homework Statement



Let ##L: R^{2} → R^{2}## be a linear map such that ##L ≠ O## but## L^{2} = L \circ L = O.##
Show that there exists a basis {##A##, ##B##} of ##R^{2}## such that:

##L(A) = B## and ##L(B) = O.##​

The Attempt at a Solution


Here's the solution my book provides :
problem.JPG

Well I have two questions:
1.Why do they say that ##aA+bB=O##???. I mean I don't understand the solution from that point until the end (Why the solutions ##a=0## and ##b=0## are enough to prove the existence of that basis??May someone please explain??
Thanks in advance :smile:. Any help would be appreciated
 

Answers and Replies

  • #2
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The solution says IF [itex]aA+bB=0[/itex], THEN [itex]a=b=0[/itex]. That is what it means for the vectors [itex]A[/itex] and [itex]B[/itex] to be linearly independent. Vectors in a basis must be linearly independent.
 
  • #3
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The solution says IF [itex]aA+bB=0[/itex], THEN [itex]a=b=0[/itex]. That is what it means for the vectors [itex]A[/itex] and [itex]B[/itex] to be linearly independent. Vectors in a basis must be linearly independent.
Thanks but why ##O=L(aA+bB)=aL(A)##?? could you please explain what they do there, please??
 
  • #4
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Thanks but why ##O=L(aA+bB)=aL(A)##?? could you please explain what they do there, please??
[itex]L[/itex] is a linear map, which means [itex]L(aA+bB) = aL(A) + bL(B)[/itex].
 
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  • #5
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[itex]L[/itex] is a linear map, which means [itex]L(aA+bB) = aL(A) + bL(B)[/itex].
Thanks :smile:
 

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