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I'm still paralyzed in this question: Calculate a linear transformation with a specific kernel

Well I understood why I'm wrong and so what I did was to express $W$ as vectors: $(2,1,0)$ and $(-1,0,1)$... I transform it into equations with zero as the equality and I got to $y=-2x$ and $z=x$. So my conclusion is that the transformation might be $T(x,y,z) = (x,-2x,x)$ But it can't be correct because it's a transformation to $\mathbb{R^2}$...

Can someone clarify this ideas to me, please?

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Could you not just define $T$ as $T:\Bbb R^3 \to \Bbb R^2$ where, $$T[x,y,z]=[x-2y+z,x-2y+z]$$

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You’ve found a basis for the kernel, which is a good start. Now, extend that basis to span all of $\mathbb R^3$. In this case that’s easy to do—the equation you were given for the kernel tells you that $(1,-2,1)$ is orthogonal to the kernel, so it’ll do nicely as a third basis vector. We want $T$ to map each of the kernel basis vectors to $0$, and the third basis vector can be mapped to anything else, so pick $(1,0)$ for simplicity. Since we know that the columns of a transformation matrix are the images of the basis vectors, you can write this matrix down directly: $$\pmatrix{0&0&1\\0&0&0}.$$ All that’s left to do is to convert this to the standard basis: $$\pmatrix{0&0&1\\0&0&0}\pmatrix{2&-1&1\\1&0&-2\\0&1&1}^{-1}=\pmatrix{\frac16&-\frac13&\frac16\\0&0&0}.$$ From this, you can read off $T(x,y,z)=\left(\frac16x-\frac13y+\frac16z,0\right)$.

Choosing a different value for $T(1,-2,1)$ will yield a different linear transformation with the same kernel. In particular, choosing $T(1,-2,1)=(6,0)$ gives $T(x,y,z)=(x-2y+z,0)$. This suggests a simpler way to solve the problem: We’re given that $x-2y+z=0$ for elements of $W$ and no other vectors in $\mathbb R^3$, so, $T(x,y,z)=(x-2y+z,0)$ will be the zero vector if and only if $(x,y,z)\in W$, which is exactly what was required.

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