define-t-mathbbp_2-rightarrow-mathbbr2-by-tmathbfpleftbeginarraycmathbfp0-mathbfp1endarrayright-for-

Question

Define $T : \mathbb{P}_{2} ightarrow \mathbb{R}^{2}$ by $T(\mathbf{p})=\left[\begin{array}{c}{\mathbf{p}(0)} \ {\mathbf{p}(1)}\end{array}ight] .$ For instance, if $\mathbf{p}(t)=3+5 t+7 t^{2},$ then $T(\mathbf{p})=\left[\begin{array}{c}{3} \ {15}\end{array}ight]$ 
a. Show that $T$ is a linear transformation. [Hint: For arbitrary polynomials $\mathbf{p}, \mathbf{q}$ in $\mathbb{P}_{2}$ , compute $T(\mathbf{p}+\mathbf{q})$ and $T(c \mathbf{p}) . ]$
b. Find a polynomial $\mathbf{p}$ in $\mathbb{P}_{2}$ that spans the kernel of $T,$ and describe the range of $T$

Michael Jacobsen · Accepted Answer

in this example, we have a transformation t and it&#x27;s quite an unusual transformation that we&#x27;re going to be investigating. What it does is it takes a degree to at most polynomial, they come from p two, and it maps it into our to. So imagine a function that takes a Paul no meal and gives you a point in the plane. That&#x27;s effectively what we&#x27;re doing. And so T R P is going to be p of zero for the first cornet and p of one for the second coordinate. Let&#x27;s now do a full investigation of this transformation. T Well, the questions we might ask for such a transformation is, Is this transformation? T linear? It turns out that it is, but we need to show this. So let&#x27;s start by showing that T is linear. Let&#x27;s pull up the definition of linearity to guide us. So for the definition of linearity, we have two equations to show. Let&#x27;s start with this equation here. So to that end, begin by saying, Let P and cue be in p two. We&#x27;re using P and Q because that&#x27;s suggestive of the typical elements from art domain. The set of Paul no meals. Then we need to consider the equation t of P plus Q, which is analogous to this left hand side. And our goal is to convert vector addition of the polynomial into vector addition of the vectors in our to So this is how we&#x27;ll do it. First, use the definition of tea, which does the following for whatever a polynomial you have evaluated at zero and one. And that&#x27;s your 1st 2 entries for the defector in our to, So we&#x27;ll be writing P plus Q Evaluate at zero, then people Askew, which is evaluated at one. The next steps are function notation and function notation. If we have the sum of two functions evaluated at a 20.0 here, it becomes P of zero plus Q of zero, then p of one plus Q of one down below. Now that we&#x27;ve used up function notation, notice that we can break down this as this sum of two vectors will be p of zero p of one for the first factor, plus que of zero and Q of one for the second vector. Now it&#x27;s a good time to pause and look at what we have from the first vector. This is really just TMP. So we can say that this is equal to T at P. The plus here copies to a plus here. And the second vector is nothing but a t at Kew. And that means we have broken down the vector addition on the inside to have the sum of vectors where we&#x27;re looking at the images to have p anti of Q. So part one is complete For the second part. I&#x27;m going to continue to work with P, but we also need a scaler see. So let&#x27;s start off by saying, Let see be in our. So now that we have a scaler, let&#x27;s look at this left hand side of the equation. We can say then t at sea Times p will be equal to and before we get too far ahead, our goal is to do the following. We start with a C on the inside. If we can convert it to the outside, our work is complete so for the first step will take will use the definition of the transformation. T This is equal to see time&#x27;s P evaluate at zero and c times P evaluate at one. So all I have done that this step is in place of the function p that we saw here and here. I&#x27;m using CPI. So they go here and here. So next will use function notation just as we did going here to here To say that multiple of a function evaluated zero is the same as C times PS zero. Similarly, the second entry ISI times p at one. So notice on the first equation, we had these elements here and we broke down into vector addition. Now we have these elements and we&#x27;re going to break this down into a statement of scaler multiplication. We can put a C on the outside and placed p of zero p of one on the inside, but we know p of zero p. A one is the vector TMP. So this is now equal to C times T at P and this equation is complete notice, though we&#x27;re not done until we stay our conclusion. So this is our conclusion. This shows the work above that t is linear. And now that we know that t is linear, there&#x27;s a lot of wonderful properties we could look at, but we&#x27;re going to cut our analysis a little bit short and just look at two more. Particular subspace is We want to look at the Colonel and the Range, starting with the colonel. So let&#x27;s go to, ah, clean sheet and investigate the Colonel of the Transformation T. First, let&#x27;s say what the colonel is. The kernel of tea is a set of all vectors. P such that t of p is the zero vector in our two. So let&#x27;s see what that would mean. If t of P is equal to the zero vector, then that would imply P of zero is equal to zero and p of one is equal to zero. And this is because t. Of p by definition is PS zero and p of one. But we&#x27;re setting this equal to zero vector. So I set both entries equal to zero at the same time. So if p of zero is equal to zero and p of one is equal to zero, then this would next imply the following Using ideas of algebra, we can say that p AT T is going to be equal to T minus zero where I&#x27;ve taken away the century times t minus one. Subtracting away this element here or to simplify p of t is equal to tee times T minus one. And this is for all t in our So this is our function and notice that we can do other things too. This function If we were to multiply by a number three then we still have t A P is equal to zero since changing a three here. If we put in a zero for tea or a one for t, we still get this equal to zero, no matter what scaler we put in front. So this has implicate implications for us. It tells us that the kernel of the transformation T is going to be equal to the span of tea times t minus one, the single vector. And this is because the span of this single vector is all in your combinations, which would just be constant multiples. Finally, the range of the transformation T is easier to look at notice that we set for T A. P. The first entry is PS zero, but that could be any real number whatsoever if we pick a generic polynomial p. Similarly, for P of one So this tells us that the range of tea is all of our two. So for this transformation, T we found that it&#x27;s linear. The colonel is the span of this one polynomial, and the range is all of our two.

Doruk Isik · Answer

in this problem were given a lenient information and were arrested by two Paul Newman&#x27;s that spent this, uh, space and river as to find the range off this transformation. So what we know is this since we&#x27;re operating and be too, it means that any any bull no milk what, their degree off listen very carefully to and it satisfies. P zero is equal to zero will be in the journal, solicits beautiful little purity lesson is a plus B t plus c d. Screw you Listen like this all the mill at zero. So that would be too right. And the transformation would then be he comes to Aye, Aye. Since reelecting this holding the leg zero point in this transformation. So we want this moment to be zero. Right? So then any pull Lemuel off the form uh, Bt Plus he describe well being the colonel, since we want this to be cooked zero. So basically here we&#x27;re taking a to B there. So many people know me. Look, this form will be in the colonel so we can in general, right this one or defined to pull no Mills s. Let&#x27;s define people on S t l s defined p two as t squared and we&#x27;re gonna say since people and then people early, nearly 10 minutes, they&#x27;ll spend the space so it gets a t you scrabbling he faces. So find a ring. Let&#x27;s again you more general form, right? Are formulas a plus B T plus C Ty&#x27;s where? So if you participles the military in art information, what we end up is this. So we&#x27;re gonna let this pole a meal at zero again, We&#x27;ll get a and again, Well, they&#x27;re like this poll number, like zero this most of you. So this is basically an image off this whole army. Once we put it in, put that in this linear confirmation. Right? So this means that range is Hey, hey, where a is any number, Any riel number. So he&#x27;s just an arbitrary number, and it is a really little

Michael Jacobsen · Answer

in this problem, we&#x27;re going to be showing that the transformation t that maps square matrices of size to buy two into the set of two by two square major sees defined by TV equals a plus and set transpose is linear now. The words we want to show t is linear To do this. The definition Lanier is a very important one in linear algebra, so it should immediately pull up that definition in the book. That definition tells us that the transformation is linear. Then it must solve two equations. Let&#x27;s look at Equation one. First. We must show that tea at a vector U Plus a Vector V is equal to U plus TV. In our case, U and V are going to be square, Major sees. So let&#x27;s denote them by A and B and for our purposes, maybe. Let&#x27;s first look at the ray and side of equation one t at a plus to yet be by the definition of our transformation, T would be equal to any place that you transposed, plus B plus be transposed, so that would be our right hand side of the equation. Let&#x27;s look at the left hand side now will manipulate the left hand side of this equation. Trying to get to this format here we can get to hear their work will be concluded that indeed, t A plus B is equal to t a plus t a p. Let&#x27;s start our solution over here. First, we need to define all of our objects in linear algebra. So will make a statement that says, Let A and Be Being in M two by two, which is the domain of the transformation team. Then, looking at the left hand side of the equation, we&#x27;re ready to begin. Well, look at the addition of the matrices A and B inside the parentheses and right T Yet a plus B equals next. We have two things to look at when we define this transformation. We took a matrix, any matrix whatsoever, output itself. Plus it&#x27;s transposed in our situation. The matrix that we&#x27;re going to be using is the sum of A and B will be equal to itself. Plus, it&#x27;s transposed. So was a basic format. This is going to be the evaluation of Tia. A plus B. Inside will have a plus B here and a plus B here a swell, so that starts it soft. The next thing to do is we&#x27;re trying to get to this format of a plus a trance, PLO&#x27;s plus people speed transposed. But as it stands, a plus B is stuck inside the parentheses due to the transpose operator. So now we need to look at properties of transposed in order to get to through this problem. One important property of transpose is is that we have a summer Metris ease and we take the transpose of that song will be equivalent to taking the transpose of eight first, the transpose of be first and then adding the result. In other words, these Prentice&#x27;s sees can be dropped by writing a transposed plus be transposed. Then we can drop these practices here to say Ta Plus B is now equal to a plus B plus a transposed plus be transposed and at this stage were well on her way to the solution. Since we do have a plus, a transposed as we&#x27;re expecting, and now we have B plus be transposed as well to work with here and here, which gives us this format. The rest of the problem is really more of a matter of substitution. So it&#x27;s right as our next step A plus, a transposed associate id together plus B plus be transposed. Associate together is equal to the prior step just by rearrangement and association. Then this whole thing will be equal to t a. So if we verified that t A results in a plus a transposed plus TFB this verifies equation one since two a plus B can now be expressed through our transitive any of the qualities as t a plus t a p. Now we&#x27;re on to the second step of verifying linearity for this problem in the step. One interesting feature we have is we&#x27;re not looking at two different vectors from the domain vector space were stead looking for a single vector to work with and a scaler or rial number C. We&#x27;re going to start our off our problem this time by saying, let see be any real number and lit in this case a matrix A be in the vector space off to buy two major sees and sub two by two. So our job is much like what we did before. We want to look at the right hand side of this equation and see just what that would look like for us it b c Times T A. That would result in the scaler see times the result of a plus a transposed. However, this time we need to indicate parentheses and right see times a plus, a transposed. This work here in the scratch work side of this solution tells us if we can get to this stage that once again, we&#x27;ll go back to see times today by substitution. So let&#x27;s start on the left hand side of that equation and right T at See Time&#x27;s A. This will be equal to the input, plus the transpose of that same input. So insert C times a here and C times eight Here is well, now we have to once again use properties of transposed. This time, one property of transpose says, If we&#x27;re dealing with a skill or multiple of a matrix and transposing the entire result, don&#x27;t be the same result as first taking a transposed and then multiplying that entire metric spicy itself in symbols, we would have see times a just copying from the first part of this term, plus C times a transposed. This is still not quite what we&#x27;re looking for. We&#x27;re trying to get See Time&#x27;s April&#x27;s a chance pose, but we&#x27;re getting very close at this stage. In fact, the next thing to do is just as an ordinary algebra. We can factor out the scaler, see and say that this is C Times A plus, a transpose now through substitution. A plus eight transposed is t A. So you could say that this is now see times t at a. That verifies part two of this solution. Whoever work is still not done since in linear algebra, it&#x27;s always very critical to state our all of our conclusions will say the following This work shows that t is linear, since we were able to verify equations one and two from the definition of linearity that finishes part A of this problem. For a party of this problem, we&#x27;re going to do something a little bit different. We&#x27;re going to assume that be is equal to be transposed. Where be is in M two by two, which just says B is once again a square matrix of size. To buy two. For this problem, we want to show or maybe a better way to say it is solved. T a equals B for A. In other words, if we know that B is equal to be transposed, meaning that matrix B a symmetric, we want to know if there&#x27;s some matrix A from the domain such that t a would happen to be equal to be before we get too far. With this problem, let&#x27;s take a look at what this equation here might mean. If t a is equal to be, then we would have a plus. A transposed equals B itself, since that&#x27;s our definition off what the transformation T does to any Matrix? A. So we&#x27;re trying to solve this equation. However, we need to find some matrix a such that a plus it&#x27;s transpose is now equal to be. This can be a tricky thing to do with the given information. But one piece of information we have not yet used in this problem is that B is equal to be transposed. If we apply this information correctly, we should be able to guess what that matrix a should be. So one thing to note is, if we put the Matrix B here and here we would come up with the equation where the be transposed would just result in the B from the given information of this problem. So it&#x27;s set a equal to be. Then, of course, we have two copies of a A and it&#x27;s transposed, so it&#x27;s put in a factor of 1/2 here as a scaler. Then if that were the case, we would get in place of a 1/2 B plus in place of a transposed 1/2 be transposed and this must be equal to be somehow some way where once again they&#x27;re trickle be be transposed. Here is itself equal to be. Of course, this is not a solution. This could should be taken only a scratch. But it gives us the thought for how to write out that full solution. So let&#x27;s begin here. We will set a equal to 1/2 times be so a scaler. Multiple of that matrix B or B satisfies b equals be transposed. We&#x27;re meant to solve t a equals B. And we are assuming that this is now our solution. But we have to verify that this is in fact, the case. Let&#x27;s right then t a would be equal to a plus, a transposed, and that&#x27;s just by the definition of this transformation. T Now we can make our substitution since we said on the onset of our solution is in fact 1/2 of B. So it&#x27;s right that in next will have 1/2 of the Matrix B plus 1/2 of the matrix be transposed. Once again, we can use a trick like we saw in part a where skill and multiple of a matrix is not affected by the transpose operator. So we can say that this is now equal to 1/2 times be plus 1/2 times be transposed now at that good position where we can use the given information of the problem that be itself is equal to be transposed and a word of warning with problems like these, typically, information like this would never be given in a problem unless it&#x27;s somehow relevant. So if we came up with a solution that never used the fact that B is equal to be transposed, we should be very suspicious of that solution. Next. In our case, let&#x27;s make the substitution here to say that this is now equal to 1/2 of B plus 1/2 of B. And of course this is equal to the Matrix B itself. So we have shown that if a is equal to 1/2 be, then fact t a would be equal to be itself. And that concludes the second part of this problem. For this problem, we&#x27;re going to be verifying the following statement about the range of this transformation. T We want to show that the range of tea is equal to the set of all vectors be in the co domain M two by two such that B is equal to be transposed. Let&#x27;s carefully walk through the definition of ranged to see just exactly what we&#x27;re going to be getting into. So the range of tea for a transformation satisfies the following. It&#x27;s equal to the set of all vectors be in the domain. W Our domain is two by two. Such that t of X equals B for some vector X in the domain V were for our case of V is m two by two as well. I&#x27;m going to color code much of this problem since if the domain and the co domain are different, then our solution has to follow the same structure that we&#x27;re going to lay out. So let&#x27;s begin. In fact, part be answered half of this question already. So let&#x27;s say that as part of our solution part be dealt with an equation t yet a equals B and we saw from part B. If b equals be transposed, then t a will be equal to be for some matrix A in two by two were m two by two comes from the domain in this case. So what have we shown? We&#x27;ve shown that we already know that B is equal to be transposed then that make tricks be must be in the range of tea. This feels like pretty good work, and a common mistake to make here is to assume the work is now complete. However, there is typically two parts. When we&#x27;re doing with set equality such as this, we now have to reverse engineer the process and say, Well, what if our matrix B came from the range of tea? Then can we show that B is equal to be transposed? If we could do that, then equality is verified and we&#x27;ll know that the range of tea is in fact a set of all of these major sees that are symmetric. In other words, be equals be transposed. So let&#x27;s begin with the second part of this solution. I will say Let be be in the range of tea. Rain is range T for short. We have to say, Well, what does that mean? Next will be is in the range of tea. Then we will be able to say t of Excell equal be equal to be for some vector X envy Whoever in our case, we&#x27;re dealing with the matrices. So X plays the role of a well, say next then for some matrix A in m two by two, which would be, in this case, the domain. We know that today will be equal to be. But now we have to say, What does this mean exactly? Remember, calls to now show that B is going to be equal to be transposed, but it could be tricky to get all the way there. Let&#x27;s use another portion of this problem that we have not used just yet. The definition of the transformation itself If t of a is equal to be. Then it follows that a plus a transposed would be equal to be itself, Since that&#x27;s the definition of the transformation. T it takes the Matrix a outputs itself. Plus it&#x27;s transposed. So you might say, Well, this doesn&#x27;t look particularly helpful. So let&#x27;s look at what the transposed might look like of that matrix B. So the scratch work board portion of this solution weaken, say, be transposed would be by substitution, taking a plus a transposed and taking the transpose of that result. Well, we already saw in part a that the transposed process will distribute across matrix addition. So it&#x27;s right out that as a very next step in our scratch work, this will be equal to a transposed, plus a transposed transposed. That&#x27;s a very interesting statement to make. So we have to wonder. What does it mean to take the transpose of a transposed matrix? Well, we already know that the transpose have made matrix will interchange the columns of rose. So if we do that interchange twice by taking the transposed twice, we should get the original Matrix A. And in fact, by looking at properties of transposed, that is the case so we can say that this is a transposed plus a itself coming from a transposed transposed. But if we look at this carefully, this is just a rearrangement of the original Matrix B itself that shows be transposed is equal to be. Let&#x27;s state this next in our official solution will say then be transposed will be a plus, a transposed transposed which is a transposed plus a transpose transposed, which is next. Hey, transpose plus a The order is different, but matrix addition is community tive, So we can now say that this is in fact, equal to be keep in mind once again that leaning algebra. We have to be very careful about her conclusions. So stay all of our conclusions. Next. Let&#x27;s also see what we&#x27;ve just shown. We&#x27;ve shown that if B is in the range of tea, then just so happens that B is equal to be transposed. But that&#x27;s what it requires to be in the set on the right hand side of this problem. Yes, we can. L say by this work. This shows that be is in the set of vectors be in m two by two such that B equals be transposed, then. This shows that the range of tea isn&#x27;t practical to that set B in m two by two such that b equals be transposed. Now that we&#x27;ve stayed all the conclusions, the second part is complete. This was a part involving set equality. So whenever you see in linear algebra a statement about quality and steps, keep in mind that typically involves two parts. The last problem. We verified a statement about the range of this transformation. T In this problem, we want to investigate with the colonel that transformation would be equal to. In other words, we want to determine the following What is the kernel of the transformation teeth for this problem? Let&#x27;s go back to the definition of what the kernel of a transformation T is equal to. It&#x27;s said to be the set of all vectors. Exit V such that t of X equals zero, where zero is the zero vector from the co domain. That&#x27;s right out. What that is your vector would be in this case, zero as a vector comes from the set of to Buy to meet Tracy&#x27;s, so it must be a square matrix with two rows two columns, and it has only zero entries. So that&#x27;s what this Euro vector would look like in our case. Now we&#x27;re ready to get started. After analyzing the definition, the underlying equation and the zero vector for that equation, we&#x27;ll start by saying the following supposed okay is in the colonel of tea immediately. Once we made this statement that a came from the kernel of tea, this equation is on the table. So let&#x27;s start by working with that equation, then t at a would be equal to is your vector, which implies that a plus a transposed is the zero vector itself. Since that&#x27;s the definition of teeth transformation teeth. Now we need to determine all maitresse sees a in that kernel of T. One way to do this is to specify what the entries of that matrix A should look like. So it&#x27;s insert some additional notation for this specific problem. Let a be equal to a B C D. Going rogue wise now that we have a form for the two by two Matrix A. We could go back to this equation, start manipulating each one of the entries. Let&#x27;s write out first a plus a transposed will say next, then a, which is a B CD plus a transposed recall that we said in the prior steps that when we take the transpose of a what we&#x27;re really doing is inter changing rose with columns. So if I go to Row One, which goes a B that will be interchange into a column, so call him one of the transposed matrix is a B. So, in other words, this just gets flipped to a column rather than a row. Similarly, CD gets transposed into CD as a column at the same time this is equal to the zero matrix itself. Let&#x27;s write it down on additional step and say, If we add together all of these and these two matrices on the left, going entry wise then in the 11 position will have to A in the 22 positions will have to D. Then we&#x27;ll have a B plus C in the one to position and in the row to call in one position. The 21 position will have C plus B, and all of this is equal to zero matrix containing Onley zero entries. The next step we can analyze is we&#x27;re dealing with not your ordinary quality, but with equality of maitresse. Ease recall that matrices air equal if, and only if each of the corresponding entries are equal. So this last equation gives us a tremendous amount of information. It tells us to A is equal to zero. We go to the next entry. It tells us also that B Plus C is equal to zero going on down the line. C plus B is also equal to zero and finally, to d corresponding to this entry tells us to D is also zero notice that we&#x27;re writing down zeros in two different ways. This would be the ordinary zero, and when we write zero this way were indicating an actual vector which came from the CO domain M two by two. Let&#x27;s come back to these equations from the 1st 2 equations on the left. Here we can conclude her right away that a must be equal to D, which must be equal to zero since two is non zero and the product of two numbers equaling zero means one of the other had to have been zero coming over to these two sets of equations. We can say immediately that be should be equal to negative C. This tells us now that that matrix A, which we identified earlier on in the solution here, can be expressed as follows eighties effect equal to a which is equal to zero in this position de which is equal to zero in this position. So the main diagonal this matrix contains only zero elements. We have a bee in the one to position. Let&#x27;s copy that in and we have a C in the one. Excuse me to one position. We can say that this C could be re expressed as C equals negative B. So it&#x27;s insert negative. Be here now that&#x27;s our matrix. A but linear algebra. Once again, if we leave our answer in this form, we may not get full credit in many classes. We have to fully state our conclusion to the problem on our conclusions should be what is the kernel of tea? So it&#x27;s right that out. I will say this work shows the kernel of tea is equal to the set of Mitrice&#x27;s that satisfy the following. They would have been the say of matrix sees X and B from the definition recall the in this case is the domain of matrices. And two by two that&#x27;s the major sees a we were working with all along and so we can say that matrix a satisfies entries zero b in row one negative B zero NRO to and after the colon. We just had to give a condition on B. The only condition is that be, is any rial number and that finally concludes our problem. We have determined that the kernel of tea is all matrices A in this form.

Henry Carnick · Answer

so the solve this problem. First thing I would like to know is that our transformation is linear, which means I can&#x27;t distribute t across to the individual terms, which means our new expression becomes tee off a X square plus t off be X plus t of seeing on our second observation. Is that because again T usual in your I can factor out or pull out the constant terms? I mean constant coefficients in our terms. In this case, I could pull out the c I could pull the be constant, make a plan of a constant. So now our new expression becomes a times You hear that square plus b times t of X plus c times to your one and knowing how how to use to find we can further right our expression asked. Eight times three x plus two Busby Times X squared minus one plus c times X plus one

Henry Carnick · Answer

So that&#x27;s all this problem. First note that t of two x equals four x And because tea is linear, I can move my constant coefficient outside so I end up with two times. TLX is equal to four x, which implies that t of axe equals two acts after dividing both sides by two. Secondly, if we look at t of three X Plus two, that&#x27;s gonna be equal to two x plus three. But T is linear. That means I can distribute my teeth overthrew the individual terms. That means we end up with t 03 x plus t of to and just being the two on the right side, we get two x Plus six similar. Easy as before. Tea is linear is like move the cost a coefficient outside of tea. So now we get three times to you vax plus two times t of one because two times one is equal to which is equal to two experts. Six. Now we actually know that TX is actually go to act so well. Something to that in and we&#x27;ll have six x plus two times t of one is equal to two x plus six It&#x27;s a fact ing things moving things across we get to attempt to you. One is equal to six minus four x drying everything by two. We end up with TF. One is equal to three minus two x. So now if we start with, uh, let&#x27;s see t a x square minus one, which is equal to X squared plus X minus three teas Linear. That means we could distribute t to the individual terms. So we we get end up getting t of X square minus two. You have one, but we really solved for what two of one is So in the next line, we can substitute to you if one for three minus two x So we end up with Kiev X square minus three minus two acts which, of course, is still equal to X squared plus X minus three being TX word by itself. We end up with the expression X squared minus X. So the salt the last part of expression If we start with key of a X square, plus BX plus C right, we know we can distribute because two years later we could distribute T to the individual terms so That means we end up with t of a X Square Krusty of Be Axe Plus to your seat. But T is linear so we can move the constant terms back throughout the concert terms outside of the of our transformations. So then we end up with a t of X square, plus b times t of X plus c times t of one because one time C is equal to see. So now we can substitute what we know what we know TLX Square to have accent here one are equal to and what we end up with is a times X squared minus x both the times to EPS plus C time three minus two acts. So if we distribute everything ah, and rearrange stuff, you end up with a X square minus a minus to B plus to see times X plus three seat

Doruk Isik · Answer

Yeah, in this trouble were given a set up. So we&#x27;ll start solution by writing the organ directors for each polynomial. So for people in the quarter, Inspector Ho. But like this 101 said, the 1st 1 is the question. But these year old 2nd 1 is to one of you two. So using this Pete, too apartment that too perfect. You could be rich. Zero. What being? And for P three, that could be, um, that would be one one negative three. So then the augmented matrix phoned my bees. Corn factors would be 101 01 negative three and 11 Negative three. Now let&#x27;s write this over mental midgets in traditional echo form. And that is he could do warn. 00010001 is you can see. We have three columns and each column has a few bits elements. So three columns, three events elements at each column. So it means that these vectors Spector&#x27;s spend our fee. It means that the given Colonials no meals spin he two. Now in part, we were given 1/4 collector for a loan Jew, and that is next to 112 And whereas to find this bologna to using this So we know that cube of native one times people plus one p two plus thio three So Dunn will be negative one times ones You&#x27;re one plus one times zero wounding three waas Two times One wonder three And we find this to be one the next 10. And using this, uh, using the information that these numbers represent corruption tease. We can ride the roller milk you is one plus B t minus 10 t spring.

Problem

Define $\quad$ a linear transformation $\quad T :…

Problem 31 Hard Difficulty

Answer

a. $T$ is closed under vector addition.
b. the range of $T$ is all of $R ^ { 2 }$

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Calculus 3 Educators

Lily A.

Anna Marie V.

Kayleah T.

Samuel H.

Calculus 3 Courses

Vectors Intro

Vector Basics - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 4

Video Transcript

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

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Kayleah T.

Samuel H.

Vectors Intro

Vector Basics - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

a. $T$ is closed under vector addition.b. the range of $T$ is all of $R ^ { 2 }$

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Kayleah T.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Anna Marie V.

Kayleah T.

Samuel H.

a. $T$ is closed under vector addition.
b. the range of $T$ is all of $R ^ { 2 }$

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

Linear Algebra and Its Applications