# Exercise 4, Section 3.1 of Hoffman’s Linear Algebra

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.

• Your proof is correct, but what about the main question? Why do you say “stronger result” since the question is not a result? Jul 11 at 19:05
• @user264745 Did you mean $T:\mathbb{R}^{3}\to\mathbb{R}^{2}$? Jul 11 at 19:12
• @azif00 You mean why I didn’t explicitly showed proof of primary question? I agree, stronger result is probably not the accurate term. Jul 11 at 19:13
• @user264745 Ok, no problems. I have fixed the title as you have suggested. Jul 11 at 19:18
• @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.” Jul 11 at 19:19

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

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

• 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 $T(1,1,1)=(0,1)$. Jul 11 at 19:37

Of course there is. It is the matrix

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