Is there a linear transformation $T$ from $\Bbb{R^3}$ into $\Bbb{R^2}$ such that $T(1,-1,1)=(1,0)$ and $T(1,1,1)=(0,1)$
My try:
According to me i think that I have to show that $(1,-1,1),(1,1,1)$ form a basis but according to my computations they don't. Now since they don't form a basis i cannot write the vector (1,0,0) in terms of these two vectors. So because of this at least there will one vector in $\Bbb{R^3}$ who wont have a linear transformation in $\Bbb{R^2}$ since by the given mapping i cannot represent the standard basis vectors. Therefore there isnt a linear transformation with such kind of a mapping. Am i right?
sorry but i found a duplicate over here
:-is there any transformation T from $\Bbb R^3 \to \Bbb R^2$ for this
Best Answer
Extend the two given vectors in $\mathbb R^3$ to a basis $B$ of $\mathbb R^3$ by arbitrarily adding a third linearly independent vector (say, $\vec e_1$) that is mapped to any other vector (say, $\vec 0$) by $T$. Express each vector in the standard basis for $\mathbb R^3$ as a linear combination of the vectors in $B$ and use the linearity of $T$ to find each column of the standard matrix for $T$. For example, notice that since: $$ \frac{1}{2}\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} + \frac{1}{2}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} - \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$ It follows that the third column of the standard matrix is: $$ \frac{1}{2}\begin{bmatrix} 1 \\0 \end{bmatrix} + \frac{1}{2}\begin{bmatrix} 0 \\ 1 \end{bmatrix} - \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1/2 \\ 1/2 \end{bmatrix} $$ Indeed, it's not too hard to verify that the linear tranformation $T\colon \mathbb R^3 \to \mathbb R^2$ defined by $T(\vec x) = A\vec x$ where: $$ A = \begin{bmatrix} 0 & -1/2 & 1/2 \\ 0 & 1/2 & 1/2 \end{bmatrix} $$ satisfies the desired property.