Question to Determine Whether a linear map $L(x) = A x$ is One to One Function.

Question

Consider a linear map $L:\mathbb{R}^3 \to \mathbb{R}^3$ with $L\left( {\bf x} \right) = \left[ {\begin{array}{*{20}{c}} 1&0&1\\ 1&1&2\\ 2&1&3 \end{array}} \right]{\bf x}$. Show that this $L$ is not one-to-one.

I know this is not one-to-one because the kernel of $L$ is not $\{{\bf 0}\}$. However, if I want to stick with the one-to-one definition, I got some trouble there. Here is my thinking: Assume $L({\bf x}) = L({\bf y})$, then I must show ${\bf x} = {\bf y}$. So pick ${\bf x} = [x_1, x_2, x_3]^T$ and ${\bf y} = [y_1, y_2, y_3]^T$ both lives in $\mathbb{R}^3$. Then from the hypothesis of $L({\bf x}) = L({\bf y})$, I got $$\left[ \begin{array}{l} {x_1} + {x_3}\\ {x_1} + {x_2} + 2{x_3}\\ 2{x_1} + {x_2} + 3{x_3} \end{array} \right] = \left[ \begin{array}{l} {y_1} + {y_3}\\ {y_1} + {y_2} + 2{y_3}\\ 2{y_1} + {y_2} + 3{y_3} \end{array} \right]$$ I think this should gives me $x_i = y_i$ for $i=1,2,3.$ which means ${\bf x} = {\bf y}$. and I got $L$ one-to-one but this contradicts to the kernel statement...

Can anyone help to pointing out which part I went wrong. Thank you.

Answer 1

you take $x=(1,1,-1) $ and $y=(0,0,0)$ and check that $Ax=Ay$ But $x\neq y$. So from the three equations that you have concluded $x=y$, that was wrong.

Answer 2

@Black -horse has already given an answer, so here let me just point out that $L$ is a linear operator and it does not have full rank. In fact, its rank is just $2$. By Rank-Nullity Theorem, $Rank(A)+Nullity(A)=3$ implying $Nullity(A)=1$ showing it is not injective.

Answer 3

The determinant of the matrix vanishes, so it is not invertible. A one to one function would have an inverse.

Your error is to assume that you can infer $x = y$ from the system. If you substract the second from the third equation and have this the new third equation, you see it repeats the first one. So you have two equations in six unknows. If you provide three extra equations to assume a certain $x$, this still leaves one unkown $y_i$ free.

Answer 4

You’ve already shown that the kernel is non-trivial. Choose $\mathbf v_1, \mathbf v_2\in\ker L, \mathbf v_1\neq\mathbf v_2$. Then $L(\mathbf v_1)=L(\mathbf v_2)=0$, so $L$ is not bijective.