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Thread: Linear transformation proof problem

  1. #1
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    Linear transformation proof problem

    Let $\displaystyle T:R^n\rightarrow R^m$ be a linear transformation.
    If $\displaystyle v_1,v_2,...,v_k$ are vectors in $\displaystyle R^n$ such that the set $\displaystyle \{T(v_1),T(v_2),...,T(v_k)\}$ is linearly independent in $\displaystyle R^m$, prove that the set $\displaystyle \{v_1,v_2,...,v_k\}$ is linearly independent.
    If $\displaystyle n=m$ and, for every linearly independent set $\displaystyle \{w_1,w_2,...,w_k\}$ in $\displaystyle R^n$, the set $\displaystyle \{T(w_1),T(w_2),...,T(w_k)\}$ is always linearly independent, prove that there is an inverse transformation $\displaystyle S:R^n\rightarrow R^n$. (i.e. $\displaystyle S\circ T, T\circ S$ are both the identity transformation).

    This one is pretty tricky, needless to say.
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  2. #2
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    Suppose the set $\displaystyle \{v_i\}$ were not linearly independent so there are $\displaystyle a^i$ not all zero s.t. $\displaystyle a^iv_i=0$. Apply T to this equation to reach a contradiction.

    For the second, try showing that $\displaystyle \mathrm{ker}\; T = 0$.
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  3. #3
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    Quote Originally Posted by maddas View Post
    Suppose the set $\displaystyle \{v_i\}$ were not linearly independent so there are $\displaystyle a^i$ not all zero s.t. $\displaystyle a^iv_i=0$. Apply T to this equation to reach a contradiction.

    For the second, try showing that $\displaystyle \mathrm{ker}\; T = 0$.
    We don't have the Kelvin function available, so that wouldn't work.

    Also, though I could probably write out all of the first part, I'm tied up with other Algebra-related material, so do you think you could show a bit more detail?
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  4. #4
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    Kelvin function? I meant the kernel lol.

    1) If $\displaystyle a^iv_i=0$ then $\displaystyle T(a^iv_i)=T(0)$ meaning $\displaystyle a^iT(v_i)=0$. But the $\displaystyle T(v_i)$ are linearly independent so the $\displaystyle a^i $ are identically zero.

    2) Let $\displaystyle v_i$ be a basis for the vector space. Let $\displaystyle x=x^iv_i$ be a vector in the kernel of T. Then $\displaystyle T(x)=x^iT(v_i)=0$. But because $\displaystyle T(v_i)$ are linearly independent, this implies that the $\displaystyle x^i$ are identically zero so x=0. So the kernel of T is zero. It follows that T is injective. Now show it is surjective and use these to deduce it is invertible.

    P.S. I use the Einstein summation convention. I'm going to get in trouble if I don't get back to work now lol. Good luck!
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  5. #5
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    Quote Originally Posted by maddas View Post
    Kelvin function? I meant the kernel lol.

    1) If $\displaystyle a^iv_i=0$ then $\displaystyle T(a^iv_i)=T(0)$ meaning $\displaystyle a^iT(v_i)=0$. But the $\displaystyle T(v_i)$ are linearly independent so the $\displaystyle a^i $ are identically zero.

    2) Let $\displaystyle v_i$ be a basis for the vector space. Let $\displaystyle x=x^iv_i$ be a vector in the kernel of T. Then $\displaystyle T(x)=x^iT(v_i)=0$. But because $\displaystyle T(v_i)$ are linearly independent, this implies that the $\displaystyle x^i$ are identically zero so x=0. So the kernel of T is zero. It follows that T is injective. Now show it is surjective and use these to deduce it is invertible.

    P.S. I use the Einstein summation convention. I'm going to get in trouble if I don't get back to work now lol. Good luck!
    Thanks for your help on the first half, but as for the second half, we don't have kernels available. I got the name mixed up last post.
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  6. #6
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    There is no reason not to use kernels. I'm not sure what you mean by them not being available..... The kernel of a map is the set of vectors it sends to zero. There is a beautiful relationship between the kernel and the image of a map called the rank-nullity theorem.

    ...but in any case, this particular proof doesn't use them in any essential way so I'll reformulate it without them:

    Lemma. If T is linear and $\displaystyle Tv=0\implies v=0$ then T is injective.

    Proof. For suppose not. Then there are distinct x and y s.t. Tx=Ty. But then T(x-y)=0, which implies that x-y=0, or x=y, contrary to our assumption.

    Then to prove your 2) Take an $\displaystyle x=x^iv_i$ and suppose that Tx=0. Then $\displaystyle Tx=x^iTv_i=0$ so by linear independence of the $\displaystyle Tv_i$, the $\displaystyle x^i$ are all zero, meaning x=0. Therefore if Tx=0, then x=0, and we may apply the lemma to see that T is injective. Surjectiveness is easy and so T is invertible.
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