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!

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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
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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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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: \(\lambda_1 = \lambda_2 = \lambda_3 = 0\).
Therefore, \(x_1 = y_1\); \(x_2 = y_2\) and \(x_3 = y_3\) making this a one-one function.
What I meant in my first post was "you can't start with two different points and end up with the same point".
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?
the matrix is linearly independant
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)).
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.
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.
For instance, \(x_3 = x - y\) from the first two equations.
Similarly, \(x_1 = \dfrac{y+z -3x_3}2 = \dfrac{y + z - 3x + 3y}{2} =\dfrac{4y - 3x + z}{2}\)
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.
If you know basic linear algebra, Dan's solution is actually WAYYYY simpler than mine is.
ok so a simple answer would be if its linearly independent its onto and if not 1-1?
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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