rsst123
  • rsst123
Let T:R3->R3 be the linear transformation such that T(x1,x2,x3) = (x1+2x2+2x3,x1+2x2+x3,x1-2x2+2x3) a) is T 1-1? Briefly explain b) is T onto? Briefly explain I know T is 1-1 if the only solution to T(x)=0 is x=0 and for onto I'm not really sure. Can someone please help me understand 1-1 and onto and how to solve this problem. Thanks!
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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dan815
  • dan815
|dw:1434823327570:dw|
dan815
  • dan815
ya
ParthKohli
  • ParthKohli
So first of all, we have to really refer to the definitions. What does a one-one transformation really mean? It tells you if there's only one way to get to a point after a transformation. You can't start with two same points and end up with the same point after the transformation. Let's assume that there *are* two such points, say, \((x_1, x_2, x_3)\) and \((y_1, y_2, y_3)\). Now \(T(x_1, x_2, x_3) = (x_1+2x_2+2x_3,~x_1+2x_2+x_3,~x_1-2x_2+2x_3) \) and \(T(y_1, y_2, y_3) = (y_1 + 2y_2 + 2y_3, ~ y_1 + 2y_2 + y_3, ~y_1 - 2y_2 + 2y_3)\). If these two points are identical for unidentical triples \((x_1, x_2, x_3); (y_1, y_2, y_3)\) then we can claim that the function is not one-one. This means:\[x_1+2x_2+2x_3 =y_1 + 2y_2 + 2y_3 \]\[x_1+2x_2+x_3 =y_1 + 2y_2 + y_3 \]\[x_1-2x_2+2x_3 = y_1 - 2y_2 + 2y_3 \]

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ParthKohli
  • ParthKohli
Let's write these equations as:\[\lambda_1 + 2\lambda_2 + 2\lambda_3 = 0\]\[\lambda_1 + 2\lambda_2 + \lambda_3 = 0\]\[\lambda_1 - 2 \lambda_3 + 2\lambda_3=0\]Where \(\lambda_1 = x_1 - y_1\) and so on. Turns out that the only solution here is the trivial one: \(\lambda_1 = \lambda_2 = \lambda_3 = 0\).
ParthKohli
  • ParthKohli
Therefore, \(x_1 = y_1\); \(x_2 = y_2\) and \(x_3 = y_3\) making this a one-one function.
ParthKohli
  • ParthKohli
What I meant in my first post was "you can't start with two different points and end up with the same point".
rsst123
  • rsst123
ok so the points for x and y have to be the same to be 1-1 right? and is onto just the opposite when two different points and end up with the same point?
dan815
  • dan815
the matrix is linearly independant
ParthKohli
  • ParthKohli
What does an onto-transformation mean? It means that the transformation covers all points. If I take any point, say (1, 1, 2), then I can find a point that transforms to give that point (here, (1, 1, 2)).
ParthKohli
  • ParthKohli
So let's say I have a point \((x,y, z)\) then I have to show that there exists a point \((k_1, k_2, k_3)\) such that \(T(k_1, k_2, k_3) = (x,y,z)\) if this is an onto function.
ParthKohli
  • ParthKohli
Assume that \(T(x_1, x_2, x_3) = (x,y,z)\) for some \((x_1, x_2, x_3)\). By the definition of our transformation, this point \((x,y,z)\) is the same as \((x_1+2x_2+2x_3,x_1+2x_2+x_3,x_1-2x_2+2x_3)\). Therefore,\[x_1 + 2x_2 + 2x_3 = x\]\[x_1 + 2x_2 + x_3 = y\]\[x_1 - 2x_2 + 2x_3 = z\]Sounds good so far. What do we really have to do here? We have to express \(x_1, x_2,x_3\) in terms of \(x, y, z\) to show that this function is onto. Can we do that? Yup, we can.
ParthKohli
  • ParthKohli
For instance, \(x_3 = x - y\) from the first two equations.
ParthKohli
  • ParthKohli
Similarly, \(x_1 = \dfrac{y+z -3x_3}2 = \dfrac{y + z - 3x + 3y}{2} =\dfrac{4y - 3x + z}{2}\)
ParthKohli
  • ParthKohli
So basically what I did here was that I actually found the point that maps to \((x,y,z)\) to show that this is an onto function.
ParthKohli
  • ParthKohli
If you know basic linear algebra, Dan's solution is actually WAYYYY simpler than mine is.
rsst123
  • rsst123
ok so a simple answer would be if its linearly independent its onto and if not 1-1?
rsst123
  • rsst123
with your help and a little research I figured it out, so this is actually both onto and 1-1 because if there is a pivot in every row T is onto and every column T is 1-1 so putting it in a matrix each row and column has a pivot making T an isomorphism.

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