How Does the Density of Rational Numbers Prove Matrix Linearity?

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SUMMARY

The discussion centers on proving that a continuous transformation F on Rn is linear if it satisfies the condition F(X+Y) = F(X) + F(Y) for all X and Y in Rn. The key hint provided is that rational numbers are dense in the reals, which is crucial for establishing linearity. The participants clarify that a transformation A is linear if it also satisfies A(aX) = aA(X) for any scalar a, alongside the additive property. This reinforces the necessity of both properties for linearity in transformations.

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Homework Statement


Show that if F is continuous on Rn and F(X+Y) = F(X) + F(Y) for all X and Y in Rn, then A is linear. Hint: Rational numbers are dense in the reals.

Homework Equations


A transformation A is linear iff A(X) = (a matrix)
[ a11x1+...+a1nxn ]
[... ... ...]
[ am1x1+...+amnxn ]

The Attempt at a Solution



F(X) = A(X) is continuous and F(X+Y) = A(X+Y) = F(X) + F(Y) = A(X)+A(Y)

I feel like this basically proves itself...since a tranformation A is linear if A(X+Y) = A(X) + A(Y)... I don't understand where the denseness of rational numbers comes in?

Any help is greatly appreciated! :)
 
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hi bobbarker! :smile:
bobbarker said:
… a tranformation A is linear if A(X+Y) = A(X) + A(Y)

nooo :redface:

a transformation A is linear if A(aX) = aA(X) for any scalar a :wink:
 

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