Problem: Let $T: \mathbb{R} ^3 \to \mathbb{R} ^3$ be a linear transformation satisfying
\[
T(1,0,0) = (0,1,1), T(1,1,0) = (1,0,1) \text{ and } T(1,1,1) = (1,1,2).
\]
Find $T(x,y,z)$ for any $(x,y,z) \in \mathbb{R} ^3$. Check whether $T$ is one-one and onto.
Solution: At first we note that $\{ (1,0,0), (1,1,0), (1,1,1) \} $ is a basis for $\mathbb{R} ^3$. Using the image of $T$ on the basis elements, we have
\begin{align*}
T(0,1,0) = T((1,1,0) - (0,1,0)) = (1,-1,0) \\
T(0,0,1) = T((1,1,1) - (1,1,0)) = (0,1,1).
\end{align*}
Therefore, for any $(x,y,z)$ we have
\begin{align*}
T(x,y,z) & = T(x(1,0,0) + y(0,1,0) + z(0,0,1)) \\
& = x (0,1,1) + y (1,-1,0) + z(0,1,1) \\
& = (y, x - y + z, x + z).
\end{align*}
Observe that $T$ is a linear map from a $3$-dimensional space to a $3$-dimensional space and hence using the Rank-Nullity theorem, it is one-one if and only if it is onto. Let us write the matrix of $T$ with respect to the standard basis $\{ e_1, e_2, e_3 \} $.
\[
[T] =
\begin{bmatrix}
0 & 1 & 0 \\
1 & -1 & 1 \\
1 & 0 & 1 \\
\end{bmatrix}.
\]
Since the determinant of $[T]$ is $0$, so $T$ is neither one-one nor onto.