LINEAR ALGEBRA  Describe the kernel of a linear transformation GEOMETRICALLYby VinnyCee Tags: algebra, geometrically, kernel, linear, transformation 

#1
Dec306, 04:54 AM

P: 492

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 [tex]T\left(\overrightarrow{x}\right)\,=\,det\left[\overrightarrow{x}\,\,\overrightarrow{v}\,\,\overrightarrow{w}\right][/tex] 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
Dec306, 05:19 AM

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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. 



#3
Dec306, 05:19 AM

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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. 



#4
Dec306, 08:22 PM

P: 492

LINEAR ALGEBRA  Describe the kernel of a linear transformation GEOMETRICALLY
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 "layperson" 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). 



#5
Dec406, 02:24 AM

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#6
Dec406, 02:53 AM

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The formula [itex]L_y : x \mapsto x \wedge y[/itex] says that L_{y} 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 L_{y} is the set of vectors {x  L_{y}(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 C_{y} 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? 



#7
Dec406, 06:32 AM

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This might help: For three vectors, [itex]\vec{x},\vec{u},\vec{v}[/itex],
[tex]det\left[\overrightarrow{x}\,\,\overrightarrow{v}\,\,\overrightarrow{w}\right][/tex] 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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