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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? –  Amit Kumar Gupta Apr 21 '11 at 1:26
    
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. –  Arturo Magidin Apr 21 '11 at 2:30
    
I was only kidding about the "making fun of me" part, sorry. I'm currently trying to understand Kernel and Ranges for linear operators. –  Cascadia Apr 21 '11 at 2:34
    
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. –  Arturo Magidin Apr 26 '11 at 4:08

3 Answers 3

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 = 1/2 –  Cascadia Apr 21 '11 at 2:43
    
@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 $R(T)$. –  Arturo Magidin Apr 21 '11 at 2:45
    
$$ \begin{matrix} 1 & -1/2 & 1/2\\ 0 & 0 & 0\\ \end{matrix} $$ –  Cascadia Apr 21 '11 at 2:52
    
that's the matrix I get after row reduction. –  Cascadia Apr 21 '11 at 2:53
    
@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. –  Arturo Magidin Apr 21 '11 at 2:54

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. –  Arturo Magidin Apr 21 '11 at 3:20
    
@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. –  user38268 Apr 21 '11 at 3:22
    
@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. –  Arturo Magidin Apr 21 '11 at 3:24
    
@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. ? –  user38268 Apr 21 '11 at 3:31
    
@David: There is no need to "sign" your posts: your name appears on the bottom right as a matter of course. –  Arturo Magidin Apr 21 '11 at 3:31

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