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I'm trying to solve an exercise where I'm asked to find this linear transformation's kernel and its dimension:

$$f(x_1,x_2,x_3,x_4)=(x_1-x_2,2x_3+x_4)$$

So, from the definition of kernel and converting to parametric equations, I found $x_1=x_2=\lambda$, $x_3=-\frac{1}{2}\mu$ and $x_4=\mu$. Therefore, I assumed: $$\ker(f)=\{(\lambda,\lambda,-\frac{1}{2}\mu,\mu): \lambda,\mu \in \mathbb{R}\}$$

And since the kernel can be described with only one vector (with two parameters), I thought $\dim(\ker(f))=1$, but my textbook says it's actually equal to $2$. Where is my reasoning flawed?

Thanks in advance!

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    $\begingroup$ A basis for the kernel is $\{(1,1,0,0), (0,0,-\frac{1}{2}, 1) \}$ $\endgroup$ – leibnewtz Jan 9 '18 at 23:20
  • $\begingroup$ @leibnewtz Hm, so that means I can only 'fix' one parameter at once? I thought it'd be possible to establish something like $\lambda=1$, $\mu=2$ so that a basis for the kernel is $\{(1,1,-1,2)\}$. Why isn't that possible? $\endgroup$ – Manuel Jan 9 '18 at 23:22
  • $\begingroup$ Of course you can do what you say...but that way you'll only get one vector, and you already know you need two lin. ind. vectors to have a basis... $\endgroup$ – DonAntonio Jan 9 '18 at 23:30
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    $\begingroup$ Since $f:\mathbb R^4\to\mathbb R^2$, its image is at most two-dimensional, so you know that its kernel must be at least two-dimensional. $\endgroup$ – amd Jan 9 '18 at 23:35
  • $\begingroup$ @Manuel Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$ – user Feb 4 '18 at 0:09
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It is all correct, the dimension of the ker(f) is 2 because you have 2 free parameters that define it $(\lambda,\mu)$.

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  • $\begingroup$ Oh, I thought the definition of dimension refered to the number of vector, but accounting for the number of parameters necessary to define those vectors makes more sense. Is it that intuitive? $\endgroup$ – Manuel Jan 9 '18 at 23:23
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    $\begingroup$ @Manuel They are equivalent concepts. As noted in the comments you can set $\lambda=1$ and $\mu=0$ and define a basis vector, then $\lambda=0$ and $\mu=1$ and define a second vector linearly independent from the first, thus the dimension is 2. This is true for any number of free parameter (EG a line or a plane in $\mathbb{R^3}$). $\endgroup$ – user Jan 9 '18 at 23:25
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    $\begingroup$ @Manuel There are two vectors in your description of the kernel: $\left(\lambda,\lambda,-\frac12\mu,\mu\right)=\lambda(1,1,0,0)+\mu\left(0,0,-\frac12,1\right)$; every vector in the kernel is a linear combination of these two linearly-independent vectors. $\endgroup$ – amd Jan 9 '18 at 23:36
  • $\begingroup$ @amd, that makes a lot of sense, thanks! $\endgroup$ – Manuel Jan 9 '18 at 23:48

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