Linear Algebra - proof of transformation

AI Thread Summary
To prove that T(0) = 0 for a linear transformation T: V -> W, it is essential to recognize that T can be expressed in terms of a transformation matrix A, where T(v) = Av. The transformation must adhere to the properties of linearity, specifically that T(u + v) = T(u) + T(v) and T(au) = aT(u). By applying these properties, substituting v with the zero vector or a with zero demonstrates that T(0) must equal 0. Thus, the proof confirms that linear transformations map the zero vector to the zero vector.
NewtonianAlch
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Homework Statement


Suppose T: V -> W is linear. Prove that T(0) = 0


The Attempt at a Solution



T(v) = Av
T(0) = A(0) = 0

Is that right?
 
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What is A? Why can you write T(v)=Av? What is your definition of linear?
 
A is the transformation matrix. For a transformation T, we need some kind of "transformer" and A is the transformation matrix I used. v is a random vector that is being transformed, in this case it's just the zero vector.
 
Essentially you are using what you are asked to prove- you can write a linear transformation as a matrix because, among other things, T(0)= 0.

A linear transformation, T, from vector space U to vector space V, must satisfy
1) T(u+ v)= T(u)+ T(v) with u and v vectors in U.
2) T(au)= aT(u) with a a member of the underlying field and u a vector in U.

Use (1) with v= 0 or (2) with a= 0.
 

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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