# linear function on $\mathbb{C}^{3}$

Practice Problem:

Define a linear transformation $f:\mathbb{C}^{3}\rightarrow \mathbb{C}^{3}$ by: $$f\left ( \left ( a_{1},a_{2},a_{3} \right ) \right )=\left ( a_{1}-a_{2}+ia_{3},2a_{1}+ia_{2},\left ( 2+i \right )a_{1}-a_{3} \right )$$ Find $\mathrm{Ker}f$ and $\mathrm{Im}f$ by giving a basis for both of them.

I found the kernel to be $\left \{ a_{1}\cdot\left ( 1,2i,2+i \right ) \right \}$ where $a_{1}\in \mathbb{C}$. Obviously dimension of the kernel is 1 and hence dimension of the image of $f$ is 2. But how can I find the image of $f$ with its basis? Thanks

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I always look at this handout of Hochster's in order to recall how to do this. The formatting isn't incredible but the idea is clear. Also, there's the Wikipedia blackout to consider :) – Dylan Moreland Jan 19 '12 at 4:17
@DylanMoreland Or, you could turn javascript off and look on Wikipedia. – Alex Becker Jan 19 '12 at 4:24
@AlexBecker Indeed, but I wanted to avoid giving computer advice. – Dylan Moreland Jan 19 '12 at 4:53

For any vector spaces $\mathbf{V}$ and $\mathbf{W}$, and for any linear transformation $f\colon\mathbf{V}\to\mathbf{W}$, if $\beta$ is a basis for $\mathbf{V}$, then $f(\beta) = \{f(\mathbf{v})\mid \mathbf{v}\in\beta\}$ spans $\mathrm{Im}(f)$. (In fact, this is true if $\beta$ is a spanning set for $\mathbf{V}$; it doesn't have to be a basis, but it may as well be, since any linear dependencies that may exist among elements of $\beta$ will necessarily exist among element of $f(\beta)$ as well...)

In particular, for your $f$, taking $\beta$ to be the standard basis we have $f(1,0,0) = (1,2,2+i)$, $f(0,1,0)=(-1,i,0)$, and $f(0,0,1)=(i,0,-1)$ will necessarily span $\mathrm{Im}(f)$.

That is, $\mathrm{Im}(f) = \mathrm{span}\Bigl( (1,2,2+i), (-1,i,0), (i,0,-1)\Bigr)$.

Now, this is a spanning set for $\mathrm{Im}(f)$, not necessarily a basis for $\mathrm{Im}(f)$. But: every spanning set contains a basis. So all we have to do is go through the three vectors, discarding any vector that is a linear combination of the previous ones. Since we know (by the Rank-Nullity Theorem) that the image of $f$ has dimension $2$, noting that $(1,2,2+i)$ and $(-1,i,0)$ are linearly independent tells us that they are a basis for the image.

(Indeed, note that $$(i,0,-1) = \left(-\frac{2}{5}+\frac{1}{5}i\right)(1,2,2+i) + \left(-\frac{2}{5}-\frac{4}{5}i\right)(-1,i,0)$$ so the third vector in the spanning set is already in the span of the first two.)

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+1. A nitpick: Calling the set of vectors a spanning set sounds slightly ambiguous to me since it could be taken to be spanning the whole space $\mathbf{W}$, instead of spanning the image of $f$. Perhaps you could explicitly state this just to make it crystal clear? – Srivatsan Jan 19 '12 at 4:35
@Srivatsan: You're absolutely right. Done. – Arturo Magidin Jan 19 '12 at 4:39
Can anyone tell me if what I wrote is mathematically correct:$\left \{ a_{1}\cdot\left ( 1,2i,2+i \right ) \right \}$ where $a_{1}$ is a complex scalar?, i.e the vector $\left ( 1,2i,2+i \right )$ a generator for the kernel. – M.Krov Jan 19 '12 at 4:49
@Zi2018Alpha: Well, $(1,2i,2+i)$ is in the kernel, and the kernel is of dimension $1$, so the vector by itself is a basis for the kernel; hence the kernel consists of all multiples of that vector. Using $a_1$ for an arbitrary scalar is technically fine, but likely to lead to confusion; I would use some other letter. Otherwise, it's fine. – Arturo Magidin Jan 19 '12 at 4:51

A general vector of Imf is the vector you have in the definition of the transformation, that is the result of aplying f to a general vector of the domain. You can descompose it as a linear combination of three vectors of image making : a1(1,2,2+i,)+a2(-1,i,0)+a3(i,0,-1) And in this way you find a generator of Imf.

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