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I have a $T\colon\mathbb{R}^2\to\mathbb{R}^2$ linear operator given by $$T(x,y) = (2x - y, -8x + 4y)$$

How can I tell if the vector $(1, -4)$ is in $R(T)$?

Ok so I set everything into a matrix:

$$\left( \begin{matrix} 2 & -1 & 1\\ -8 & 4 & -4\\ \end{matrix}\right) $$

I row reduced it and found that it was linearly dependent. So I'm going to assume that that means it's is in R(T).

Edit: I'm not sure if I did that right because it isn't set to 0.

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    What linear algebra do you know so far?2011-04-21
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    He's not making fun of you; he's trying to find out how much linear algebra you know in order to know how to answer your question. There are "low tech" and "high tech" ways of solving this problem, depending on how much linear algebra (and at what level) you know. For example, one can answer by invoking the Dimension Theorem to deduce that the range has dimension $1$, and then easily deduce whether the vector $(1,-4)$ is in the that one dimensional subspace. Or it can be done by solving a system of linear equations. He's trying to help at an appropriate level.2011-04-21
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    I was only kidding about the "making fun of me" part, sorry. I'm currently trying to understand Kernel and Ranges for linear operators.2011-04-21
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    I am adding this comment here (for now) because you deleted the other question. If what you have is a problem taken from a book/assignemnt, it is perfectly fine to *quote it* (say, in a quote box by preceding the first character in the paragraph with `> `, and then write down your own comments and describe your doubts. If you are confused about what a problem is asking you to do, then that's the way to go. But your recently deleted question was an absolute mess: it started halfway through, and you never even said what "the problem" you were trying to solve *was*.2011-04-26

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Preface added given the confusion exhibited by the OP in the comments.

We are trying to figure out whether or not there exist real numbers $x$ and $y$ such that $$T(x,y) = (1,-4).$$ (That's what it means for $(1,-4)$ to be in $R(T)$; it is in the range if there do exist such $x$ and $y$; it is not in the range if no such $x$ and $y$ exist). This is the same as asking whether or not there exist real numbers $x$ and $y$ such that $$(2x - y , -8x + 4y) = (1,-4),$$ which is the same as trying to solve the system of linear equations $$\begin{array}{rcccl} 2x & - & y & = & 1\\ -8x & + & 4y & = & -4. \end{array}$$

That is: the linear transformation can be represented by the $2\times 2$ matrix $$\left(\begin{array}{rr} 2 & -1\\ -8 & 4 \end{array}\right),$$ in the sense that $T(x,y) = (a,b)$ if and only if $$\left(\begin{array}{rr} 2 & -1\\ -8&4\end{array}\right)\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}a\\b\end{array}\right).$$

So, $(1,-4)$ is in the range of $T$ if and only if there exists $x$ and $y$ such that $$\left(\begin{array}{rr} 2 & -1\\ -8 & 4\end{array}\right)\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{r}1\\-4\end{array}\right),$$ which as you write is equivalent to solving the system with augmented matrix $$\left(\begin{array}{rr|r} 2 & -1 & 1\\ -8 & 4 & -4 \end{array}\right).$$ You will not be able to solve this by "inverting the matrix", because the matrix that corresponds to the linear transformation is not invertible. But you should be able to use Gauss-Jordan elimination (row reduction) to figure out whether or not this system has a solution.

Why are we trying to solve the system? Because finding a solution $(x,y)$ to this system is equivalent to finding a vector $(x,y)$ such that $T(x,y)=(1,-4)$. Any solution gives such a vector. And such a vector is a "witness" to the fact that $(1,-4)$ is in $R(T)$ (because $R(T)$ is the collection of all elements $(a,b)$ of $\mathbb{R}^2$ for which there exist a vector $(x,y)$ such that $T(x,y)=(a,b)$).

So: if the system has a solution, then $(1,-4)$ is in the range and the solution of the system is a witness to the fact that it is in the range. If you plug any solution to the system into $T$, you should get $(1,-4)$.

If the system is inconsistent, then that means there are no solutions, so $(1,-4)$ is not in the range.

Added. So you figured out by row reduction that the system is equivalent to the system with augmented matrix $$\left(\begin{array}{cr|c} 1 & -\frac{1}{2} & \frac{1}{2}\\ 0 & 0 & 0 \end{array}\right).$$ This is the matrix of a system of linear equations; you should be able to find the solutions to this system. Find a solution to this system: that means finding a specific value of $x$ and a specific value of $y$ that is a solution to this (and therefore to the original) system of linear equations. Those specific values of $x$ and $y$ that solve this system, which also solve the original system, should (if you've done everything right), when plugged into $T$, give you $(1,-4)$ as the output. And that will prove explicitly that $(1,-4)$ lies in $R(T)$.

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    After row reducing it I get: x - 1/2y = 1/22011-04-21
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    @Cascadia: I don't understand what you wrote (try using mark-up and proper spacing). But note that this system will either have *no* solutions, or *infinitely many* solutions (because the matrix is not invertible). However, you are not really interested in finding *all* solutions, you are only interested in finding *one* solution (or showing no solution exists). Once you find a solution, plug those values into $T$; if you did everything right, applying $T$ to the solution should give you $(1,-4)$, thus proving ("witnessing") that $(1,-4)$ is in $R(T)$.2011-04-21
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    $$ \begin{matrix} 1 & -1/2 & 1/2\\ 0 & 0 & 0\\ \end{matrix} $$2011-04-21
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    that's the matrix I get after row reduction.2011-04-21
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    @Cascadia: So that tells you that the system is equivalent to the *single* equation$$x - \frac{1}{2}y = \frac{1}{2}.$$ Does this equation have any solutions? Yes; just pick one, and plug those values into $T$ to double check that those values do indeed show you that $(1,-4)$ is in the range.2011-04-21
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    So do I just pick a random vector like (5, 4) and plug it in?2011-04-21
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    @Cascadia: No, you don't pick "a random vector". You pick **a solution to the system**. You want a vector that *solves* the system you just finished row-reducing. You pick a solution to the equation $x - \frac{1}{2}y=\frac{1}{2}$. You pick a value for the free variable (in this case, $y$), and you obtain the corresponding value for the dependent variable (in this case $x$), and you get a solution to the system. *That* is what you plug into $T$. **Not** a "random vector."2011-04-21
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    So it would be 1/2? I stil do not understand, sorry.2011-04-21
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    @Cascadia: Please stop and think. You are almost there, but *think things through*. You need to pick **a solution to the equation.** "1/2" is not a solution to the equation. A solution to the equation is something of the form "$x=$`blah` and $y=$`blablah`". You need to find a value of $x$ and of $y$ that makes the equation $x - \frac{1}{2}y = \frac{1}{2}$ **true**. Because finding a vector $(x,y)$ that satisfies $T(x,y) = (1,-4)$ is **equivalent** (thanks to the row reduction you did) to finding a vector $(x,y)$ that satisfies $x - \frac{1}{2}y = \frac{1}{2}$.2011-04-21
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    I'm trying to think this through but I'm stuck. I've tried plugging (1, -4) back into the original transformation and (6, -24) but that doesn't work because if I plug into the equation I don't get 1/2.2011-04-21
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    @Cascadia: You are trying to find an $x$ and a $y$ such that $T(x,y) = (1,-4)$. Since $T(x,y) = (2x-y,-8x+4y)$, this is the same as trying to solve the system $$\begin{align*}2x-y &= 1\\ -8x+4y &= -4.\end{align*}$$ So now we are trying to solve the system. We do this by row reduction. We end up with the matrix $$\left(\begin{array}{rr|r} 1 & -\frac{1}{2}&\frac{1}{2}\\0&0&0\end{array}\right).$$ *Find a solution to the system using this.* Once you find a solution to the equations, **that's** what you plug into $T$ to get $(1,-4)$, because the system is a way of figuring out what to plug into $T$2011-04-21
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    I still don't understand what I'm doing but thank you for your help and patience.2011-04-21
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    @Cascadia: I've rewritten the answer. But tell me this: **Why** were you trying to find the inverse of the matrix? To find a solution to the system. Why were you trying to find a solution to the system? Because a solution to the system would give you values which, when plugged into $T$, give you $(1,-4)$. The fact that you couldn't solve it using a matrix inverse does *not* change that this is a way to solve the problem. It just changes *how* you can find solutions to the system of equations you were looking at.2011-04-21
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    @Cascadia: Let's pretend that you are *not* taking Linear Algebra. Note that the second equation is obtained by multiplying every term in the first equation by $-4$. So the second equation gives no further information, imposes no further constraints. If you find a solution of the first equation, it will *automatically* be a solution of the second. Now look at the first equation, and (for example) set $x=0$. Then $y=-1$. So $(0,-1)$ works. Recall that it is automatically a solution of the second equation, but if you are worried, check by plugging in.2011-04-21
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Check whether there is a solution to the system of $2$ equations $2x-y=1$, $-8x+4y=-4$.

In response to a comment, there are many ways to try to solve a system of equations. In this case one can easily spot a solution. You may also have learned about row reduction, though that is serious overkill here.

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Notice that this is equivalent to asking if the vector $<1,-4>$ is in the span of the columns of $T$.

If row reduce this system to the reduced row echelon form of the augmented matrix, you will get all zeros in the last row, so the system is consistent.

Because there are no pivots in the last row you then conclude that there are an infinite number of solutions (as any numbers $x$ and $y$ multiplied by zero will result in zero).

Also if you notice the second equation is really the first equation multiplied by -4 so really it is just the same linear equation.

Ben

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    @David: "as any numbers $x$ and $y$ multiplied by zero will result in zero" does not make sense here in so far as giving you infinitely many solutions. Also, no, you aren't not "just talking about $\mathbb{R}$." $\mathbb{R}$ and "1 one dimensional subspace of $\mathbb{R}^2$" are not the same thing (though they are isomorphic). Three floors in a 10 story building are not a three story building.2011-04-21
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    @ArturoMagidin what I really meant to say was the last row of the augmented matrix is 0x + 0y = 0, that's why there are an infinite number of $x's$ and $y's$ that fit this linear combination. Ok sorry my bad the second bit was not very clear as well, what I sort of meant to say is those two lines are really the same line.2011-04-21
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    @David: Still not accurate (or correct). The reason there are infinitely many solutions is because there are infinitely many solutions to the associated *homogeneous* system; or equivalently, because you have a consistent system with free variables (fewer pivots than rows). Just having a row of zeros does not guarantee, in and of itself, infinitely many solutions. The fact that the last row corresponds to the equation $0x+0y=0$ tells you that the row does not provide any constraints on the solutions, not that there are infinitely many solutions.2011-04-21
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    @Arturo Ok what about this: The columns of $T$ are linearly dependent so there is not only the trivial solution to the homogeneous system which means that $T$ is not one to one? I.e. there is more than 1 solution to a particular R.H.S. ?2011-04-21
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    @David: There is no need to "sign" your posts: your name appears on the bottom right as a matter of course.2011-04-21
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    @David: Closer and much better: there are either *no* solutions or infinitely many solutions to a particular right hand side.2011-04-21
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    @Arturo Also the last row of the coefficient matrix reduces to all zeros so $T$ is not onto, i.e. there exists a R.H.S in $\mathbb{R}^2$ such that no vector in the domain maps to this vector.2011-04-21
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    @David: True, but irrelevant to the problem at hand.2011-04-21
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    @Arturo I just stated this because if one realises this fact then without calculating the determinant there is the possibility that there may be no solutions for a given R.H.S.2011-04-21
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    OK so if I get something that linearly possible like x-1/2y = 1/2 then it is in the range?2011-04-21
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    @Cascadia Regarding your answer just because the columns of $T$ are linearly dependent does not tell you anything about a specific R.H.S.; it just says that $T$ is not one to one.2011-04-21
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    @David: There are plenty of ways one can realize it without computing the determinant. Still irrelevant to the question at hand.2011-04-21
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    @Cascadia: No; what you write does not even parse. What you found was that the set of all solutions to $T(x,y) = (1,-4)$ is the same as the set of all solutions to the equation $x - \frac{1}{2}y = \frac{1}{2}$ (**do** stop writing "1/2y", which means $\frac{1}{2y}$; you can use LaTeX by using dollar signs, e.g., `$\frac{1}{2}$` to create $\frac{1}{2}$). *Find solutions to that equation*. That equation is **trivial to solve**, so solve it.2011-04-21
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    David: I am not sure I understand your question, but note that a system with more rows than columns may still have a unique solution: consider a 4x3-system in which the fourth row is dependent on the other three, and the first three rows are independent of each other. Check that rank-nullity agrees with this.2011-04-21
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    @gary You're saying that it's possible to have more equations than unknowns and still have a unique solution? Yeah that's right I just take the lines $y=1$, $x=1$ and $y=x$ and these meet at $(1,1)$. How's this related to what I posted earlier?2011-04-21
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    David: I was thinking of the comment in your answer saying that a column of all zeros then the system is inconsistent. You may have been referring to the particular conditions of this matrix, tho, and not making a general statement about matrix systems. If this last is so, or you meant something else, please ignore, and sorry.2011-04-21