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rsst123
 one year ago
Let T:R3>R3 be the linear transformation such that
T(x1,x2,x3) = (x1+2x2+2x3,x1+2x2+x3,x12x2+2x3)
a) is T 11? Briefly explain
b) is T onto? Briefly explain
I know T is 11 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 11 and onto and how to solve this problem. Thanks!
rsst123
 one year ago
Let T:R3>R3 be the linear transformation such that T(x1,x2,x3) = (x1+2x2+2x3,x1+2x2+x3,x12x2+2x3) a) is T 11? Briefly explain b) is T onto? Briefly explain I know T is 11 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 11 and onto and how to solve this problem. Thanks!

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3So first of all, we have to really refer to the definitions. What does a oneone 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 \(T(x_1, x_2, x_3) = (x_1+2x_2+2x_3,~x_1+2x_2+x_3,~x_12x_2+2x_3) \) 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 oneone. This means:\[x_1+2x_2+2x_3 =y_1 + 2y_2 + 2y_3 \]\[x_1+2x_2+x_3 =y_1 + 2y_2 + y_3 \]\[x_12x_2+2x_3 = y_1  2y_2 + 2y_3 \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Let'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
 one year ago
Best ResponseYou've already chosen the best response.3Therefore, \(x_1 = y_1\); \(x_2 = y_2\) and \(x_3 = y_3\) making this a oneone function.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3What I meant in my first post was "you can't start with two different points and end up with the same point".

rsst123
 one year ago
Best ResponseYou've already chosen the best response.0ok so the points for x and y have to be the same to be 11 right? and is onto just the opposite when two different points and end up with the same point?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the matrix is linearly independant

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3What does an ontotransformation 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
 one year ago
Best ResponseYou've already chosen the best response.3So 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
 one year ago
Best ResponseYou've already chosen the best response.3Assume 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 \((x_1+2x_2+2x_3,x_1+2x_2+x_3,x_12x_2+2x_3)\). 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
 one year ago
Best ResponseYou've already chosen the best response.3For instance, \(x_3 = x  y\) from the first two equations.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Similarly, \(x_1 = \dfrac{y+z 3x_3}2 = \dfrac{y + z  3x + 3y}{2} =\dfrac{4y  3x + z}{2}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3So 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
 one year ago
Best ResponseYou've already chosen the best response.3If you know basic linear algebra, Dan's solution is actually WAYYYY simpler than mine is.

rsst123
 one year ago
Best ResponseYou've already chosen the best response.0ok so a simple answer would be if its linearly independent its onto and if not 11?

rsst123
 one year ago
Best ResponseYou've already chosen the best response.0with your help and a little research I figured it out, so this is actually both onto and 11 because if there is a pivot in every row T is onto and every column T is 11 so putting it in a matrix each row and column has a pivot making T an isomorphism.
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