in-exercises-17-20-show-that-t-is-a-linear-transformation-by-finding-a-matrix-that-implements-the-ma

Question

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. 
$$
T\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\left(0, x_{1}+x_{2}, x_{2}+x_{3}, x_{3}+x_{4}\right)
$$

Michael Jacobsen · Accepted Answer

in this video, we&#x27;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&#x27;m going to let x be the vector containing X one through X four as components and I&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s a zero x four here zero x four here zero x four here to more cleanly explained why we picked 0001 where there&#x27;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&#x27;m going to say t of X is equal to some matrix, which I&#x27;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&#x27;ll have 0100 0110 then 0011 Finally, 0001 Now, this is our matrix A. This is the Vector X, and we&#x27;ve written the transformation t of X in the form of a Times X. It&#x27;s no different than this transformation were just using a new notation a matrix, times a vector and this proves t is linear.

Ashley Boni · Answer

so we won&#x27;t show that he is a linear transformation by finding a matrix, um, such that when we multiply it by vector X with four components, we get this as an output. So this transformation is mapping a vector in our four to vector or well, just the number in are a real number because this is just a number. If we plug in ex one next three x for so we know we need a matrix to multiply x one x two x today, that&#x27;s four. So that means we need to have four different columns one to multiply each of these, uh, vectors. But it&#x27;s being mapped to just are so we only need one room so we can stop. So we need four different components that multiply at claw on x two x three x for and we need it to give us to x one plus three x three minus or explore. So the coefficient for X one is two coefficient. Rex to zero coefficient for X three is three and the coefficient we need on export is negative. For so ignoring these lines. This is our matrix that we need to multiply

Ashley Boni · Answer

So we want to, uh, find a vectors. We&#x27;ll call it a such that when we multiply it by ex one next to it. Maps it too. This vector here on the right. So we want this to be written as a X to show that it&#x27;s ah, linear transformation. So we have a matrix A and we don&#x27;t necessarily know the size yet. We&#x27;ll figure that out. No, we want to multiply it by x one x two and we want to get a vector with first component two x to minus three x one x one minus four x two zero in x two. So since our two component factor used to be mapped to a four component vector, this needs to be a four by two matrix. So we don&#x27;t usually draw lines like Listen, Major sees. But I&#x27;m gonna do it just so we know how many spots we have the fill in. So we know that this whole first column is what&#x27;s multiplying the X one and I could call on his multiplying the X to coordinate. So to get this to x two minus three x one x one has to be multiplied by negative three and X two has to be multiplied by two For the second component, X one has to be one supplied by one. And then we have minus four times x two. So you 20 We need to multiply both components by zero to ensure that we always have a zero in this third spot. Lastly, Thio, just get the output output. The second component, Um, our vector x for this fourth row here, we need to multiply X to buy one, and we need to get rid of X one. So we&#x27;re going to multiply that by zero. So this here is our major X a.

Michael Jacobsen · Answer

in this example, we have a transformation. T that maps the three dimensional space are three into the two dimensional space are too, and the way the transformation is implemented is it takes x one x two x three and maps it to this for century. And then the second entry is X two, minus six x three. Let&#x27;s re express what this notation could be. Also written as if we let x be the vector containing x one x two and x three. Then this becomes t of X equals. Let&#x27;s also change the notation here and write this as a column vector will be x one minus five x two plus four x three for the first entry, and the second entry is next to minus six x three. Now let&#x27;s see what we can do with this particular vector. Here we have three variables to consider. X one x two x three. Let me say that this is equal to x one times a vector plus x two times a vector plus x three times 1/3 vector. So what we&#x27;re doing here in this step in green is we&#x27;re reversing the process of vector additions and scaler multiplication To do that for X one, go to the column here that contains X one and pull coefficients will have a 10 then for X to its negative 51 And for X three coefficients are four and negative six. Now that we&#x27;ve taken this step, we also note that a linear combination in this form can also be written as a matrix times a vector. The vector would be the vector containing x one, x two and X three and the matrix is formed by just pulling each one of these vectors. So it&#x27;s 10 negative. 51 and four negative six. So if we let a be 10 negative 51 four negative six. The same matrix we found then t of X, where we started with is going to be equal to eight times X where this is our matrix. A. This proves that t is linear since we have a standard matrix A and since we can write t of X equals a X where the Vector X was completely arbitrary

Ashley Boni · Answer

So we want to check if the following transformation is linear. Um, so we need to check some properties of oven, your transformation and try to disprove one s. So if we can do that, we can show that it&#x27;s not linear. So seeing the absolute value in the second coordinate here tells me that I might want to pick on X two. That is one positive one negative to tryto mess up the signs. Um, so let&#x27;s pick. Ah, vector X one x two take you for, uh, say 01 and let&#x27;s pick another one. Why warn why, too? So the No. One negative warm. And let&#x27;s try to check the property t of X plus y has to be equal to t of X plus t. Of what? I don&#x27;t see if this is true t of X plus y. So this one I&#x27;m calling X This woman calling why is equal to t of first coordinate? Zero plus one is one. Ah, second one minus one is zero So t of 10 Let&#x27;s just plug this into the formula were given. So we have four times one minus two times zero comma three times the absolute value of zero, which gives us a point of fours. So now let&#x27;s calculate ft of accent E of Walk So to your ex, plus to you, why is equal to so first let&#x27;s do to your ex. Um, we have four times zero minus two times one comma, three times absolute value of one plus t of y is equal to four times one minus two times negative, one comma, three times absolute value of minus one. So if we add these two things together, um, let&#x27;s do this in another color. Here we have zero minus two, which is negative to calm. The three plus four plus two gives us six and three times at survey of negative one mystery. So if we add these two points together, six minus two is four and three close three of six, which is not equal to our original t of expose. Why so, therefore not what year

Ashley Boni · Answer

eso were given a vector X view one and three to and we have a defined, uh, function t of x where X is a vector is equal Thio x one be one plus x to be to go where X is just the vector x one next to V one is you quarter negative to five and he too is equal to seven Negative three. So it&#x27;s really write this in matrix for so Tia Vex can be written as the vectors If you want be too Times X one extra right, cause thes air columns So this is a two by two matrix and this is our vector X, which has the following. Uh, he&#x27;s underlined components so we can rewrite to you Vex like this. And so now let&#x27;s Ah plugin RV wannabe too. The one was minus 25 He, too was seven minus three. Then we have x one x two. So, um, we&#x27;ve done everything to answer the question. So if we call this matrix A, we found a matrix. A such that t of X is equal to a X, So the system matrix we need

Problem

In Exercises $17-20,$ show that $T$ is a linear t…

Problem 17 Hard Difficulty

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.
$$
T\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\left(0, x_{1}+x_{2}, x_{2}+x_{3}, x_{3}+x_{4}\right)
$$

Answer

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

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Grace H.

Catherine R.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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

Linear Algebra and Its Applications

Grace H.

Catherine R.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Grace H.

Catherine R.

Caleb E.

Maria G.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications