5

If $L$ is a linear transformation, show that $left.d Light|_{p}=L$ at every point $p$...

Question

If $L$ is a linear transformation, show that $left.d Light|_{p}=L$ at every point $p$

If $L$ is a linear transformation, show that $left.d L ight|_{p}=L$ at every point $p$



Answers

Suppose vectors $\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}$ span $\mathbb{R}^{n},$ and let $T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ be a linear transformation. Suppose $T\left(\mathbf{v}_{i}\right)=\mathbf{0}$ for $i=1, \ldots, p$ Show that $T$ is the zero transformation. That is, show that if $\mathbf{x}$ is any vector in $\mathbb{R}^{n},$ then $T(\mathbf{x})=\mathbf{0} .$

Can discussion that given vectors V and be in are and we is not because 20 and the Parametric equation X equals two p plus t v and the transformation t is our on to our and okay. And now we have to find a way to show that the given transformation maps given line onto another line or another point. Okay, so we will use the properties off linear transformation toe map. XTO on another line. Why so X equals toe B plus TV sub D eggs it caused took t p plus T V and the X equals to t p class de Devi. Okay, now no, let t x equals two wife and d. P has took you and we request ow! Then why it was took you plus d w and another things will be one equation and the second will be when then w equals to zero. Then second will be Wyche was took you Okay, so this will be this will be degenerated degenerated case or Kevin W past 20 So we will get out result as why was too Q plus d w or y questo que Okay when we w has some value that it is why it was two Q plus the blue. And when w goes to zero, then why it was took you. So the conclusion is we have proved that that the eggs on to Why by transformation de okay, and why was two q plus dw? Thank you.

In this video, we're gonna be solving problems over 26 of section 2.3, which is based on linear independence. Ah, transformation or not, linear independence. I'm sorry. It's based on determinants and Cramer's rule and, uh, linear transformations and volumes. So here we're given that Tia's transformation, um from our himto are in and we know that p is a vector and s is a set in r r m s as is a set in, um our, um So let we can let this is a proof so we can let the vector v be any vector any vector from the set s which is on our M. And by definition, we know that people s is a set of all vectors because s a set of all vectors, um, better of the form p plus V that we did find your love. Uh and we know that V isn't s so applying t to ah, typical vector and people cess. We have, um t and we apply people. Savi innit would give us tp plus t v. And this is this vector is in the said denoted by t plus s her teepee plus t s because we know that he is then s This proves that t maps the set people says onto the set T p plus t s. So this proves that, um, t as t maps people us us onto t p plus t s was what the initial problems asking which stated that we need to show that the image of people says under T a translated set t put people's t s in rn. So there's just a simple proof so shit.

Were given a linear transformation. Well, we're giving a transformation, I should say from space of second degree polynomial is to the space of third degree polynomial that maps a polynomial p t into the polynomial t plus five times PT in part a were asked to find the image of a polynomial of degree to So we're given that the panem you'll degree to pft is to minus t plus t squared. So we have that the image of this polynomial pft is going to be the image of the polynomial tu minus T plus T squared which according to our transformation, this is the polynomial T plus five times two minus T plus t squared and foiling So yet to teen, then minus t squared and then plus T cube. So the tea cubed here we also have 10. I guess the best way to start with the constant. So we have a 10 and then we have to t we have a negative five t so negative three t and five t squared minus two squared is plus fourty squared and plus t cute. So we get 10 minus three t plus four t squared plus t cute in Part B were asked to show that this transformation is, in fact, a linear transformation. As I alluded to earlier Do this. We're going to let p of tea Q of t the polynomial of degree to and we're going to let Alfa and data the scale, er's assuming these are point no meals over the real numbers. Really, if it could be any elements of a ring, in fact, actually, I'll make this simpler instead of trying to prove everything at once. Why don't we just prove one step at a time so we'll just take one one scaler? See, then we have that. The transformation of pft plus q of t the image of this polynomial well, we have this since P and Q are both polynomial of degree to it follows that there. Some is also a polynomial of degree to since the degree of P plus Q is at most two. Therefore, it follows that by the transformation, this is going to be T plus five times the polynomial pft plus Q of T and then using the distributive property we get. This is equal to T plus five times pft plus T plus five times Q of t. And we see that this is the same as the image of PFT, plus the image of Q of T and therefore we've shown part of linearity. Now we have also the t of C times pft Since C is a scaler, it follows that C times pft is still a second degree polynomial. All right, Paula, no emulation say of a degree at most two, since we're not multiplying by a factor of X and so the image of this polynomial is going to be the polynomial t plus five times see times pft and then using communicative ity of scale Ear's. This is the same as C Times T plus five times p of tea, which is the same as C times, since PFT is a polynomial of degree at most two, the image of pft and therefore it follows that t is a linear transformation. In Part C were asked to find The Matrix 40 relative to the bases, won t t squared, repeat two and won t t squared cubed four p three. So I guess one way to do this is staffed by finding the images of the bases so we have that t of one. Well, this is a polynomial which is t plus five times ones. This simply t plus five. And therefore it follows that the vector t one sub. And I guess I was called this he for this basis of p two, this is going to be the column vector five one zero. And because the basis for P three has four basis specters, I've had another zero here. Likewise, we have that t of t. This is going to be t plus five times T, which is the same as T squared plus five t and therefore t of T subi is the column vector zero five one zero and finally t of t squared. This is going to be t plus five times t squared, which is the same as it cute plus five t Square. And so we have that t of t squared. Sophie is the column vector with entries 00 five one. And therefore we have that The matrix for T relative to these two bases is going to be the matrix. Who's calling? Victors are t of one. So be t of tea said Be and t of t squared. So be I. I really instead of using be it probably could have used to see if it doesn't really matter here. And so our matrix is the Matrix with entries 5100 Columns 5100 0510 0051

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


Similar Solved Questions

5 answers
20. If the value of the SS remains constant, state whether each of the following will increase, decrease, or have no effect on the sample variance. (a) The sample size increases_ (b) The degrees of freedom decrease. (c) The size of the population increases.in ht1g al
20. If the value of the SS remains constant, state whether each of the following will increase, decrease, or have no effect on the sample variance. (a) The sample size increases_ (b) The degrees of freedom decrease. (c) The size of the population increases. in ht 1g al...
5 answers
Sodium reacts with water to yield hydrogen gas. Why is this reaction not used in the laboratory preparation of hydrogen?
Sodium reacts with water to yield hydrogen gas. Why is this reaction not used in the laboratory preparation of hydrogen?...
5 answers
Given A = S7 + 160 + 230.At P(3,1 rad,2.46 rad), the component of A perpendicular to surface 0=1 radians is:23.00 ar + 5.00 aq b. 5.00 ar + 16.00 a0 c16.00 a0 d.5.00 ar 2.23.00 ad None is correct
Given A = S7 + 160 + 230.At P(3,1 rad,2.46 rad), the component of A perpendicular to surface 0=1 radians is: 23.00 ar + 5.00 aq b. 5.00 ar + 16.00 a0 c16.00 a0 d.5.00 ar 2.23.00 ad None is correct...
5 answers
T4pe maufaCturten ta Tana -haro 3 Compar& of 4k2 The numbe 'F waden +bslen 6 tablen both tape 40 Paaduce of hauan newvined numben maimun aLond with +6e beLooJ Tabie availablem 4ne awleGf @lle Max Hoo7 Tape ~v(Hlableweldia Finiokixx Admin1 1250 30 02 2 6 8 40 pr F; + (EvR 25 55 Paaduc tim Plan +kat fird +he aptimaL the pnafit To +k & anawen-bx Mot mizcn Fnbent 4ke ameunf of_Tke Type / 4of mun + b e; PAaqluceal Tcbl e 1 mnimize 4he PacFit
t4pe maufaCturten ta Tana -haro 3 Compar& of 4k2 The numbe 'F waden +bslen 6 tablen both tape 40 Paaduce of hauan newvined numben maimun aLond with +6e beLooJ Tabie availablem 4ne awle Gf @lle Max Hoo7 Tape ~v(Hlable weldia Finiokixx Admin 1 1 250 30 0 2 2 6 8 40 pr F; + (EvR 25 55 Paaduc t...
5 answers
Fnd y where V = CO8 (ze21 _ 2) 128ecAnswzr: ov = -21 sin (ze21 4Jeex 12sec sec tan In (12) ov = sin (re21 v) (221 1 +21 re2l z) 12sec 1 sec 1 tan Oy SI (ze21 v) (e21 I+21 ze21 ?) 12se€ In (12)sec Itan Ov =COS (re21 z) (e21 ~ + 21 Tc21 7) 12se€ T sec Itan
Fnd y where V = CO8 (ze21 _ 2) 128ec Answzr: ov = -21 sin (ze21 4Jeex 12sec sec tan In (12) ov = sin (re21 v) (221 1 +21 re2l z) 12sec 1 sec 1 tan Oy SI (ze21 v) (e21 I+21 ze21 ?) 12se€ In (12)sec Itan Ov =COS (re21 z) (e21 ~ + 21 Tc21 7) 12se€ T sec Itan...
5 answers
16. A radius of a circle is increasing at the rate of 3 feet pet minute . Find the rate at which thc 4ic4 incteusing when the radius IS Iccl;
16. A radius of a circle is increasing at the rate of 3 feet pet minute . Find the rate at which thc 4ic4 incteusing when the radius IS Iccl;...
5 answers
For ech of tlie following scrics; use appropriate test to determine whether the xerits cunergent Ieed and how the cotiditions of the test Were Iet . Show aIl divcrgent. Cletuly stnto which tcst is work(6 poins) 2 (-)" (b) (5 pouts) 2piuts
For ech of tlie following scrics; use appropriate test to determine whether the xerits cunergent Ieed and how the cotiditions of the test Were Iet . Show aIl divcrgent. Cletuly stnto which tcst is work (6 poins) 2 (-)" (b) (5 pouts) 2 piuts...
5 answers
Sohethe following Derccnt problcms 21) Mr: Velk has t0 write 500 word essay. He has already written 235 words; what % of the essay did he write?
Sohethe following Derccnt problcms 21) Mr: Velk has t0 write 500 word essay. He has already written 235 words; what % of the essay did he write?...
1 answers
Finding Values of an Inverse from a Graph A graph of a function is given. Use the graph to find the indicated values. $$ \begin{array}{lll}{\text { (a) } f^{-1}(2)} & {\text { (b) } f^{-1}(5)} & {\text { (c) } f^{-1}(6)}\end{array} $$
Finding Values of an Inverse from a Graph A graph of a function is given. Use the graph to find the indicated values. $$ \begin{array}{lll}{\text { (a) } f^{-1}(2)} & {\text { (b) } f^{-1}(5)} & {\text { (c) } f^{-1}(6)}\end{array} $$...
5 answers
1. A 5000kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 2000 kg load initially at rest, is dropped onto the car. What will be the car’s new speed? Ball (sphere): 2/5 MRˆ2. Ring:MRˆ2. Cilinder: 1/2 MRˆ2
1. A 5000kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 2000 kg load initially at rest, is dropped onto the car. What will be the car’s new speed? Ball (sphere): 2/5 MRˆ2. Ring:MRˆ2. Cilinder: 1/2 MRˆ2...
5 answers
42. Calculate the equilibrium constant for the spontaneous redox reaction at 298 K that would occur using the following two half-reactions. Hint: the total moles of electrons transferred in the reaction 5_CIOs" (aq) + 6 Ht (aq) + 5 e ~ 'Yz Clz (g) + 3 HzO () Eo =+1.47V HCIO (aq) + Ht (aq) + e = Yz Clz (s) + HzO () Eo =+1.63 V5.04 x 102 6) 0.996 1.98 10 3 3.26 x 1013 2.32 x 1052
42. Calculate the equilibrium constant for the spontaneous redox reaction at 298 K that would occur using the following two half-reactions. Hint: the total moles of electrons transferred in the reaction 5_ CIOs" (aq) + 6 Ht (aq) + 5 e ~ 'Yz Clz (g) + 3 HzO () Eo =+1.47V HCIO (aq) + Ht (aq)...
5 answers
For the [eaL-yalued functions f (x) V2r+6 andg6)=x- findcomposloand speafy its domain using intccxa notndion8 0P 066.g)6) =Dipi (0,0) @,2]CUD (O,0] [D,ODomn Jir of f + g
For the [eaL-yalued functions f (x) V2r+6 andg6)=x- find composlo and speafy its domain using intccxa notndion 8 0P 06 6.g)6) = Dipi (0,0) @,2] CUD (O,0] [D,O Domn Jir of f + g...
5 answers
Question 75 ptsWhat is the magnitude of the ball's Acceleration when it reaches its maximum height and has stopped moving for an instant?9.8
Question 7 5 pts What is the magnitude of the ball's Acceleration when it reaches its maximum height and has stopped moving for an instant? 9.8...
5 answers
Lim (22 + 1) sin( - +1)+3 2x sin( 4)]
lim (22 + 1) sin( - +1)+3 2x sin( 4)]...
5 answers
QUESTiOn 16what the ground stato oloctron configuralilon for the phosphido Ion (P)Dnta shdoLand Benod Tebla [Ne] 352 [Ne] 352 '3p3 [Ne] 3p3 [Ne] 352 3p6
QUESTiOn 16 what the ground stato oloctron configuralilon for the phosphido Ion (P) Dnta shdoLand Benod Tebla [Ne] 352 [Ne] 352 '3p3 [Ne] 3p3 [Ne] 352 3p6...

-- 0.022864--