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Alright so I have this Transformation that I know isn't one to one transformation, but I'm not sure why.

A Transformation is defined as $f(x,y)=(x+y, 2x+2y)$.

Now my knowledge is that you need to fulfill the 2 conditions: Additivity and the scalar multiplication one. I tried both of them and somehow both of them are met perfectly.

However, the transformation is NOT linear. This is because the column vectors of the transformation are linearly dependent.

So how am I supposed to relate these 2 seemingly unrelated conjectures to check the one-one transformation ?

Arnaldo
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user432184
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    It is linear. I think you are confuse about the definition of linear transformation and of linearly independent. – R.W Apr 03 '17 at 20:48
  • The failure of injectivity is measured by the kernel of the linear map, however the definition of a linear map has nothing to do with injectivity. – Mathematician 42 Apr 03 '17 at 20:50
  • The transformation most certainly IS linear. It's not one-to-one, but it is linear. – 5xum Apr 03 '17 at 20:50
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    Thanks a ton !! I just realized I was confusing the linearity with the transformation being one to one. – user432184 Apr 03 '17 at 20:53

4 Answers4

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HINT

Note you can write $$ f(x,y) = \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}, $$ and since $f$ can be represented by matrix multiplication, it must be linear, since matrix multiplication is linear...

As for it being one-to-one, you need to make sure for each $\begin{bmatrix} v \\ w \end{bmatrix} \in \mathbb{R}^2$, there is at most one solution to $f(x,y) = \begin{bmatrix} v \\ w \end{bmatrix}$. You are right, the fact that the underlying matrix is not invertible must play a crucial role in that.

Elizabeth S. Q. Goodman
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gt6989b
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hint

$$f (1,-1)=(0,0) $$ and

$$f (0,0)=(0,0) $$

hamam_Abdallah
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Hint: What is $f^{-1}(0,0)$? For example $f(0,0)=(0,0)$, so $(0,0)\in f^{-1}(0,0)$. Can you find (m)any other pair(s) $(x,y)$ such that $f(x,y)=(0,0)$?

Jimmy R.
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The transformation $f$ is linear indeed, but it is not one-to-one transformation. One-to-one means that the system \begin{equation*} x + y = u \\ 2(x + y) = v \end{equation*} has a unique solution for variables $(x, y)$, i.e. they can be expressed trough the variables $u$ and $v$. But the determinant of this linear system is equal to 0 (the rows are linearly dependent), which means that the transfomation is not one-to-one.

Swistack
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