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Math Help - [SOLVED] Basis of the nullspace and image of a function

  1. #1
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    [SOLVED] Basis of the nullspace and image of a function

    Let f: M_{22} (\mathbb{R}) \rightarrow \mathbb{R}[T] defined by f \left[\begin{matrix} a & b\\ c & d \end{matrix}\right] = (a+b) + (a + b)T + (a + b + c + d)T^2 for all \left[\begin{matrix} a & b\\ c & d \end{matrix}\right] \in M_{22}.

    1. Prove that f is linear.
    2. Calculate the basis of the null-space of f and of the image of f.

    How do I even prove that f is linear?
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  2. #2
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    Quote Originally Posted by bmp05 View Post
    Let f: M_{22} (\mathbb{R}) \rightarrow \mathbb{R}[T] defined by f \left[\begin{matrix} a & b\\ c & d \end{matrix}\right] = (a+b) + (a + b)T + (a + b + c + d)T^2 for all \left[\begin{matrix} a & b\\ c & d \end{matrix}\right] \in M_{22}.

    1. Prove that f is linear.
    2. Calculate the basis of the null-space of f and of the image of f.

    How do I even prove that f is linear?
    for 1) you need to show that for any matrices A,B \in M_2(\mathbb{R}), \ \alpha \in \mathbb{R}: \ f(\alpha A + B)=\alpha f(A) + f(B). this is straightforward.

    for 2): A=\begin{bmatrix} a & b\\ c & d \end{bmatrix} is in the null-space of f if and only if a+b=a+b+c+d=0, which gives us: b=-a, \ d=-c. thus:

    A=\begin{bmatrix} a & -a\\ c & -c \end{bmatrix}=a \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}+ c \begin{bmatrix} 0 & 0 \\ 1 & -1 \end{bmatrix}. so a basis for the null-space of f is ... ?
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  3. #3
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    Thanks NonCommAlg!

    So to prove that f is linear you have to show that the elements in f are scalar multiples of each other? As in y = mx

    And the null space are derived by taking the coefficients of f so that: (a+b) + (a + b)T + (a + b + c + d)T^2 = 0

    Wow. That really did help. Sometimes, I feel like stunned rabbit when confronted with new terminology/ presentation of a question . Thank you very very much!

    Can you recommend a good text book, for linear algebra?
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  4. #4
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    Quote Originally Posted by bmp05 View Post
    Thanks NonCommAlg!

    So to prove that f is linear you have to show that the elements in f are scalar multiples of each other? As in y = mx
    that's not what i said! i said choose any two matrices in M_2(\mathbb{R}): \ A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}, \ B=\begin{bmatrix} a' & b'\\ c' & d' \end{bmatrix}, and a scalar \alpha \in \mathbb{R}. then:

    \alpha A + B=\begin{bmatrix} \alpha a + a' & \alpha b + b' \\ \alpha c + c'& \alpha d + d'\end{bmatrix}. then from the definition of f we have: f(A)=a+b+(a+b)T + (a+b+c+d)T^2 and

    f(B)=a'+b'+(a'+b')T + (a'+b'+c'+d')T^2. now you have everything ready to find both \alpha f(A) + f(B) and f(\alpha A + B)

    and see that they're equal, which exactly means that f is linear.
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  5. #5
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    Ok- well, what does linearity mean, exactly? Is there anything special about linearity? Are there vector-spaces which are for example not linear?
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