Show that there is exactly one linear map $ \Phi: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} $

Question

In the real vector space $ \mathbb{R}^{3} $ the following vectors are given: $ v_{1}=\left(\begin{array}{c}{-2} \\ {-2} \\ {1}\end{array}\right), \quad v_{2}=\left(\begin{array}{c}{0} \\ {-1} \\ {-2}\end{array}\right), \quad v_{3}=\left(\begin{array}{c}{1} \\ {1} \\ {1}\end{array}\right) \quad $ and $ \quad w_{1}=\left(\begin{array}{c}{-1} \\ {-2} \\ {3}\end{array}\right), \quad w_{2}=\left(\begin{array}{c}{1} \\ {-1} \\ {0}\end{array}\right), \quad w_{3}=\left(\begin{array}{c}{0} \\ {1} \\ {-1}\end{array}\right) $

i) Show that there is exactly one linear map $ \Phi: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} $ with $ \Phi\left(v_{i}\right)=w_{i} $ for $i \in\{1,2,3\}$.

ii) Determine $\operatorname{ker}(\Phi), $ im$ (\Phi) $ and their dimensions.

iii) Determine $ A \in \mathbb{R}^{3 \times 3} $ with $ \Phi_{A}=\Phi $.

iv) Determine $det(\Phi)$.

How can I show that there is exactly one linear map? Thanks in advance!

Answer 1

First note that the vectors $v_1,v_2,v_3$ are linearly independent so they form a basis of $\mathbb{R}^3$. In particular, every vector $x\in\mathbb{R}^3$ can be expressed uniquely as $$ x = av_1 + bv_2 + cv_3 $$ for some constants $a,b,c\in\mathbb{R}$. As $\Phi$ must be linear, we may define $\Phi$ uniquely as $$ \Phi(x) = \Phi(av_1 + bv_2 + cv_3) = a\Phi(v_1) + b\Phi(v_2) + c\Phi(v_3) = aw_1 + bw_2 + cw_3 $$ for each $x\in\mathbb{R}$. It is now evident that $\mathrm{im}(\Phi)=\mathrm{span}\{w_1,w_2,w_3\}$. Note, however, that the vectors $w_1,w_2,w_3$ are not linearly independent as $$w_1 +w_2+3w_3 = 0. \tag{$\ast$} $$ In particular, we see that $w_1$ is in the span of $w_2$ and $w_3$ and that $w_2$ and $w_3$ are linearly independent. We may conclude that $$ \mathrm{im}(\Phi) = \mathrm{span}\{w_2,w_3\} = \left\{aw_2+bw_3\,\middle|\,a,b\in\mathbb{R}\right\}=\left\{\begin{pmatrix}a\\b-a\\-b\end{pmatrix}\,\middle|\,a,b\in\mathbb{R}\right\} $$ and that $\mathrm{dim}(\mathrm{im}(\Phi))=2$. By the rank-nullity theorem, we may conclude that $\mathrm{dim}(\mathrm{ker}(\Phi)) = 3-\mathrm{dim}(\mathrm{im}(\Phi)) = 1$. To find the kernel of $\Phi$, note from ($\ast$) that $\Phi(v_1+v_2+3v_3)=0$, and thus $$ \mathrm{ker}(\Phi) = \mathrm{span}\{v_1+v_2+3v_3\} = \left\{ \begin{pmatrix}a\\0\\2a\end{pmatrix}\,\middle|\, a\in\mathbb{R}\right\}. $$

I'll leave it to you to determine the matrix form of $\Phi$. However, as $\Phi$ does not have trivial kernel, its matrix form $A$ is not invertible, so we can immediately conclude that $\mathrm{det}(A)=0$.

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Edit: A hint for finding the matrix $A$.

Let $V$ and $W$ denote the matrices whose columns are $v_1,v_2,v_3$ and $w_1,w_2,w_3$ respectively. It is the case that $$ V\begin{pmatrix} a\\b\\c\end{pmatrix} = av_1 + bv_2 + cv_3 $$ and that $$ W\begin{pmatrix} a\\b\\c\end{pmatrix} = aw_1 + bw_2 + cw_3. $$ hold for each $a,b,c\in\mathbb{R}$. You need to find a matrix $A$ such that $$ A(av_1 + bv_2 + cv_3) = aw_1 + bw_2 + cw_3. $$ How can you find $A$ from $V$ and $W$?