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LINEAR ALGEBRA - Describe the kernel of a linear transformation GEOMETRICALLY

  1. Dec 3, 2006 #1
    1. The problem statement, all variables and given/known data

    For two nonparallel vectors [itex]\overrightarrow{v}[/itex] and [itex]\overrightarrow{w}[/itex] in [itex]\mathbb{R}^3[/itex], consider the linear transformation


    from [itex]\mathbb{R}^3[/itex] to [itex]\mathbb{R}[/itex]. Describe the kernel of T geometrically. What is the image of T?

    2. Relevant equations

    I have no idea. Maybe the equations on how to find a kernel and image?

    3. The attempt at a solution

    I don't know where to even start this exercise! How does one "describe geometrically"?
  2. jcsd
  3. Dec 3, 2006 #2

    matt grime

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    Given v and w, when does it vanish? No equations, nothing like that, just a simple statement of what it means when det vanishes. If you just use words, you'll be describing it geometrically. For instance, fix a y, and take the linear map

    L_y : x--> x /\y

    which takes x and sends it to the vector product of x and y, then the kernel is the set of x that are parallel to y (or the line spanned by y). That is a geometrical description of the kernel.

    The point is that you could let x=(x_1,x_2,x_3) and v=(v_1,v_2,v_3) etc and write down an equation f(x_1,x_2,x_3)=0 with coefficients the v_i, w_i which parametrizes the kernel, but it would be incredibly unhelpful when there is a far simpler description.
    Last edited: Dec 3, 2006
  4. Dec 3, 2006 #3


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    What is the definition of a kernel? How does that apply in your case? What is the geometrical representation of that?

    Edit: too late, again.
  5. Dec 3, 2006 #4
    Thank you for trying to explain this concept to me, however, I still do not understand!

    Can you explain the formula [tex]L_y: x\,->\,x\,\bigwedge\,y[/tex]?

    Is that also expressed as the "dot product"? The left of the equation reads "Linear transformation of y", right? [tex]L_y:\,x\,->\,\overrightarrow{x}\,\cdot\,\overrightarrow{y}[/tex]

    Maybe if you just explain it in very precise terms that a "lay-person" would understand? I always have trouble with these dang kernels!

    I know that a kernel is the functions or vectors that cause the transformation to be equal to zero (at least, I hope it is).
    Last edited: Dec 3, 2006
  6. Dec 4, 2006 #5


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    More precisely, the kernelis the set of vectors that 'cause' the transformation to be equal to zero.
  7. Dec 4, 2006 #6


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    The formula [itex]L_y : x \mapsto x \wedge y[/itex] says that Ly is a function that maps x to [itex]x \wedge y[/itex]. The wedge product is a generalization of the cross product, not the dot product. The kernel of Ly is the set of vectors {x | Ly(x) = 0}, which is exactly the same thing as [itex]\{ x\, |\, x \wedge y = 0\}[/itex]. Like I said, the wedge product is just a generalization of the cross product, so it's probably easier for you to consider instead the function Cy defined by [itex]C_y : x \mapsto x \times y[/itex]. Then:

    [tex]\mbox{Ker}(C_y) = \{ x\, |\, C_y(x) = 0\} = \{ x\, |\, x \times y = 0\}[/tex]

    This set is obviously just the set of vectors perpendicular to y, because [itex]x \times y = 0[/itex] iff x and y are perpendicular. You know that, right?
    Last edited: Dec 4, 2006
  8. Dec 4, 2006 #7


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    This might help: For three vectors, [itex]\vec{x},\vec{u},\vec{v}[/itex],
    also called the "triple" product, is [itex]\vec{x}\cdot\left(\vec{u} X \vec{v}\right)[/itex].
    Of course, the dot product of two vectors is 0 if and only if they are perpendicular, and the cross product of two vectors is perpendicular to both of them. What does that tell you about the geometric relationship between [itex]\vec{x}[/itex] and [itex]\vec{u},\vec{v}[/itex] if this is equal to 0?
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