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In Exercises $17-20,$ show that $T$ is a linear transformation by finding a matrix that implements the mapping. Note that $x_{1}, x_{2}, \ldots$ are not vectors but are entries in vectors. $$T\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\left(0, x_{1}+x_{2}, x_{2}+x_{3}, x_{3}+x_{4}\right)$$

$T(\mathbf{x})=\left[\begin{array}{llll}{0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} \\ {0} & {1} & {1} & {0} \\ {0} & {0} & {1} & {1}\end{array}\right]\left[\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{3}} \\ {x_{4}}\end{array}\right]$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 9

The Matrix of a Linear Transformation

Introduction to Matrices

Missouri State University

Baylor University

Lectures

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in this video, we're going to be showing this transformation t displayed here is linear. Well, the first step in writing out the solution to this would possibly be to re express this in a different notation. I'm going to let x be the vector containing X one through X four as components and I'm going to write this rage inside in our standard vector notation So vertically will have zero then x one plus x two. Then we have x two plus x three and finally x three plus x four. So this is specifically what T does to a vector X and when we see this, this is really just nothing but a four by one vector. But if we reverse engineer what matrix was multiplied by A with the right vector will be able to determine a standard make tricks A of t. So let's begin by running the following. We can put in an X one times a specific vector, Then down the line Take x two times a vector plus x three times a vector plus x four times some specific vector. So what we're reversing is a scaler multiplication. Here, here, here and here along with all these vector additions that we see here. In order to reverse engineer these operations, we just go teach column here at the way we organized and pull coefficients. So for the X one variable, we have a zero in entry one or Row one and Row two, we have a coefficient of one. Then x one never appears again. So the next coefficients are all zero. Now for X two, we have coefficients zero first, then 11 corresponding from here and then zero again. Now for X three. We have zeros here, so I'm going to start with a 00 Then the coefficients coming from here are both one in one. Likewise, for X four, we have 000 and one. If we really would like, we could have said, there's a zero x four here zero x four here zero x four here to more cleanly explained why we picked 0001 where there's a coefficient one here. But I would consider that optional and more for seeing the process. Well, now that we have this format, which is a linear combination, we also know that linear combinations can be written in the form of a matrix times a vector. So the next step, I'm going to say t of X is equal to some matrix, which I'll place here times a vector which contains the values x one, x two x three and x four. Now the way we form the Matrix is we just grab each one of these columns that we see displayed here. So we'll have 0100 0110 then 0011 Finally, 0001 Now, this is our matrix A. This is the Vector X, and we've written the transformation t of X in the form of a Times X. It's no different than this transformation were just using a new notation a matrix, times a vector and this proves t is linear.

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