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Math Help - linear transformations

  1. #1
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    linear transformations

    consider the linear transformation T:R4--> R4 defined by
    [w] [w+z]
    T [x] = [y]
    [y] [x-y]
    [z] [w]

    for each i have to justify the answer
    a. is T injective?

    b. is T surjective?

    c.is T invertible?

    can anybody please help?!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by linearalgebradunce View Post
    consider the linear transformation T:R4--> R4 defined by
    [w] [w+z]
    T [x] = [y]
    [y] [x-y]
    [z] [w]

    for each i have to justify the answer
    a. is T injective?
    T is injective if \ker T = \{\vec{0}\}. Is this true ?
    b. is T surjective?
    T is surjective if, given any vector v\in\mathbb{R}^4, one can find u\in\mathbb{R}^4 such that T(u)=v. In other words, given v=\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\in\mathbb  {R}^4, can you find at least a 4-tuple (w,x,y,z) such that \begin{cases} w+z=a\\y=b\\x-y=c\\w=d\end{cases} ?
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  3. #3
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    Ok I understand the definitions but am still struggling to decide if T is injective or not, for the kernel(T) obviously we know this equals the null space(T) so if I find one vector in R4 such that Tv=the zero vector does this mean it is injective.
    eg. if I choose the vector
    [1]
    [0]
    [0]
    [-1]
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by linearalgebradunce View Post
    Ok I understand the definitions but am still struggling to decide if T is injective or not, for the kernel(T) obviously we know this equals the null space(T) so if I find one vector in R4 such that Tv=the zero vector does this mean it is injective.
    eg. if I choose the vector
    [1]
    [0]
    [0]
    [-1]
    To show that T is injective you have to show that the only vector v such that Tv=0 is v=0; that's not what you're doing. Anyway, this idea is often used to show that a linear application is not injective. For if you find a non-zero vector v such that Tv=0 then \ker T contains both 0 and v hence \ker T\neq \{0\} and T can't be injective.

    EDIT: by the way, for v=\begin{bmatrix}1\\0\\0\\-1\end{bmatrix},\, Tv\neq 0
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  5. #5
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    Oh yeah I see that that is incorrect. So T(0) = 0 so T is injective!? Is that what you're saying?
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by linearalgebradunce View Post
    Oh yeah I see that that is incorrect. So T(0) = 0 so T is injective!? Is that what you're saying?
    No! T(0)=0 is true for any linear application T, even if T is not injective (this result comes from T(0)=T(x-x)=T(x)+T(-x)=T(x)-T(x)=0). A linear application is injective if T(v)=0 \implies v=0. You are saying that a linear application is injective if v=0\implies T(v)=0. These two sentences are very different and only the former is correct.

    Now, how can one show that T(v)=0\implies v=0? Well, the only thing we have to do is solve T(v)=0 for v. Let v=\begin{bmatrix}w\\x\\y\\z \end{bmatrix}.
    T(v)=0\implies \begin{cases} w+z=0\\y=0\\x-y=0\\w=0\end{cases}\implies\begin{cases} z=-w=0\\y=0\\x=y=0\\w=0\end{cases}\implies v=0
    hence T is injective.
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