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Math Help - Proof for a basis of a linear transformation

  1. #1
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    Proof for a basis of a linear transformation

    Suppose that T,S:R^n \rightarrow R^n are inverses.

    If { v_1 ,v_2 ,..., v_k } is a basis for a subspace V of R^n and w_1 = T(v_1), w_2 = T(v_2),..., w_k = T(v_k), prove that { w_1, w_2,..., w_k} is a basis for T(V).

    In addition, give an example to show that this need not be true if T does not have an inverse.
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  2. #2
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    Quote Originally Posted by Runty View Post
    Suppose that T,S:R^n \rightarrow R^n are inverses.

    If { v_1 ,v_2 ,..., v_k } is a basis for a subspace V of R^n and w_1 = T(v_1), w_2 = T(v_2),..., w_k = T(v_k), prove that { w_1, w_2,..., w_k} is a basis for T(V).

    In addition, give an example to show that this need not be true if T does not have an inverse.
    First, show that \{w_1,...,w_k\} spans T(V). Hint:
    we can write any v \in V as v = \sum_{i=1}^k a_iv_i, ~ a_i \in \mathbb{R}, therefore: T(v) = T(\sum_{i=1}^k a_iv_i) = ...

    Now, show that \{w_1,...,w_k\} is linearly independent: Assume it is not, and reach a contradiction.

    Therefore it is a basis.
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    First, show that \{w_1,...,w_k\} spans T(V). Hint:
    we can write any v \in V as v = \sum_{i=1}^k a_iv_i, ~ a_i \in \mathbb{R}, therefore: T(v) = T(\sum_{i=1}^k a_iv_i) = ...

    Now, show that \{w_1,...,w_k\} is linearly independent: Assume it is not, and reach a contradiction.

    Therefore it is a basis.
    I suppose this answer could work, but I'd like to, if possible, avoid using summation notation.
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    Quote Originally Posted by Runty View Post
    I suppose this answer could work, but I'd like to, if possible, avoid using summation notation.
    v = a_1v_1 + ... + a_kv_k

    T(v) = T(a_1v_1 + ... + a_kv_k) = a_1T(v_1) + ... + a_kT(v_k)
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    v = a_1v_1 + ... + a_kv_k

    T(v) = T(a_1v_1 + ... + a_kv_k) = a_1T(v_1) + ... + a_kT(v_k)

    Since T,S : R^n --> R^n  which implies it's an nxn matrix, you can just apply the big theorem, to prove it
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  6. #6
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    Quote Originally Posted by Defunkt View Post
    v = a_1v_1 + ... + a_kv_k

    T(v) = T(a_1v_1 + ... + a_kv_k) = a_1T(v_1) + ... + a_kT(v_k)
    Okay, that solves the first part. But I still need an example to show that this isn't necessarily true, provided that T does not have an inverse.

    Honestly, I find this whole question to be pretty obscure.
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  7. #7
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    Quote Originally Posted by Runty View Post
    Okay, that solves the first part. But I still need an example to show that this isn't necessarily true, provided that T does not have an inverse.

    Honestly, I find this whole question to be pretty obscure.
    Take any transformation that is not invertible; for example, T:\mathbb{R}^2 \to \mathbb{R}^2 defined by T((x,y)) = (x,0). Since T is not invertible, Ker T \neq \{0\}. In fact, Ker T = \{(x,y) \in \mathbb{R}^2 : x = 0\}.

    Then, Ker T is spanned by (0, 1), however T(0,1) = (0,0) which is not a base.

    This will work for any transformation that is not invertible:
    Since it is not invertible, Ker T \neq \{0\} but for any w \in Ker T, Tw = 0, and therefore the image of any base of Ker T will be mapped to the zero vector.
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  8. #8
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    Quote Originally Posted by Defunkt View Post
    Take any transformation that is not invertible; for example, T:\mathbb{R}^2 \to \mathbb{R}^2 defined by T((x,y)) = (x,0). Since T is not invertible, Ker T \neq \{0\}. In fact, Ker T = \{(x,y) \in \mathbb{R}^2 : x = 0\}.

    Then, Ker T is spanned by (0, 1), however T(0,1) = (0,0) which is not a base.

    This will work for any transformation that is not invertible:
    Since it is not invertible, Ker T \neq \{0\} but for any w \in Ker T, Tw = 0, and therefore the image of any base of Ker T will be mapped to the zero vector.
    By KerT, do you mean determinant? I've never seen the term Ker used before.
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  9. #9
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    <br />
c_1\vec{w}_1+\cdots +c_k\vec{w}_k=0\rightarrow c_1T(\vec{v}_1)+\cdots +c_kT(\vec{v}_k)=0 <br />

    <br />
\rightarrow c_1T(\vec{v}_1)+\cdots  +c_kT(\vec{v}_k)=0 <br />
\rightarrow T(c_1\vec{v}_1)+\cdots  +T(c_k\vec{v}_k)=0 <br />
\rightarrow T(c_1\vec{v}_1+\cdots  +c_k\vec{v}_k)=0 <br />

    <br />
=A(c_1\vec{v}_1+\cdots  +c_k\vec{v}_k)=0<br />
\rightarrow c_1\vec{v}_1+\cdots  +c_k\vec{v}_k=A^{-1}0=0<br />

    <br />
\rightarrow c_1=\cdots =c_k=0 \rightarrow \{\vec{w}_1,\cdots,\vec{w}_k\}<br />
is linearly independent

    <br />
dim \{\vec{w}_1,\cdots,\vec{w}_k\} = k = dim(V)~ \therefore<br />
it's basis of V
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  10. #10
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    Quote Originally Posted by Runty View Post
    By KerT, do you mean determinant? I've never seen the term Ker used before.
    "ker(T)" is the "kernel" of T, also called the "null space" of T. It is the subspace of all vectors, v, such that Tv= 0.

    If T is invertible, then Tv= 0 gives T^{-1}T(v)= T^{-1}(0) or v= 0. That is, if T is invertible, its kernel (null space) consists only of the 0 vector.

    In fact, you can also prove the other way: if the 0 vector is the only vector in the kernel of T, T is invertible.

    "null space" is used exclusively in linear algebra. "kernel" of an operator is also used in group theory, ring theory, etc.
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