C20 The linear transformation T : \(C^4 → C^3\) is not injective. Find two inputs \(x, y ∈ C^4\) that yield the same output (that is \(T (x) = T (y)\)).

\[ T\left(\begin{array}{cc} \left(\begin{array}{cc} x_{1}\\ x_{2}\\ x_{3}\\ x_{4} \end{array}\right) \end{array}\right) = \left(\begin{array}{cc} 2x_{1}+x_{2}+x_{3}\\ -x_{1}+3x_{2}+x_{3}-x_{4}\\ 3x_{1}+x_{2}+2x_{3}-2x_{4} \end{array}\right) \]
General Explanation:
* A function T: V -> W is one-to-one (injective) if and only if every vector in the range of T has a single (unique) preimage. An equivalent definition is that T is one-to one if and only if T(u) = T(v) implies that u = v.
* A function T: V -> W is onto (surjective) if and only if the range of T is W (the codomain of T ). And equivalent definition is that that T is one-to one if and only if every element in W has a preimage in V.

If T : \(C^4 → C^3\) is a linear transformation and not injective, then:

Let \(n ∈ C^4\) \[ T\left(\begin{array}{cc} n \end{array}\right) = 0 = \left(\begin{array}{cc} 0\\ 0\\ 0 \end{array}\right) = \left(\begin{array}{cc} 2n_{1}+n_{2}+n_{3} & | & 0\\ -n_{1}+3n_{2}+n_{3}-n_{4} & | & 0\\ 3n_{1}+n_{2}+2n_{3}-2n_{4} & | & 0 \end{array}\right) \]

=> \(\begin{cases} 2n_{1}+n_{2}+n_{3}=0 \\ -n_{1}+3n_{2}+n_{3}-n_{4}=0 \\ 3n_{1}+n_{2}+2n_{3}-2n_{4}=0 \end{cases}\)

\[ \left(\begin{array}{cc} 2 & 1 & 1 & 0\\ -1 & 3 & 1 & -1\\ 3 & 1 & 2 & -2 \end{array}\right) = \left(\begin{array}{cc} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & -3 \end{array}\right) = \left(\begin{array}{cc} n_{1} & 0 & 0 & n_{4} & | & 0\\ 0 & n_{2} & 0 & n_{4} & | & 0\\ 0 & 0 & n_{3} & -3n_{4} & | & 0 \end{array}\right) = \left(\begin{array}{cc} n_{1} = -n_{4}\\ n_{2} = -n_{4}\\ n_{3} = 3n_{4} \end{array}\right) = n_{4}\left(\begin{array}{cc} -1\\ -1\\ 3\\ 1 \end{array}\right) \]

At random \(x=\left(\begin{array}{cc}1\\1\\1\\1\end{array}\right)\), \(y = x + n = \left(\begin{array}{cc}1\\1\\1\\1\end{array}\right)+\left(\begin{array}{cc}-1\\-1\\3\\1\end{array}\right)=\left(\begin{array}{cc}0\\0\\4\\2\end{array}\right)\)

=> \(T(y) = \left(\begin{array}{cc}4\\2\\4\end{array}\right) = T(x)\)