Proving a linear transformation is isomorphic Let $T: \mathbb{C}^3 \rightarrow \mathbb{C}^3$ be such that
$$T(1, 0, 0) = (1, 0, i) \\ T(0, 1, 0) = (0, 1, 1) \\
T(0, 0, 1) = (i, 1, 0)$$
I was requested to decide whether $T$ is isomorphic or not. I understand this is equivalent to deciding on whether the mapping defined by $T$ is bijective, or one-to-one. I was able to deduce the range of $T$, since it is straigthforward to notice that
$$\text{Im}(T) = \Big\{\lambda_1(1, 0, i) + \lambda_2(0, 1, 1) + \lambda_3(i, 1, 0) \mid \lambda_i \in \mathbb{C} \Big\}$$
However, I don't know how to prove (or disprove) the one-to-one correspondence between $\mathbb{C}^3$ and $\text{Im}(t)$. Any help is much appreciated.
 A: To determine if $\operatorname{im}(T) = \mathbb{C}^3$ (i.e. if $T$ is surjective) you can consider the rank of the matrix
$$
\begin{pmatrix}
T(1,0,0) & T(0,1,0) & T(0,0,1)
\end{pmatrix}
= \begin{pmatrix}
1 & 0 & i \\
0 & 1 & 1 \\
i & 1 & 0
\end{pmatrix}
$$
This will have full rank if and only if $T$ is an isomorphism. So you just need to row reduce this matrix and determine how many leading $1$'s there are in the reduced row echelon form.
A: Recall that we say that two vector space they are isomorphic if and only if there exists a isomorphism between them. Also, a result says that a linear transformation $T\colon V\to W$ is a isomorphism between vector spaces if and only if ${\rm ker}(T)=0_{V}$ (in words "$T$ is one-to-one") and ${\rm  im}(T)=W$ (in words "$T$ is onto"). Finally, it is important to remembet that an useful result is that the vector spaces $V$ and $W$ they are isomorphic vector space if and only if $\dim V=\dim W$.
Now, notice that in your problem both dimension they are same, then the vector spaces they are isomorphics. However, in order to determinate if $T$ is a isomorphism in your problem, that is another story, because we need to see if $T$ is one-to-one and if $T$ is onto (both condition). Of course, there are many ways to find this out quickly, to give a simple example. One can see that $\det ([T])=0$ then $T$ is not invertible, then $T$ is not bijective and then $T$ is not isomorphim. However, I will write a work-based approach on the definitions and results I mentioned in the first paragraph, since this may be for future readers more interesting.
Using the hypothesis of the problem and considering ${\bf C}^{3}$ as a ${\bf C}$-vector space, you can check that the rule for $T$ is given by
$$\forall \begin{pmatrix} z_{1}\\z_{2}\\z_{3}\end{pmatrix}\in {\bf C}^{3}\colon \quad  T\begin{pmatrix} z_{1}\\z_{2}\\z_{3}\end{pmatrix}=\begin{pmatrix}z_{1}+iz_{3}\\z_{2}+z_{3}\\iz_{1}+z_{2}
 \end{pmatrix}=\begin{pmatrix}
1 & 0 & i \\
0 & 1 & 1 \\
i & 1 & 0
\end{pmatrix}\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix} $$
Starting with the image of $T$.

*

*By definition,
\begin{align*}
{\rm im}(T)&:=\{w\in W\colon \exists v\in V\colon T(v)=w\}\\
&=\left\{\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}\in {\bf C}^{3}\colon \exists \begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}\in {\bf C}^{3}\colon T\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}=\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix} \right\}\\
&=\left\{\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}\in {\bf C}^{3}\colon \exists \begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}\in {\bf C}^{3}\colon \underbrace{\begin{pmatrix}z_{1}+iz_{3}\\z_{2}+z_{3}\\iz_{1}+z_{2}
 \end{pmatrix}=\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}_{(*)}\right\}
\end{align*}
Solving the linear system of equations $(\ast)$ we get
$$\begin{pmatrix}
1 & 0 & i & | & w_{1}\\
0 & 1 & 1 & | &w_{2} \\
i & 1 & 0 & |& w_{3}\\
\end{pmatrix}\cdots \sim \cdots \begin{pmatrix}1&0&i&|&w_{1}\\0&1&1&|&w_{2}\\0&0&0&|&-iw_{1}-w_{2}+w_{3} \end{pmatrix}.$$
Thus, for the image of $T$ we have
\begin{align*}
{\rm im}(T)&=\left\{\begin{pmatrix}w_{1}\\w_{2}\\w_{3} \end{pmatrix}\in {\bf C}^{3}\colon -iw_{1}-w_{2}+w_{3}=0\right\}\\
&={\rm span}\left\{\begin{pmatrix}1\\0\\i\end{pmatrix},\begin{pmatrix}0\\1\\1\end{pmatrix} \right\},
\end{align*}
and since both vector they are independent linearly we have $2=\dim {\rm im}(T)\not=\dim {\bf C}^{3}=3$, then ${\rm im}(T)\not={\bf C}^{3}$ and so $T$ is not onto. You can stop here and review the result of the first paragraph to conclude that $T$ is not an isomorphism since $T$ does not satisfy one of the conditions of the result above, but let's continue to study what happens to the kernel of $T$.


*By definition,
\begin{align*}
{\rm ker}(T)&:=\{v\in V\colon T(v)=0_{W}\}\\
&=\left\{\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}\in {\bf C}^{3}\colon \quad T\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\right\}\\
&=\left\{\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}\in {\bf C}^{3}\colon  T\begin{pmatrix} z_{1}\\z_{2}\\z_{3}\end{pmatrix}\in {\bf C}^{3}
=\begin{pmatrix}0\\0\\0\end{pmatrix}\right\}\\
&=\left\{\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}\in  {\bf C}^{3}\colon \underbrace{\begin{pmatrix}z_{1}+iz_{3}\\z_{2}+z_{3}\\iz_{1}+z_{2}
 \end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}}_{(**)}\right\}
\end{align*}
Solving the linear system of equations $(**)$ we get
$$\begin{pmatrix}
1 & 0 & i & | & w_{1}\\
0 & 1 & 1 & | &w_{2} \\
i & 1 & 0 & |& w_{3}\\
\end{pmatrix}\cdots \sim \cdots \begin{pmatrix}1&0&i&|&0\\0&1&1&|&0\\0&0&0&|&0 \end{pmatrix},$$
where of course I reused the row reduction for the image case with the changes quite transparent.
Thus, for the kernel of $T$ we have
\begin{align*}
{\rm ker}(T)&=\left\{\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}\in{\bf C}^{3}\colon z_{1}+iz_{3}=0,z_{2}+z_{3}=0 \right\}\\
&={\rm span}\left\{\begin{pmatrix}-i\\-1\\1\end{pmatrix} \right\},
\end{align*}
and since the onlyn non-zero vector is independent linearly, then we have $1=\dim {\rm ker}(T)\not=\dim ({\rm span}\{0\})$, then ${\rm ker}(T)\not=0_{{\bf C}^{3}}$ and so $T$ is not one-to-one. Thus, again we can claim that $T$ is not isomorphism.
Of course, another way is using the dimension theorem $\dim V=\dim {\rm im}(T)+\dim {\rm ker}(T)$ and then we can use the fact $\dim {\bf C}^{3}=3$ and notice that the dimension theorem holds.
