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LINEAR ALGEBRA: Given a linear transformaiton from R^2 to R^2 & parallelogram

  1. Oct 1, 2006 #1
    LINEAR ALGEBRA: Given a linear transformation from R^2 to R^2 & parallelogram...

    Let T be an invertible linear transformation from [itex]R^2[/itex] to [itex]R^2[/itex]. Let P be a parallelogram in [itex]R^2[/itex] with one vertex at the origin. Is the image of P a parallelogram as well? Explain. Draw a sketch of the image.

    My answer:

    Yes. I don't have an explanation however, because I don't understand what an "image" is or how to do a linear transformation from R2 to R2. They are the same spaces, so the parallelogram would not change dimensions, right? A drawing? I would just guess an identical parallelgram.
    Last edited: Oct 1, 2006
  2. jcsd
  3. Oct 1, 2006 #2
    Here is the parallelogram supplied with the problem. The points are arbitrary, so I just made them x = 0, 1, 3, 2 and y = 0, 2, 3, 1:

    Last edited: Oct 1, 2006
  4. Oct 1, 2006 #3
    I am going to use the following scaling transformation on the parallelogram:

    [tex]\mathbf{T}\,=\,\left[ \begin{array}{cc}
    2 & 0 \\
    0 & 2 \\
    \end{array} \right][/tex]

    Here is a picture of the transformation:


    Would this picture count as a "sketch of the image"? What are they implying by "image" anyways? There is an Im(v) function in linear algebra as well, right? This all seems very confusing, how would I start a proof of the line segment thing?

    One other question: projections are NOT invertible, correct?
    Last edited: Oct 1, 2006
  5. Oct 1, 2006 #4


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    So I'll guess you know what a linear transformation is. Usually this takes a vector to another vector. Now, a vector is an abstract thing, but is usually thought of geometrically as a line segment from the origin to a point in R^2, with a little arrow at the end. It's usually written (a,b), or maybe [itex]a \hat x + b \hat y [/itex]. But it can just as easily be thought of as the point at (a,b), and we can think of a linear transformation as moving points around.

    The image of a certain set of points (like those in a parellelogram) is just the set of points that the linear transformation maps them to. So for example, if the transformation was a rotation by 90 degrees, each point would be mapped to another point via a 90 degree rotation about the origin, and so the parallelogram in the picture would rotate the same way.

    An invertible transformation is one which is a bijection, ie, one-to-one and onto, and they can be represented by invertible matrices. These are rotations, stretches, and skews. To show the image of the parallelogram is still a parallelogram, just show that if two line segments are parallel in R^2, then their images under a linear transformation are also parallel.
  6. Oct 1, 2006 #5
    I've read a little about 'bijection' from a google search, but I still do not understand the concept of "onto". Can you explain please?

    Rotations, stretches and skews are the ONLY types of transformations that are invertible?

    How would one go about proving that two lines are parallel? Maybe set arbitrary points on two vectors to be compared and see if a line through both points is perpendicular to the two vectors being compared? Sounds all good, but how to start?
    Last edited: Oct 1, 2006
  7. Oct 1, 2006 #6


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    The easiest, though maybe not the nicest, way is to look at the slopes of the lines. It's easy to work out what happens to points on the line y=mx+b, and it turns out that b won't affect the final slope, so all lines with the same m (slope) are taken to lines with the same slope.

    The only thing you have to worry about as far as invertible/non-invertible is that a non-invertible transformation will collapse your parallelogram to a line segment or a point. This is because it collapses all of R^2 to a line or point. In fact, the invertible transformations are exactly those which map R^2 to all of R^2 (which is what "onto" means, ie, the image of the domain is all of the range).
  8. Oct 1, 2006 #7
    How do I start out proving this using the slope of the two lines?

    One can prove that a "point" is parallel if it is a scalar multipule of the point in question, right?

    For example, comparing each of the parallelograms 2nd points above [i.e. - (1, 2) & (2, 4)]:

    [tex]\left[ \begin{array}{c} 1 \\ 2 \\\end{array} \right][/tex]

    [tex]\left[ \begin{array}{c} 2 \\ 4 \\\end{array} \right][/tex]

    [tex]k\,\left[ \begin{array}{c} 1 \\ 2 \\\end{array} \right]\,=\,\left[ \begin{array}{c} 2 \\ 4 \\\end{array} \right][/tex]


    It also says in the book that the im(A) (image of A) is all real numbers if A is invertible. Does this mean that the "image" of the parallelogram is all of the real numbers in [itex]R^2[/itex], a plane?
    Last edited: Oct 1, 2006
  9. Oct 1, 2006 #8


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    Actually, there's a really easy way to do this I overlooked. There's 4 vertices on the parallelogram. The one at the origin will remain at the origin (easy to show). The two that are adjacent to the origin will go where they go. Then for the new shape to be a parallelogram, where must the final point go? (thinking of the points as vectors, then what must be the vector corresponding to the final point in terms of those corresponding to the other points?) Does a linear transformation necessarily put it in there?

    The only other things you need to worry about are that the two adjacent points aren't put on the same line (this has to do with the transformation being invertible) and to verify that a line segment between points a and b is mapped by a linear transformation T to the line segment between T(a) and T(b).
    Last edited: Oct 1, 2006
  10. Oct 1, 2006 #9
    I still do not know where to start out in terms of "proving" that a parallelogram, including the origin as one of its veritces, that under goes an invertible linear transformation is still a parallelogram. The answer in the b.o.b. is "Yes".

    Is a "skew" the same thing as a "shear" regarding linear transformations?
    Last edited: Oct 1, 2006
  11. Oct 1, 2006 #10


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    What didn't you understand about my last post? And yes, by skew I meant shear.
  12. Oct 1, 2006 #11
    I don't know how to put it into writing that the last point has to undergo the same linear transformation as the two adjacent points to the origin. Maybe like this?:


    [tex]A_{rot}\,=\,\left[\begin{array}{cc} a & -b \\ b & a \\\end{array}\right][/tex]

    [tex]\overrightarrow{x}\,=\,\left[\begin{array}{c} x_n \\ x_m \\\end{array}\right][/tex]

    [tex]T\left(\overrightarrow{x}\right)\,=\,A_{rot}\,\overrightarrow{x}\,=\,\left[\begin{array}{cc} a & -b \\ b & a \\\end{array}\right]\,\left[\begin{array}{c} x_n \\ x_m \\\end{array}\right][/tex]

    [tex]T\left(\overrightarrow{x}\right)\,=\,\left[\begin{array}{c} a\,x_n\,-\,b\,x_m \\ b\,x_n\,+\,a\,x_m \\\end{array}\right][/tex]


    That only proves (for one case) that the transformation is a subspace though.
    Last edited: Oct 1, 2006
  13. Oct 1, 2006 #12


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    What does that mean? Of course they're all undergoing the same transformation by assumption.
  14. Oct 1, 2006 #13
    What I don't understand how to put into writing is "verify that a line segment between points a and b is mapped by a linear transformation T to the line segment between T(a) and T(b).".

    How do I produce such a proof?
  15. Oct 1, 2006 #14


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    Parametrize the line segment, ie, show it consists of the points a+(b-a)t as t ranges from 0 to 1. Then apply the transformation to this equation.
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