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Thread: 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?
    $\displaystyle T$ is injective if $\displaystyle \ker T = \{\vec{0}\}$. Is this true ?
    b. is T surjective?
    $\displaystyle T$ is surjective if, given any vector $\displaystyle v\in\mathbb{R}^4$, one can find $\displaystyle u\in\mathbb{R}^4$ such that $\displaystyle T(u)=v$. In other words, given $\displaystyle v=\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\in\mathbb {R}^4$, can you find at least a 4-tuple $\displaystyle (w,x,y,z)$ such that $\displaystyle \begin{cases} w+z=a\\y=b\\x-y=c\\w=d\end{cases}$ ?
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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 $\displaystyle T$ is injective you have to show that the only vector $\displaystyle v$ such that $\displaystyle Tv=0$ is $\displaystyle v=0$; 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 $\displaystyle v$ such that $\displaystyle Tv=0$ then $\displaystyle \ker T$ contains both 0 and $\displaystyle v$ hence $\displaystyle \ker T\neq \{0\}$ and $\displaystyle T$ can't be injective.

    EDIT: by the way, for $\displaystyle v=\begin{bmatrix}1\\0\\0\\-1\end{bmatrix},\, Tv\neq 0$
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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! $\displaystyle T(0)=0$ is true for any linear application $\displaystyle T$, even if $\displaystyle T$ is not injective (this result comes from $\displaystyle T(0)=T(x-x)=T(x)+T(-x)=T(x)-T(x)=0$). A linear application is injective if $\displaystyle T(v)=0 \implies v=0$. You are saying that a linear application is injective if $\displaystyle v=0\implies T(v)=0$. These two sentences are very different and only the former is correct.

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