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Which of the following is a linear transformation from R2 to R3?a) T(a,b) = (a,0,b2) b) T(a,b) = (1,a +b,-b) c) T(a,b) = (a - b,ab,a + b) d) T(,b) = (a,a +b,-a)...

Question

Which of the following is a linear transformation from R2 to R3?a) T(a,b) = (a,0,b2) b) T(a,b) = (1,a +b,-b) c) T(a,b) = (a - b,ab,a + b) d) T(,b) = (a,a +b,-a)

Which of the following is a linear transformation from R2 to R3? a) T(a,b) = (a,0,b2) b) T(a,b) = (1,a +b,-b) c) T(a,b) = (a - b,ab,a + b) d) T(,b) = (a,a +b,-a)



Answers

The transformation T maps [x // y] to [a & b // c & d][x // y]
Show that invariant points other than the origin exist if ad - bc = a + d - 1

Were given a transformation t from the set of polynomial of degree at most two from the set of polynomial of degree at most four, the maps pointing Will PT into the polynomial p T plus T squared PT. In part, they were asked to find the image of a certain polynomial. P f t equals two minus t plus t squared. We have that the image of this polynomial pft. Well, since it is a polynomial of degree at most two, it's going to be by definition, pft plus T squared pft and calculating. We have that This is the same as Tu minus T plus t squared plus t squared times two minus T plus T squared, which is the same is we have the constant to we have negative t we have t squared plus two T squared sets plus three t squared. Then we have minus T cubed and plus T to the fourth. So we get the polynomial to minus T plus three t squared minus t cubed plus T to the fourth, which is, in fact a polynomial of degree at most four in Part B were asked to show that this transformation is a linear transformation. So we're going toe. Let PFT and Q of T bee pollen or mules of degree at most to and see be any scaler. So will prove this in two steps. So first we have that t of PT plus q of t What we have that since both P and Q are polynomial of degree at most two, it follows that the degree of P plus Q is that most too, and therefore that the transformation is well defined for this point. Oh mule and gives us P f T plus Q of t plus T squared times, pft plus Q of T. This could be written as pft plus t squared pft After using the distributive property plus Q of T plus T squared Q of t. And we see these expressions in the brackets are simply t of p of tea plus t of q of t. So we've shown activity. Now we have that t of C times p of tea. Well, since he is a scaler, it follows that since the maximum degree of P is to maximum degree of C times, P. M T is still too. So this expression makes sense and is equal to see Time's pft plus T squared times see times p of tea and this could be written factoring out. See, we get C Times, pft plus T squared times Pft using the distributive property, which recognizes the Samos See Times t of PFT and therefore it follows that t is a linear transformation. In Part C were asked to find the Matrix for T relative to the bases, won t t squared and won t t squared cubed T to the fourth. So this time make it more explicit will have to. The basis be is going to be one t and t squared. The basis see is going to be won t he squared and cubed and then t to the fourth. So first, let's calculate the image of each of the basis vectors for from B So we have t of one is going to be one plus t squared times. One just simply t squared or one plus t squared. And so it follows that in terms of the basis factors, see, this is the column vector with entries one zero one 00 Likewise, we have that tea of tea is going to be t plus t squared times t which is t plus tty cute. And so we have that t f t respect to the basis. See, this is the column vector with entries 0101 zero t of t squared is equal to t squared plus t squared Times T squared, which is equal to t squared plus t to the fourth. And so we had the t f t squared with respect to See is the column vector with entries 00 101 Mhm. And so we have that The Matrix for T relative to these two bases. This is going to be the Matrix whose column vectors are t of one with respect to see t of tea with respect to see and t of t squared with respect to see which using our previous calculations. This is the Matrix, which has column vectors. 101 00 01010 and 00101

In this problem, we have been given to linear transformations. Now in the first problem, what we need to determine is T. One composition T two of E. Now using the definition of composition, this is equal to T one or T two of E. So this will be T one of D two of the matrix E. And the matrix E is given to be the matrix E B C E. So this is what we need to tell me now T two of a is given to be equal to transport. So T two of this matrix will be the transpose of this matrix. So we have E C B. Yeah. And T one of these metrics will be the trace of this mattress because T one of them is going to be the trace of E. So this is what we need. Now the trace of a matrix is the some of the elements on its principal diagonals. This will become equal to eight plus. Hence required. Rather anyone composition T two of a will be equal to a plus. Next we need to the german whether or not it is possible to find T. Two composition T one of E. Now by the definition of composition this will be key to of T one of 8 And that will be T. two of the trace of the matrix. Because do you want to is equal to Tracy? Now T two of a is equal to a transposed. So T two of Tracy should be the transpose of trace. However, the trace of a is not a matrix. The trace of a matrix is a real number and thus it is not possible to determine the transpose of it. Hence it will not be possible To determine T two composition T one of 8. So the question is can we find this composition? The answer is no. And the reason is that this will not be defined because the one of a is a real number and T. Two of is defined on the set of mattresses. But you wanna way is not a matrix.

So the Sol this suit solve this problem? The first thing I would like to know is that the transformations are linear, which means I can just shooting the tea over to to be one distributed over 23 v to assume Iraq to distribute to your the one and distribute over B two. So our expression to come Teoh to be one musty of three V two, which is able to be Willen plus B two. In a similar vein, it's equal to the want TV one post here, the to which is equal to three, the one minus V two and then secondly, because he is later, I can pull out my cost, a term, the two and the cost term of three. So this expression becomes two times T O. V. One because three times t o V two, which is unequal to be one plus it be to So in order, stall for the system of equations. I'm going to subtract from this equation from this one by multiplying it by two. So it goes like this. So it's a fact. Two times t of the one plus trv to equal to minus two times three V one minus B two. So this expression becomes TRV to is equal to the one plus V two minus two times three v one minus feature. So what TV to it. One with simple I it comes, is equal to minus four b one plus three be to so now defined TRV one. We just take one of our equations and substitute to the two into it, the simplest one being the second equation. So it turns out TV one plus TV too equals Ah three V one. Mine isn't you to substituting for to be too. We end up with TLV one plus minus four. Do you want plus three? The two is equal to three. B one amounted to be to getting TV one by itself into collecting like terms we end up with TV one is equal to 71 minus four to be to

The transformation T maps The Vector X. Y. two. The product of the matrix times victor. The metrics is A. B C. D. The vector X. Y. That is the metrics ap city times to victor X. Y. Is the transformation of the picture X. Y. Through the map. T. Or its information. T. Show that in variant points other than the urgent exists if eight times the minus B times C equal A plus the minus one. So these information T goes from victor's of two components are too From Victor's of two components because the Metrics A, B, C&D. These two times or two x 2. And that multiplied by two times one victor give us a two times one picture. Anti of victor, X. Y. Is to find a symmetric ap C and D times a vector ex wife. That is. We multiply the metrics here times to give him victor and that result that result in victor is the image of the victor X. Y. Given. Now we talk about in variant points also called fixed points. These are points that has the property of having as the image through its information T equal to the same point. That is very important X. Y. In the plane as Birdie. That T of the point is it will the same point. There is the point is not modified when we apply the transformation T. It's called variant point. It is clear that the origin is an environment point. Because if we multiply the matrix times 00 victory. That is the origin. We get again to Syria zero victory. So 00. The origin is an in variant vector or any variant point. So we want to write the condition that need to be needs to be fulfilled in order to have in variant points other than these. Seriously a picture. So let's say that T of X. Y is equal to X. Y. If and only if T F X Y A B C D times X. Y. That's the same. Sorry, equal to X. Y. That is because of X. Y is defined this way. This is equivalent to let's multiply the metric stein's victor. We get a X plus B. Y. His first component, second component C. X plus dy. And that can be equal to X. Y. And equality of vectors means that correspondent components can be equal. So we got to have two equalities, E X plus B. Y equal X. And see eggs plus D. Y equals Y. And put in these terms to the left of the equations. We get a -1 Time six. That's when we put eggs to the left and take a factor eggs plus B. Y equals zero. And here with the same with Y we get C X Plus The -1. Y. people soup. Mhm. Okay. And this is the same as a matrix vector system. That is a linear system in X&Y. Which can be written a metrics form as a -1. Me first role of the metrics than C. The minus one times X Y equal serious. Ooh so this is a metric form of this linear system here. That is we want to find X. Y. Different from the version which satisfies his equations. So to have a non null solution to this equation. This linear system, it is uh equivalent to saying that the metrics is not invariable because if the metrics a no swan pc and even this one is variable multiplied by its inverse to the left and the only solution will be xy consider so. Okay, let's say that this way here linear system Amen, swan, E C&D -1 times X. Y equals serious. You as solutions different from 0000. Sorry if and only if The Metrics A Menace one, B. The U -1 is not either do has not been verse and that's the same. And this is equivalent. Two. The determinant of the matrix equals zero. Okay, equivalent to the fact that t determinant of these metrics be equal 20 And the determinant is not zero. The metrics is invariable. And as we said here, above the system, we have only the serious, serious solution. So we need this determinant to be equal to zero. That is the case in that case mattress has no reverse And this system get to have solutions infra France 00. There is determinant of the matrix. Amen is one. B C T -1 equals zero. Mhm. And we know this determinant is equal to the product of the elements on the main diagonal, A -1 times t minus one minus the product of the elements on the second, f minus B C. And that gotta be equal zero. And so we do this multiplication, hearing at a d minus a minus D plus one minus bc. He got zero. And so a d minus bc. That is left is to terms on the left of the equation equal. And the three terms go to the right of the equation. We get a plus the -1 which is just the condition said uh Name in the statement, there is this one here. So if this relationship between the interests of the metrics, A B c D defining the linear transformation is satisfied, then there will be in variant points of this transformation or to this transformation different from the origin. And that is because uh when we stayed the fact that it victor, x wire point X, Y. It's an environment point which is this equation here. This lead us to a linear system which will have non serious solutions. When the determinant of the matrix is equal to zero. That is when the metrics is invariable, it's not over. So the determinant is calculated and leads to this condition between abc and the Okay


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