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Is there is a linear transformation $T$ from $\Bbb{R}^3$ into $\Bbb{R}^2$ such that $T(1,-1,1)=(1,0)$ and $T(1,1,1)=(0,1)$?

We can prove a stronger result:

Let $V$ be a finite-dimensional vector space over field $F$ with $\mathrm{dim}(V)=n\in \Bbb{N}$ and Let $W$ be a vector space over field $F$. If $\{\alpha_1,…,\alpha_m\}\subseteq V$ is linearly independent and $(\beta_1,…,\beta_m)$ is sequence in $W$, then $\exists T :V\to W$ such that $T$ is linear map and $T(\alpha_j)=\beta_j$, $\forall j\in J_m$.

My attempt: $\{\alpha_1,…,\alpha_m\}\subseteq V$ is linearly independent. By theorem 5 corollary 2 section 2.3, $\exists B\subseteq V$ such that $B$ is finite basis of $V$ and $\{\alpha_1,…,\alpha_m\}\subseteq B$. Since $\mathrm{dim}(V)=n$, $|B|=n$. Let $B=\{\alpha_1,…,\alpha_n\}$. Define $\beta_j=0_W$, $\forall j\in J_n\setminus J_m$. By theorem 1 section 3.1, $\exists !$ $T\in L(V,W)$ such that $T(\alpha_j)=\beta_j$, $\forall j\in J_n$. Hence $T(\alpha_j)=\beta_j$, $\forall j\in J_m$. Is my proof correct? Proof is basically corollary of theorem 1 section 3.1.

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  • $\begingroup$ Your proof is correct, but what about the main question? Why do you say “stronger result” since the question is not a result? $\endgroup$
    – azif00
    Jul 11 at 19:05
  • $\begingroup$ @user264745 Did you mean $T:\mathbb{R}^{3}\to\mathbb{R}^{2}$? $\endgroup$ Jul 11 at 19:12
  • $\begingroup$ @azif00 You mean why I didn’t explicitly showed proof of primary question? I agree, stronger result is probably not the accurate term. $\endgroup$
    – user264745
    Jul 11 at 19:13
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    $\begingroup$ @user264745 Ok, no problems. I have fixed the title as you have suggested. $\endgroup$ Jul 11 at 19:18
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    $\begingroup$ @user264745 No, I mean, you should finish the problem with something like: “So, the answer to the main question is yes, since $(1,-1,1)$ and $(1,1,1)$ are linearly independent.” $\endgroup$
    – azif00
    Jul 11 at 19:19

2 Answers 2

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Let $ T(x) = A x = y $

Then

$ A \begin{bmatrix} 1 && 1 \\ -1 && 1 \\ 1 && 1 \end{bmatrix} = \begin{bmatrix}1 && 0 \\ 0 && 1 \end{bmatrix} $

Matrix $A$ has three columns $A_1, A_2, A_3$, and we want

$A_1 - A_2 + A_3 = e_1 $

$A_1 + A_2 + A_3 = e_2 $

Subtracting, we get

$A_2 = \dfrac{1}{2} (e_2 - e_1) $

And then we must choose $A_1$ and $A_3$ such that

$ A_1 + A_3 = e_1 + A_2 = e_2 - A_2 = \dfrac{1}{2} (e_1 + e_2) $

So matrix $A$ is of the form

$ A = \begin{bmatrix} a && -0.5 && 0.5 - a \\ b && 0.5 && 0.5 - b \end{bmatrix} \hspace{25pt} a, b \in \mathbb{R} $

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  • $\begingroup$ Thank you so much for the answer. You gave explicit construction of linear map from $\Bbb{R}^3$ into $\Bbb{R}^2$ such that $T(1,-1,1)=(1,0)$ and $T(1,1,1)=(0,1)$. $\endgroup$
    – user264745
    Jul 11 at 19:37
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Of course there is. It is the matrix

$A=\begin{pmatrix} 0& -1/2 &1/2 \\ 0& 1/2 & 1/2 \\ \end{pmatrix}$

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