# find the bases for the range of a linear operator and the null space

I need to find the bases for a linear operator.

Here is the question given to me:

Consider $V=\mathbb{C}_{1\times 2}$ as a vector space over the real numbers.

let the linear operator $\tau : V \rightarrow V$ be defined by $\tau (z_{1},z_{2})=(z_{1}-\overline{z_{1}}, z_{2}-\overline{z_{2}})$

Find bases for $\tau(V)$ and $NS(V)$ and determine whether $V=\tau(V) \bigoplus NS(\tau)$

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Linear operators don't have bases. Vector spaces have bases. The problem states that you are looking for bases of the range and the nullspace of the linear operator, which is something different. –  Arturo Magidin Jun 19 '12 at 21:01

First: linear operators don't have bases. Vector spaces (and subspaces) have bases. You'll note that the question does not ask for a basis for the linear operator, it asks for a basis of the range of $\tau$, and for a basis of the nullspace of $\tau$; and it so happens that both of those are vector spaces, so we can talk about bases for them.

Now, $V$ is $4$-dimensional as a vector space over $\mathbb{R}$; a possible basis is $\{(1,0), (i,0), (0,1), (0,i)\}$.

You should also check and make sure that $\tau$ is a linear transformation if you haven't done so.

Let's consider first the nullspace of $\tau$ (it is simpler than the range). When is $(z_1,z_2)$ in the nullspace of $\tau$?

$(z_1,z_2)\in\mathbf{N}(\tau)$ if and only if $\tau(z_1,z_2)=(0,0)$, if and only if $z_1-\overline{z_1}=0$ and $z_2-\overline{z_2}=0$, if and only if $z_1=\overline{z_1}$ and $z_2=\overline{z_2}$, if and only if $z_1$ is real and $z_2$ is real. That is, the nullspace is the span of $(1,0)$ and $(0,1)$, and they give you a basis.

What is the range of $\tau$? Note that if $z = a+bi$ is a complex number, then $z-\overline{z} = (a+bi) - (a-bi) = 2bi$. So another way of writing $\tau$ is: $$\tau(z_1,z_2) = \Bigl( 2i\mathrm{Im}(z_1), 2i\mathrm{Im}(z_2)\Bigr).$$ So you should notice that the image of $\tau$ is contained in the span of $(i,0)$ and $(0,i)$.

Is the image of $\tau$ exactly equal to $\mathrm{span}((i,0),(0,i))$?

Given the answer to the latter question, can you find the answer to your question of deciding whether $V = \tau(V)\oplus \mathbf{N}(\tau)$?

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thanks a mil if the definition of $\tau$ changes to $\tau(z_{1},z_{2})=(z_{1}-z'_{1}, z_{2}+z'_{2})$ am I correct in saying $NS(\tau)=span{(1,0), (0,i)}$ and that $\tau(V)=span{(i,0), (0,1)}$ ? by looking at these two definitions am I correct in concluding that $V=\tau(V) \bigoplus NS(\tau)$? –  sarah jamal Jun 19 '12 at 21:32
sorry, can you tell me how to type the complement with the correct symbol? Im still trying to figure out how to use latex. –  sarah jamal Jun 19 '12 at 21:33
@sarah: Yes, if $\tau$ is defined as in your comment, then $\tau(z_1,z_2) = (2i\mathrm{Im}(z_1), 2\mathrm{Re}(z_2))$, so the image is spanned by $(i,0)$ and $(0,1)$, and the kernel is spanned by $(1,0)$ and $(0,i)$. Yes, you are concluding correctly that the space is the direct sum, since the Rank-Nullity theorem tells you that the rank plus the nullity equals the dimension of $V$, and we have $\tau(V)\cap \mathbf{N}(\tau)=\{\mathbf{0}\}$. I don't know what you mean by "symbol for complement", but you can use deTexify to try to find it. –  Arturo Magidin Jun 20 '12 at 0:07