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Math Help - Linear Transformation question

  1. #1
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    Linear Transformation question

    According to my book,

    There exists a linear transformation T:R^3 \rightarrow R^3 that has (1,-1,2),(2,1,-1),(1,2,-3) \in KerT and (1,-1,2) \in ImT

    They explain that (1,-1,2),(2,1,-1),(1,2,-3) are dependent vectors.

    My question is: If you are dealing with the standard basis (or any set of 3 independent vectors), does there exist a linear transformation where T(1,0,0)=T(0,1,0)=T(0,0,1) =(0,0,0,) and have (1,0,0) \in ImT
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  2. #2
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    ...does there exist a linear transformation where...
    I would say no. Not if T:\mathbb{R}^{3}\to\mathbb{R}^{3}. The vectors you have exhibited there are a basis. So, take any vector \mathbf{x} in \mathbb{R}^{3}. Then \mathbf{x}=a\hat{i}+b\hat{j}+c\hat{k}, where the hatted vectors are the unit vectors you mentioned. Then

    T\mathbf{x}=T(a\hat{i}+b\hat{j}+c\hat{k})=aT\hat{i  }+bT\hat{j}+cT\hat{k}=0. Therefore, (1,0,0)\not\in\text{Image}(T).

    However, this is not the situation you're dealing with in the first part of your post. In the first part, the book says that the three vectors there are dependent. A basis is, by definition, independent. Therefore, all bets are off.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    I would say no. Not if T:\mathbb{R}^{3}\to\mathbb{R}^{3}. The vectors you have exhibited there are a basis. So, take any vector \mathbf{x} in \mathbb{R}^{3}. Then \mathbf{x}=a\hat{i}+b\hat{j}+c\hat{k}, where the hatted vectors are the unit vectors you mentioned. Then

    T\mathbf{x}=T(a\hat{i}+b\hat{j}+c\hat{k})=aT\hat{i  }+bT\hat{j}+cT\hat{k}=0. Therefore, (1,0,0)\not\in\text{Image}(T).
    So that means that ImageT= (0,0,0) = KernelT ?

    However, this is not the situation you're dealing with in the first part of your post. In the first part, the book says that the three vectors there are dependent. A basis is, by definition, independent. Therefore, all bets are off.
    I know! That's why I was asking. (:
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  4. #4
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    So that means that ImageT= (0,0,0) = KernelT ?
    No, that's a mistake in categories. Let's say I had a linear transformation T:V\to W. Then \text{Image}(T)\subseteq W, but \text{Ker}(T)\subseteq V.

    In words, the image is the set of all vectors that T can "reach" by acting on vectors in its domain. The kernel is the set of all vectors in the domain that get killed by T (sent to the zero vector).

    I know! That's why I was asking. (:
    Been doing some computations. Let's start with our three vectors in the kernel of T. According to my computations, this implies that T looks like this:

    T=\begin{bmatrix}-a/3 &5a/3 &a\\ -b/3 &5b/3 &b\\ -c/3 &5c/3 &c\end{bmatrix}.

    Setting T\mathbf{x}=(1,-1,2)^{T} yields the augmented matrix

    \left[\begin{matrix}-a/3 &5a/3 &a\\ -b/3 &5b/3 &b\\ -c/3 &5c/3 &c\end{matrix}\left|\begin{matrix}1\\-1\\2\end{matrix}\right].

    The three rows of T are obviously linearly dependent. So you can do an elementary row operation that kills the second and third lines. You get

    \left[\begin{matrix}-a/3 &5a/3 &a\\ 0 &0 &0\\ 0 &0 &0\end{matrix}\left|\begin{matrix}1\\-(b/a)-1\\-(c/a)+2\end{matrix}\right].

    Therefore, in order for this system to have a solution at all, you must have -b/a=1 and c/a=2. Thus, b=-a and c=2a. Now then. Your remaining equation is

    -(a/3)x+(5a/3)y+az=1. This will be a two-parameter solution. Set y=s and z=t. Then -(a/3)x=1-(5a/3)s-at, from which it follows that x=-3/a+5s+3t. Assuming a\not=0, you can definitely find values of s and t to make this work. Hence, any matrix of the form

    T=\begin{bmatrix}-a/3 &5a/3 &a\\ a/3 &-5a/3 &-a\\ -2a/3 &10a/3 &2a\end{bmatrix}=a\begin{bmatrix}-1/3 &5/3 &1\\ 1/3 &-5/3 &-1\\ -2/3 &10/3 &2\end{bmatrix}

    will work. You can check that it still kills all the vectors in the kernel. Let's say you set a=s=t=1. Then the vector

    \mathbf{x}=\begin{bmatrix}-3+5+3\\1\\1\end{bmatrix}

    gets sent to (1,-1,2)^{T}.

    So the book is correct.
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