How to determine (quickly) whether a linear map with certain given mappings can exist?

Question

Let's jump into an example. Assume we want to establish a linear mapping $f : \mathbb{R}^3 \to \mathbb{R}^4$ with the following traits: $$ f(\begin{pmatrix}1\\2\\3\end{pmatrix}) = \begin{pmatrix}2\\4\\0\\1\end{pmatrix} ,\quad f(\begin{pmatrix}4\\5\\6\end{pmatrix}) = \begin{pmatrix}0\\2\\-3\\1\end{pmatrix} ,\quad f(\begin{pmatrix}7\\8\\9\end{pmatrix}) = \begin{pmatrix}-2\\0\\6\\0\end{pmatrix} $$

Can such a linear mapping possibly exist?

My solution

I've come to the conclusion that this particular linear mapping can not exist. As per usual, I've done this using relatively complicated techniques. The idea is to try to determine the matrix $A$ which represents $f$ using the given traits. I'll try to keep it short.

Let $$ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \\ \end{pmatrix} $$

be the matrix representing $f$. Then there's four systems of equations which need to be solvable, each corresponding to a row of the image vectors. For instance, for the last row we'd have: $$ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{pmatrix} \cdot \begin{pmatrix} a_{41} \\ a_{42} \\ a_{43} \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} $$

Let's bring that matrix on the left into row echolon form using elementary matrix manipulations. We'll receive the following equation: $$ \begin{pmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & 0 & 0 \\ \end{pmatrix} \cdot \begin{pmatrix} a_{41} \\ a_{42} \\ a_{43} \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix} $$

This equation contains a contradiction in the last row and thus can't be solved (while the other three equation systems can be solved, by the way). Thus no representing matrix $A$ for $f$ can be determined and as such $f$ cannot exist.

The question

Is there an easier way to show that the above linear mapping can not exist (assuming my solution is correct)? I did notice that the argument vectors given above are not linearly independent (rank 2), whereas the resulting image vectors are (rank 3). Could this be used as a simple proof?

Answer 1

Yes, if $v_1,v_2,v_3$ are linearly dependent vectors and $w_1,w_2,w_3$ are linearly independent vectors, there is no linear $f$ such that $f(v_i)=w_i$ for $i=1,2,3.$

Also, if the dimension of the span of the $v_i$ is strictly less than the dimension of the span of the $w_i,$ then there is no $f.$

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You can do less work than it takes to show that the $w_i$ are independent, though.

Since the $v_i$ are dependent, you must have $a,b,c$ not all zero so that:

$$0=av_1+bv_2+cv_3.\tag{1}$$

But if $f$ is linearly and $f(v_i)=w_i,$ this means that:

$$0=f(0)=f(av_1+bv_2+cv_3)=aw_1+bw_2+cw_3\tag{2}$$

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In the case $v_1=(1,2,3),v_2=(4,5,6),v_3=(7,8,9)$ you have $(a,b,c)=(1,-2,1).$

If we know the $w_i$ are independent, then no such $a,b,c$ exists for (2.)

But, rather than check that the $w_i$ are independent, you can check specifically if $w_1-2w_2+w_3=0.$ Since it is not, we know there is not such $f.$

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This shows that for $f$ to exist, you need, for any linear dependence $(1)$ amongst the $v_i,$ we must have the same dependence $(2)$ on the $w_i.$

This is actually necessary and sufficient:

Theorem: Given two (finite-dimensional) vector space $V,W$ and $v_1,\dots,v_n\in V,\, w_1,\dots,w_n\in W$ there is a linear map $f:V\to W$ with $f(v_i)=w_i$ for $i=1,\dots,n$ if and only if, for all $a_1,\dots,a_n:$ $$\sum_{i=1}^{n} a_iv_i=0\implies \sum_{i=1}^n a_iw_i=0.$$

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[You don't necessarily need the vector spaces to be finite-dimensional in this theorem, but it requires a harder proof to prove that $f$ exists in this case, requiring the axiom of choice.]

Answer 2

Such $f$ couldn't exist.

Suppose for contradiction there exist such $f$. Then we must have $$f(\begin{pmatrix}1\\2\\3\end{pmatrix}+\begin{pmatrix}7\\8\\9\end{pmatrix}-2\begin{pmatrix}4\\5\\6\end{pmatrix})=f(\begin{pmatrix}1\\2\\3\end{pmatrix})+f(\begin{pmatrix}7\\8\\9\end{pmatrix})-2f(\begin{pmatrix}4\\5\\6\end{pmatrix}) = f(\begin{pmatrix}0\\0\\0\end{pmatrix})=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}$$ Contradiction.

Answer 3

The reason your map does not exist is because the vectors $(1,2,3), (4,5,6), (7,8,9)$ and their images do not satisfy the same dependencies.

In general, if you pick three linearly independent vectors and their images you have a well-defined map. So let $v_1, v_2, v_3$ be independent, and pick any $w_1, w_2,w_3 \in \Bbb R^4$, then you can finda linear map $f$ such that $f(v_i) = w_i$.

How to determine quickly if a map exists? Given vectors $v_i$ and images $f(v_i)$, compare the span of the $v_i$ and the span of the $f(v_i)$. If the span of the former has lower dimension than the latter, we have a clear impossibility.

Answer 4

the argument vectors given above are not linearly independent (rank 2), whereas the resulting image vectors are (rank 3)

This is a proof of nothing but the fact that if it exists, then it may not be unique. In my opinion, another proof would be to write the relation you found between the 3-dimensional vectors and verify if the image of that linear combination is zero (it must be zero, if it is a linear application...).