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(a)Is there a linear mapping $f : \mathbb{R}^{3} \rightarrow \mathbb{R}^3$ with $ f(1,1,1)=(3,2,7), f(0,2,1)=(2,1,-1), f(2,0,1)=(1,0,0)$?

(b)Is there a linear mapping $f :\mathbb{R}^{3} \rightarrow \mathbb{R}^3$ with $ f(0,2,-1)=(3,-1,3), f(1,0,2)=(0,-2,1), f(-2,2,-5)=(3,3,1)$?

Substantiate your statements and if it exists, provide such a linear mapping.

My attempt so far:

(a) is it first important to determine whether $(1,1,1), (0,2,-1), (2,0,1)$ (called vectors, right??) are a basis for $\mathbb{R}^{3}$? in that case my understanding is a basis is linear independent vectors which span a vectorspace. Since $2(1,1,1)-(0,2,1)=(2,0,1)$ i would like to conclude that these vectors are linearly dependent $\Rightarrow$ do not form a basis for $\mathbb{R}^{3} \Rightarrow$ there is no linear map for these vectors. In case my reasoning is correct, I just wanted to double check is that i could have reached the linear dependence conclusion by turning the vectors into a matrix and seeing if that yielded a row of zeros or a free variable, right? (i happened to just look at it and recognize that this time).

(b)actually after writing my (a) attempt i realize that (b) would also be linearly dependent, correct? i still haven't learned to create a matrix with latex yet, but in case someone is skeptical that it is linearly dependent i could show you my steps. however, first i'd prefer to know if i am even on the right track ;) as a note i'm always happy to hear if i am using terminology improperly. thanks in advance for all your help!

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It isn't enough to say that the vectors you are defining your map with are not linearly independent. As a silly example, is there a linear map with $f(1,1,1)=(1,0,0)$, $f(2,2,2)=(2,0,0)$, and $f(3,3,3)=(3,0,0)$? Of course! The map $f(a,b,c)=(a,0,0)$ is just such a linear map. You are trying to use the fact that where a basis is sent determines a linear map completely. If you don't have a basis, then you can't use this fact, but it certainly doesn't imply that such a map does not exist; just that you cannot use that fact! – user1306 Nov 20 '10 at 18:36
up vote 4 down vote accepted

(i) You are sort of on the right track, but you are trying to (1) prove too much; and (2) drawing the wrong conclusion.

If you have a bunch of linearly independent vectors, then you can always find (at least one) linear transformation that maps them to whatever you want. This because we can always do it with a basis (decide where you want the basis vectors to go, and that gives you a linear transformation), and we can always take any collection of linearly independent vectors and complete it to a basis.

If your three vectors were linearly independent (hence a basis for $\mathbb{R}^3$, then you would know that there is a linear transformation with the desired values, because you can always find a linear transformation that does whatever you want to a basis.

However, if the three vectors are not linearly independent, this does not preclude the possibility that there is a linear transformation with the desired properties! For instance, $(1,0,0)$, $(0,1,0)$, and $(1,1,0)$ do not form a basis for $\mathbb{R}^3$, but I can nonetheless find a linear transformation such that $f(1,0,0)=(2,1,-1)$, $f(0,1,0) = (0,1,7)$, and $f(1,1,0)=(2,2,6)$ (in fact, many); for instance, $f(x,y,z) = (2x,x+y,7y-x)$ will do it.

The problem with your argument is that you misusing statement. You know that "If they are a basis, then a linear transformation exists". You are trying to use this by saying "If they are not a basis, then there is no linear transformation", but that is false. What you can conclude is that "if there is no linear transformation, then they are not a basis" (this is called the "contrapositive"). In general, if you know that "If P then Q", then from "P is false" you cannot conclude anything. Q may or may not hold. For instance, "If I fall into the pool, I will get wet." Does that mean that if I don't fall into the pool, I will be dry? No; I could get caught in a rainshower. If I know I didn't fall into the pool, I just don't know whether I will be wet or dry.

For this problem, thoguh, you have actually put your finger on exactly what you need to put your finger on: the fact that one of your vectors is a linear combination of the other two. Since $(2,0,1) = 2(1,1,1)-(0,2,1)$, that means that for any linear transformation $T$, you must have $$T(2,0,1) = T\bigl(2(1,1,1)-(0,2,1)\bigr) = 2T(1,1,1) - T(0,2,1).$$ So you would need $f(2,0,1)=(1,0,0)$ to be equal to $$2f(1,1,1)-f(0,2,1)=2(3,2,7)-(2,1,-1) = (6-2, 4-1, 14+1) = (4,3,15).$$ Since $(1,0,0)\neq (4,3,15)$, there can be no linear transformation with the desired properties.

Added: The key here is that if you have vectors that satisfy some linear dependence, then any linear transformation must also respect that linear dependence. Since $(2,0,1) = 2(1,1,1)-(0,2,1)$, then any linear transformation $f$ will necessarily satisfy that $f(2,0,1) = 2f(1,1,1)-f(0,2,1)$. Since the values you have in (a) don't satisfy this, then you know there is no such linear transformation.

What if they do satisfy it, as in (b)? Here you have that $(0,2,-1)=2(1,0,2)+(-2,2,-5)$, so if $f$ is a linear transformation, then you must have $$f(0,2,-1) = 2f(1,0,2) + f(-2,2,-5).$$ Here, this does happen, since $$2f(1,0,2) + f(-2,2,-5) = 2(0,-2,1) + (3,3,1) = (3, -1, 3) = f(0,2,-1).$$ So, since they satisfy the equations, does there exist a linear transformation that does this? Yes! In fact, there are infinitely many different linear transformations that do this.

How can you find one? Well, this is where you can try doing systems of linear equations. You want to find a linear transformation $$f(x,y,z) = (ax+by+cz, \ell x+my+nz, rx+sy+tz)$$ such that if you plug in $x=0$, $y=2$, and $z=-1$, you get $(3,-1,3)$; and if you plug in $x=1$, $y=0$, $z=2$, you get $(0,-2,1)$. This gives you six equations in nine unknowns (the unknowns being $a$, $b$, $c$, $\ell$, $m$, $n$, $r$, $s$, and $t$), and it will have a solution. (The other equation you get when you plug in $x=-2$, $y=2$, and $z=-5$ is actually a consequence of the six we already had; taht is what we checked to begin with). So you can try solving that system and getting one possible map.

(When the three vectors you have are a basis, you will get nine equations in nine unknowns with a single possible solution, so that will be the linear transformation you want.)

But there is a way here to save yourself a lot of work: take your favorite two vectors from this set of three, and then find a fourth vector that, together with the first two, makes up a basis. Then define the linear transformation by picking an arbitrary value for the image of that one.

For instance, start with $(0,2,-1)$ and $(1,0,2)$. Notice that $(1,0,0)$ is not a linear combination of these two, so $(0,2,-1)$, $(1,0,2)$, and $(1,0,0)$ makes a basis. Since we can always find a linear transformation that maps $(0,2,-1)$ to $(3,-1,3)$, $(1,0,2)$ to $(0,-2,1)$, and $(1,0,0)$ to anything (say $(0,0,0)$ because it makes things easier), that one will do. Notice that, if $f(1,0,0) = (0,0,0)$, that tells you that $a=\ell=r=0$, so now we only have three more unknowns. In fact, if we can figure out what $f(0,1,0)$ must be, that will gives us automatically $b$, $m$, and $s$; and $f(0,0,1)$ will give us $c$, $n$, and $t$. Can we do that? Yes: because $(0,2,-1)$, $(1,0,2)$, and $(1,0,0)$ are a basis: we have $$(0,1,0) = \frac{1}{2}(0,2,-1) + \frac{1}{4}(1,0,2) -\frac{1}{4}(1,0,0).$$ So $$f(0,1,0) = \frac{1}{2}(3,-1,3) + \frac{1}{4}(0,-2,1) - \frac{1}{4}(0,0,0).$$ Then to find the values of $c$, $n$, and $t$, we use the fact that $$(0,0,1) = \frac{1}{2}(1,0,2) - \frac{1}{2}(1,0,0).$$

This also works when you are trying to find your function $f$ explicitly when the three vectors you have are a basis: try to write $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ as linear combinations of the three vectors (this amounts to solving three systems of three linear equations with three unknowns first; or to finding the inverse of a matrix; so it looks like the same amount of work, but because there are a lot of zeros it is often easier to solve these than the general one with $a$, $b,\ldots,t$). And then use the value that $f$ must have at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ to figure out the "formula" for $f$.

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@Arturo i think i'm understanding what you're saying. definitely the part about concluding things incorrectly. my only problem is that i am having trouble understanding the general approach to such a problem... before your response i was about to post: linear combination: $(a,b,c) = x \left( \begin{array}{c} 1\\ 1\\ 1 \end{array} \right) + y \left (\begin{array}{c} 0\\2\\1 \end{array} \right) + z \left ( \begin{array}{c} 2\\0\\1 \end{array} \right) = \left ( \begin{array}{c} x + 2z \\ x + 2y \\ x + y + z \end{array} \right)$ – ghshtalt Nov 20 '10 at 22:19
$\begin{array}{rrrr} x & +0y& + 2z &=a \\ x& +2y &+0z &=b \\x& + y& +z&=c \end{array}$ i then use the following equalities and substitutions: $x=-2z+a$ $y=\frac{1}{2} (b-x)$ $z=-(-2z+a)-[\frac{1}{2}(b-x)]+c$ $z=2z-a-\frac{1}{2}b+\frac{1}{2}x+c$ $z=2z-a-\frac{1}{2}b+\frac{1}{2}(-2z+a)+c$ $z=2z-z-\frac{1}{2}a-\frac{1}{2}b+c$ $z=z-\frac{1}{2}a-\frac{1}{2}b+c$ $0=-\frac{1}{2}a-\frac{1}{2}b+c$ but would this have been valid and sufficient? (i don't know if this actually shows that the system is inconsistent.....) – ghshtalt Nov 20 '10 at 22:19
@user3711: (cont) As to what you wrote, this doesn't really help, as far as I can tell, because I can certainly find values of $a$, $b$, and $c$ for which the system will have solutions. The system cannot be "inconsistent in the abstract". Now, for specific values of $a$, $b$, and $c$ you might be in better luck, but I think this is not the right way to approach this problem at first. It might be a good way to go once you know there is a linear transformation and you need to figure it out, but there are better ways of doing it. I'll add some words about it to the answer shortly. – Arturo Magidin Nov 20 '10 at 22:24
@user3711: Sorry: I misread your comment as "I think I'm not understanding", hence my harping on it... I can't edit any more, so sorry. – Arturo Magidin Nov 20 '10 at 22:36
@user3711: I hope the answer actually settles your questions; if this is not the case and you still have doubts, then don't accept yet and keep asking... – Arturo Magidin Nov 20 '10 at 22:36

Hint: Suppose we have a linear map $f:\mathbb{R^3}\to \mathbb{R^3}$. Say, $f_1(x,y,z)=a_1x+b_1y+c_1z$, $f_2(x,y,z)=a_2x+b_2y+c_2z$, $f_3(x,y,z)=a_3x+b_3y+c_3z$ where $f_1,f_2,f_3$ are the $x,y,z$ components of $f$. Substituting the values $(1,1,1),(0,2,1),(2,0,1)$ in this map, we get the corresponding values of the output if such a function exists. Now this gives us $9$ equations in $9$ unknowns, $a_i,b_i,c_i$ with $i=1,2,3$. If these equations form a determined system, we have a unique solution and hence a map. If the system is overdetermined, i.e. one of the equations is a linear combination of the others, then we have several such maps. If the system is inconsistent, i.e. the equations cannot be satisfied simulatenously, then we have no solution and no such map. So it just boils down to solving these equations simultaneously.

Edit: Adrian makes a useful remark, see his comment below.

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Your map $f$ doesn't produce vectors in $\mathbb{R}^3$. – Adrián Barquero Nov 20 '10 at 19:15
@Adrian: Sorry I misread the question. I have modified my answer. Hope it's fine now. – Timothy Wagner Nov 20 '10 at 19:27
Please you don't have to apologize. Also, the 9 equations in 9 unknowns that you get are just 3 systems of 3 equations in 3 unknowns, which is a little bit less daunting than the 9 by 9 system. =) – Adrián Barquero Nov 20 '10 at 19:34
ok i'm sorry for being so clueless, but is this in the right direction? as a linear combination i have: $(a,b,c) = x \left( \begin{array}{c} 1\\ 1\\ 1 \end{array} \right) + y \left (\begin{array}{c} 0\\2\\1 \end{array} \right) + z \left ( \begin{array}{c} 2\\0\\1 \end{array} \right) = \left ( \begin{array}{c} x + 2z \\ x + 2y \\ x + y + z \end{array} \right)$ – ghshtalt Nov 20 '10 at 19:37
@Adrian: Thanks, edited my post to reference your comment. – Timothy Wagner Nov 20 '10 at 19:37

For (a), your reasoning is incomplete. For example consider the problem of finding $f$ when

$f(1,0,0) = (0,0,4)$, $f(2,0,0) = (0,0,8)$ and $f(3,0,0) = (0,0,12)$

There is clearly a linear function $f$ that satisfies this (infinitely many, in fact), even though the vectors $(1,0,0), (2,0,0), (3,0,0)$ are linearly dependent.

If you show that the "output" vectors are linearly independent, however, your justification would be good.

Timothy Wagner's answer is slightly incomplete: you have to solve the 3 equations in 3 unknowns problem 3 times: once for each output dimension.

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If you define $f(x,y,z)=A\cdot \left( \begin{array}{} x \\ y \\ z \\ \end{array} \right)$, where the matrix $A$ is given for the 9 ecuations as:

$$ A\cdot \left( \begin{array}{} 1 \\ 1 \\ 1 \\ \end{array} \right) = \left( \begin{array}{} 3 \\ 2 \\ 7 \\ \end{array} \right)$$

You can find all coefficients of the matrix $A$.

So each relation as the previous we find 3 ecuation, in total 9 ecuations with 9 unknowns, then you problem is solves .

But remember this depends of that ecuations system can be solved !!!!

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There are three equations for each problem. One for part a is shown, the other two are similar using the other given relations. – Ross Millikan Nov 20 '10 at 20:06

Given the linear dependence you observed in the input vectors in (a), you could check if the same dependence applies to the output vectors. As a linear map requires that f(ax+b)=af(x)+f(b), if the output vectors do not satisfy the same dependence there is no map. This would save you trying to solve the equations.

On the other hand, if you prove the input vectors independent, there is guaranteed to be a linear map.

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A brute-force approach: try to write using different notation: e.g., say you are searching for a matrix $A$ that satisfies $AX=Y$, where the columns of $X$ are given by the different vectors that are arguments to your $f$, and the columns of $Y$ are the different vectors that you obtain on applying $f$. Then, see if you can "solve" to get $A=XY^{-1}$.

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