线性代数Linear讲解课件.ppt

1、2023/5/221線性代數Linear Algebra東吳大學數學系 葉麗娜2023/5/222第七 章 Linear Transformations(線性轉換)n7.1 The idea of a linear Transformationn7.2 The Matrix of a Linear Transformationn7.3 Change of Basisn7.4 Diagonalization and the Pseudoinverse2023/5/2237.1 The idea of a linear Transformationn定義：qA transformation T a

2、ssigns an output T(v)in output space W to each input vector v in input space V.The transformation is linear if it meets these requirements for all v and w:(a)T(v+w)=T(v)+T(w)(b)T(cv)=c T(v)for all c in Rqlinearity:T(cv+dw)=c T(v)+d T(w)n依據”linear”的性質 T(0)=0n例子：矩陣乘法運算T(v)=Av就是一個線性轉換n例子：T(v)=v+u0 不是一個

3、線性轉換，除非u0=0 T(v)=v 稱為“identity transformation”2023/5/2247.1 The idea of a linear Transformationn例子：T(v)=Av+u0”linear-plus-shift transformation”不是一個線性轉換n例題1：假設a=(1,3,4),T(v)=a v(inner product)是一個線性轉換。q解：令 A=1 3 4，則 T(v)=a v=A vn例題2：The length T(v)=|v|不是一個線性轉換。q解：因為|v+w|v|+|w|，而且 T(-v)=|-v|=|-v|-|v|n例

4、題3：T(v)=旋轉v向量30o(xy平面)是一個線性轉換。q解：我們將整個平面旋轉30o，會使得這個轉換滿足線性關係 。在這裡不用提及矩陣。2023/5/2257.1 The idea of a linear Transformation:Lines to Lines,Triangles to Trianglesn下圖說明input線上的點對應到output線上的點，其間保持等距關係。1111 ()()()2222TTTuvwuvw2023/5/2267.1 The idea of a linear Transformationq註：”transformation”有自己定義使用的語言，雖然

5、可能沒有用到矩陣，但相同的觀念依然可使用。qRange of T=set of all outputs T(v):corresponds to column space qKernel of T=set of all inputs for which T(v)=0:corresponds to nullspaceqThe range is in the output space WqThe kernel is in the input space VnThe rule of linearity extends to combinations of three vectors or n vect

6、ors.n例題4：T(v)=投影v向量(in R3)到 xy平面。q解：The range is the xy plane,the kernel is the z-axis.Let v=(v1,v2,v3),then T(v)=(v1,v2,0)=0,hence v1=0=v2,the transformation(projection)is linear.1 1221122.()()().()nnnncccTcTc Tc Tuvvvuvvv2023/5/2277.1 The idea of a linear Transformationn例題5：T(v)=投影v向量(in R3)到 z=1平

7、面。q解：Let v=(v1,v2,v3),then T(v)=(v1,v2,1),T(cv)=(cv1,cv2,1)cv hence the transformation is not linear.n例題6：假設A是可逆矩陣，而且T(v)=Av則存 在逆線性轉換(inverse transform)T-1，使得T-1(T(v)=v。q解：Let T-1(w)=A-1w,for all w in the range of T then T-1 is linear and T-1(T(v)=T-1(w)=A-1w=A-1Av=v,for w=T(v)=Av 2023/5/2287.1 The

8、idea of a linear Transformation:Linear Transformations of the planen假設平面上一間”房子”有11個頂點 vi=(xi,yi),i=1,11.我們做一個線性轉換，將這11個頂點對應到頂點，而且他們之間的直線對應到直線，來產生新的”房子”。n觀察不同的矩陣所產生的效果.6 6 7 0 7 6 6 3 3 0 0 67 2 1 8 1 2 7 7 2 2 7 7H2023/5/2297.1 The idea of a linear Transformation:Linear Transformations of the plane6

9、 6 7 0 7 6 6 3 3 0 0 67 2 1 8 1 2 7 7 2 2 7 7H2023/5/22107.1 The idea of a linear Transformation:worked examplesn7.1B Nonlinear transformationsqFor x in R(實數),下列T都不是線性 T1(x)=x2 ,T2(x)=x3 ,T3(x)=x+9 T4(x)=ex ,T5(x)=1/x ,x 0qHowever,T2 ,T3 ,T5 are invertible.2023/5/22117.1 The idea of a linear Transfo

10、rmationn問題：每個線性轉換都可經由矩陣產生嗎?q解：如果線性轉換的定義域(domain)V=Rn，值域(range)W=Rm，則每個m by n的矩陣會產生 一個線性轉換。T(v)=Av 反之，每個線性轉換都隱含一個矩陣嗎？這就是我們要問的問題：例如：”projection”,“rotation”,解：是的(V,W基底向量之對應，以後說明)。所以 一般線性代數可以不必從矩陣開始講起。2023/5/22127.2 The matrix of a linear transformationn本節中說明：每一個線性轉換都有對應之矩陣q例如：v1=(1,0),v2=(0,1),w1=(1,0,

11、0),w2=(0,0,1),w3=(0,0,1)假如 T(v1)=(2,3,4),T(v2)=(5,5,5)則每個向量v=(c1,c2)in R2 T(v)=T(c1 v1+c2 v2)=c1 T(v1)+c2 T(v2)=c1(2,3,4)+c2(5,5,5)=Av 其中nKey idea:當我們知道基底向量v1,v2,vn的對應值T(v1),T(v2),T(vn)，則每個向量v的T(v)也可確定 122 53 5 ,4 5ccAvStandard basisA is a standard matrix for T2023/5/22137.2 The matrix of a linear t

12、ransformationnSuppose v=c1 v1+c2 v2+cn vn .Then linearity requires T(v)=c1 T(v1)+cn T(vn)q例題：input space V=cubic polynomials(degree3)則 1,x,x2,x3 是一組基底，令 ，依照微分的法則，T是一個線性轉換 T(v)=T(a+bx+cx2+dx3)=aT(1)+bT(x)+cT(x2)+dT(x3)=b+2c x+3d x2假如Output space W=cubic polynomials，則 range(T)=quadratic polynomials(de

13、gree2),a subspace of W。假如Output space W=quadratic polynomials，則 range(T)=W。qkernel(T)contains v with T(v)=0,kernel(T)=constant polynomials()dTdxvv2023/5/22147.2 The matrix of a linear transformationn(續上頁)q假如 input space V=cubics,W=quadratics ，則dim(V)=4,dim(W)=3,對應於 之矩陣 T(v)=T(a+bx+cx2+dx3)=aT(1)+bT(

14、x)+cT(x2)+dT(x3)=b+2c x+3d x2,以1,x,x2基底組合之係數=A v 即 T(v)=Av,當v寫為向量形式 v=(a,b,c,d)n(dimension of range of T)+(dimension of Kernel of T)=dimension of Vq3+1=4()dTdxvv230 1 0 00 0 2 0(1)()()()0 0 0 3TT x T xT xAA(3 by 4)is a standard matrix for T2023/5/22157.2 The matrix of a linear transformationn例題：考慮下列

15、的積分(微分的逆向)令 w1=1,w2=x,w3=x2，T-1 將 對應到則T-1是線性轉換而且T-1(w)使得 W=quadratics對回到V=cubics，其所應之矩陣(4 by 3)表示如右：(dimension of range of T-1)+(dimension of Kernel of T-1)=dimension of W remark:3+0=3 (Kernel of T-1 contains only 0 vector)2BCxDxw223000111 ,23xxxdxxx dxxx dxx12311()23TBxCxDxw112211330 0 0 01 0 0 0 0

16、0 0 BBCCDD2023/5/22167.2 The matrix of a linear transformation:Matrices for the Derivative and Integraln(續上頁)對應於微分與積分線性轉換(T/T-1)的矩陣A/A-1滿足n因為A不是方陣所以僅有單邊的反矩陣(one-sided inverse)n積分再微分回到自己，但是微分再積分會喪失常數項qTT-1(w)=w,however T-1T(v)vnT-1T(1)=integral of zero function=0qA-1A的第一行全為零。110 0 0 01 0 00 1 0 00 1

17、0 ,0 0 1 00 0 10 0 0 1butAAA A2023/5/22177.2 The matrix of a linear transformation:Construction of the Matrixn本節主題：針對每個線性轉換T 建立對應的矩陣Aq假設T是從V(n-dimensional)對到W(m-dimensional)的線性轉換，v1,v2,vn與w1,w2,wm分別為V與W的一組基底，則A為m by n矩陣。A的第一行是T(v1)：T(v1)=a11 w1+a21 w2+am1 wm是W基底的組合 v1對應到T(v1)，針對w1,w2,wn 這組基底，T(v1)之係

18、數可視為A矩陣乘上向量(1,0,0)n7A Each linear transformation T from V to W is represented by a matrix A(after the bases are chosen for V and W).The jth column of A is found by applying T to the jth basis vector vj:T(vj)=a1j w1+a2j w2+amj wm=combination of basis vectors of W2023/5/22187.2 The matrix of a linear

19、transformation:Construction of the Matrixn當每個基底向量vj的對應值T(vj)j=1,2,n確定後，一般向量v=(c1,c2,cn)的對應值T(v)可寫成 T(v)=c1T(v1)+c2T(v2)+cnT(vn)=T(v1)T(vn)v針對以w1,w2,wn為基底的線性組合，T(v)之係數=Avn同樣的線性轉換T選擇不同的基底，其對應之矩陣A也不同q例如：V=cubics(多項式 degree 3)我們變更基底的順序為 x,x2,x3,1，而W=quadratics不變，則新矩陣如下：1 0 0 00 2 0 00 0 3 0newAMatrix fo

20、r the derivative T when the basis of V change to x,x2,x3,12023/5/22197.2 The matrix of a linear transformation:Construction of the Matrixn例題：考慮V=W=R2,線性轉換T(v)=v，要找出T所對應之矩陣。這個矩陣隨基底選擇而變。q解：如果選擇標準基底 v1=(1,0),v2=(0,1)且 w1=(1,0),w2=(0,1)則T(v1)=v1=w1=1w1+0 w2 ,T(v2)=v2=w2=0w1+1w2。矩陣為單位矩陣(ii)如果選擇基底v1=(-1,0

21、),v2=(0,1)且 w1,w2不變 則T(v1)=v1=-w1=-w1+0 w2 ,T(v2)=v2=w2=0w1+1w2。矩陣A變成為(iii)如果選擇基底v1,v2不變且 w1=(-1,0),w2=(0,-1)則T(v1)=v1=-w1=-w1+0 w2 ,T(v2)=v2=-w2=0w1-w2。矩陣A變成為1 0 0 1A1 00 1A1 0 0 1A2023/5/22207.2 The matrix of a linear transformation:Construction of the Matrixn例題：考慮V=W=R2，T旋轉平面的向量一個角度，要找出對應之旋轉矩陣。這個

22、矩陣會隨基底選擇而變。q解：如果選擇標準基底 v1=(1,0),v2=(0,1)且w1=v1,w2=v2 則 T(v1)=(cos,sin)=cosw1+sinw2 T(v2)=(-sin,cos)=-sinw1+cosw2 cos sinsin cosAStandard matrix for T2023/5/22217.2 The matrix of a linear transformation:Construction of the Matrixn例題：考慮V=W=R2，線性轉換T投影平面的向量到45o的直線，要找出2個不同基底T所對應之投影矩陣。q解：(i)如果選擇標準基底 v1=(1

23、,0),v2=(0,1)且w1=v1,w2=v2，則 同樣 T(v2)=T(v1)=1/2(w1+w2)，對應之矩陣(ii)如果選擇基底 v1=(1,1),v2=(-1,1)且w1=v1,w2=v2，則 T(v1)=(1,1)=w1,T(v2)=0=0 w1+0 w2 對應之矩陣T11T11 211()=,11 212T v uvuuu u1 2 1 21 2 1 2A1 00 0ARemark:T2=T,A2=A v1v2u2023/5/22227.2 The matrix of a linear transformation:Products AB Match Transformation

24、s TSnS是從U到V的線性轉換，u1,u2,up與v1,v2,vn 分別為U與V的一組基底，B為n by p矩陣。T是從V到W的線性轉換，v1,v2,vn與w1,w2,wm 分別為V與W的一組基底，A為m by n矩陣。n7B(Multiplication)The linear transformation TS starts with any vector u in U,goes to S(u)in and then to T(S(u)in W.The matrix AB starts any x in Rp,goes to Bn in Rn and then to ABx in Rm.T

25、he matrix AB correctly represents TS:TS:U V W AB:(m by n)(n by p)=(m by p)qu=x1u1+x2u2+xpup ,x=(x1,x2,xp)2023/5/22237.2 The matrix of a linear transformation:Products AB Match Transformations TSqTS:U V W AB:(m by n)(n by p)=(m by p)qu=x1u1+x2u2+xpup ,x=(x1,x2,xp)qTS(u)=T(S(u)=T(x1S(u1)+xpS(up)=T(x1(

26、b11v1+bn1vn)+xp(b1pv1+bnpvn)=(x1b11+xp b1p)T(v1)+(x1bn1+xp bnp)T(vn)=(x1b11+xp b1p)(a11w1+am1wm)+(x1bn1+xp bnp)(a1nw1+amnwm)=(x1b11+xp b1p)a11+(x1bn1+xp bnp)a1n w1+(x1b11+xp b1p)am1+(x1bn1+xp bnp)amn wm=x1(a11b11+a1nbn1)+xp(a11b1p+a1nbnp)w1+x1(am1b11+amn bn1)+xp(am1b1p+amnbnp)wm ABx2023/5/22247.2 Th

27、e matrix of a linear transformation:Products AB Match Transformations TSn例題：S,T旋轉平面的向量一個角度，則 T S旋轉平面的向量2，要找出對應之旋轉矩陣。q解：選擇標準基底，u1=(1,0),u2=(0,1)且u1=v1=w1,u2=v2=w2 則 S(u1)=(cos,sin)=cosv1+sinv2 ,S(u2)=(-sin,cos)=-sinv1+cosv2 T(S(u1)=cosT(v1)+sinT(v2)=cos(cosw1+sinw2)+sin(-sinw1+cosw2)=(cos2-sin2)w1+2

28、sincosw2 =cos2w1+sin2w2 同理T(S(u2)=-sin2w1+cos2w2cos sinsin cosA22222cossin 2sincoscos2 sin2sin2 cos22sincos cossinA T=S TS2023/5/22257.2 The matrix of a linear transformation:Products AB Match Transformations TSn例題：假設T是對45o線做映射(reflection)，S是對y-軸做映射，假如v=(2,1)要找出 S(T(v)與 T(S(v)。q解：選擇標準基底，u1=(1,0),u2=

29、(0,1)且u1=v1=w1,u2=v2=w2 則 T(v)=(1,2),S(T(v)=S(1,2)=(-1,2)S(v)=(-2,1),T(S(v)=T(-2,1)=(1,-2)ST TS 另一方面，T(u1)=(0,1)且T(u2)=(1,0)S(u1)=(-1,0)且S(u2)=(0,1)standard matrix for T and S:A and B0 11 0 ,1 0 0 1AB請同學自行驗証：standard matrix for TS and ST:AB and BA,ABBAv(2,1)(1,2)2023/5/22267.2 The matrix of a linear

30、 transformation:The Identity Transformation and Change of Basisn本節要說明特別的線性轉換(Identity transformation)T(v)=v，所對應的矩陣M：q當輸入與輸出空間(V and W)選擇相同的基底，即T(vj)=vj=wj，則T對應的矩陣為單位矩陣I.q當輸入與輸出空間使用不同的基底，T(vj)=vj=m1jw1+m2jw2+mnj wn，則T對應的矩陣為M,“change of basis matrix”。v1 v2 vn=w1 w2 wnM111212122212 .nnnnnnmmmmmmmmmM202

31、3/5/22277.2 The matrix of a linear transformation:The Identity Transformation and Change of Basisn例題：T(v)=v，v1=(3,7),v2=(2,5)且w1=(1,0),w2=(0,1)，找出T所對應的矩陣M。q解：T(v1)=v1=3w1+7w2，T(v2)=v2=2w1+5w2 則T對應的矩陣n相反地，T(v)=v，v1=(1,0),v2=(0,1)且w1=(3,7),w2=(2,5)，則T對應的矩陣為何?q解：T(v1)=v1=a11w1+a21w2 T(v2)=v2=a12w1+a22w

32、2 故矩陣A為上面M的反矩陣nThe 2 change of basis matrices are inverses.3 27 5M12121 0 0 1wwAvv1 5 27 3M2023/5/22287.3 Change of BasisnIn Rn,qstandard basis:e1,e2,en,ej=(0,1,0)each vector v=(v1,v2,vn)qnew basis:w1,w2,wn,and each vector v=(c1,c2,cn)=c1w1+c2w2+cnwn =Wc or c=W-1v W is the“basis matrix”n7C The coord

33、inates c=(c1,c2,cn)of v in the basis w1,w2,wn,are given by c=W-1v.The change of basis matrix W-1 is the inverse of the basis matrix W.qThe standard basis has W=I.jth2023/5/22297.3 Change of Basisn例題：(wavelet basis)for image compressionnChoose a better basis of ws,5%of basis vectors can come very clo

34、se to the original data.nInput v coefficients c compressed compressed12341 1 1 01 11 0 11 0 111 01 wwww c vLosslessc=W-1v lossyReconstructvWc2023/5/22307.3 Change of Basisn簡介：the Haar transformationqFor vector v=(v1,v2),w=Hv=(w1,w2)w1,w2分別是兩個分量v1,v2的和與差再除以qH是正交矩陣，因此 v=H-1w=HTw(表示v可經由w復得)qFor 2 by 2

35、matrix x and y,y=HxHT If 1 111 12H2 1 2a babcdabcdthenc dabcdabcd xy2023/5/22317.3 Change of Basis:the Haar transformationImage,XTransform samples,Y 1 2a babcdabcdthenc dabcdabcd xy這個轉換對應於下列濾波器的處理：1.Top left:a+b+c+d=4-point average or 2-D lowpass(Lo-Lo)filter2.Top right:ab+cd=Average horizontal grad

36、ient or horizontal highpass and vertical lowpass(Hi-Lo)filter.3.Lower left:a+bcd=Average vertical gradient or horizontal lowpass and vertical highpass(Lo-Hi)filter.4.Lower right:abc+d=Diagonal curvature or 2-D highpass(Hi-Hi)filter.a+b+c+d2023/5/22327.3 Change of Basis:the Haar transformationn我們將一張圖

37、片(X)劃分為2by2的方塊，應用這個轉換，得到的結果(Y)經過整合後(將Lo-Lo,Hi-Lo,Lo-Hi,Hi-Hi部分分別放在一起)如下：100200300400500100200300400500100200300400500501001502002503003504004505002023/5/22337.3 Change of Basis:The Dual Basisnbasis vectors w1,w2,wn,let W=w1 w2 wn to find the coefficients of a vector in this basis,use W-1nThe rows of

38、 W-1 are the“dual basis”nThe two bases are“biorthogonal”nW-1W=I and WW-1=InIf W is an orthogonal matrix,the us are the same as ws.T11T nuWuT11T1 120 T nijnifijifijsouW Ww wwIu wu2023/5/22347.3 Change of Basis:The Dual BasisnThe coefficients arenIf c=W-1v then reconstruct v=Wc=WW-1vT iic u vT11 ()nni

39、iiiandthe vectorcvww u v2023/5/22357.4 Diagonalization and the Pseudoinverse (A+)nDiagonalization(對角化)：qS-1AS=when the input and output bases are eigenvectors of A.nInput 與 output 空間用相同的基底，矩陣A是方陣。qU-1AV=when the input and output bases are eigenvectors of ATA and AAT.nInput 與 output 空間用不同的基底，矩陣A不必是方陣

40、。nGram-Schmidt factorization A=QRqThe input uses the standard basis given by I.qThe orthogonal output basis given by Q.(僅output基底改變)qA=QRI,output基底矩陣Q在左邊，input基底矩陣I在右邊。2023/5/22367.4 Diagonalization and the Pseudoinverse :Similar Matrices:A and S-1AS andW-1AWn當Input V=Rn與 output W=Rn，用相同的標準基底且T(v)=A

41、v：q如果改變input basis 則基底變換矩陣(change of basis matrix)為M，線性轉換T 所對應的矩陣為AM。q如果改變output basis 則基底變換矩陣為M-1，線性轉換T 所對應的矩陣為M-1A。q如果同時改變input basis與 output basis 則線性轉換T 所對應的矩陣為M-1AM。q令A的特徵向量(線性獨立)為新基底，是M矩陣的行向量，則M-1AM=S-1AS=。2023/5/22377.4 Diagonalization and the Pseudoinverse :Similar Matrices:A and S-1AS andW-

42、1AWn7D When the basis contains the eigenvectors x1,x2,xn,the matrix for T becomes.qT(xj)=Axj=jxj=0 x1+jxj+0 xn 00j2023/5/22387.4 Diagonalization and the Pseudoinverse :Similar Matrices:A and S-1AS andW-1AWn例題：投影平面的向量至135o直線 y=-x，找出對應之對角矩陣。標準基底e1=(1,0),e2=(0,1)q解：若採用標準基底，則 A的特徵值與特徵向量：1=1,x1=(1,-1)2=0

43、,x2=(1,1)因此，若採用特徵向量為基底，則對角矩陣為12 11 1 1 0.5 0.5111(),()=1 11 10.5 0.5222TTeeA1 00 02023/5/22397.4 Diagonalization and the Pseudoinverse :Similar Matrices:A and S-1AS andW-1AWn續上頁：如果基底換成v1=w1=(2,0),v2=w2=(1,1)，則 q另一方面：基底的改變可表示一個identity transformation I 如果令基底矩陣 則代表標準基底與w1,w2之間的轉換，因此線性轉換 ITI 所對應之相乘的矩陣為

44、W-1AW，W-1AW=B(B is similar to A)n請同學自行驗證 W-1AW=B112212 10 1 0(),()00 =101 0TT vwwvwwB122 1 0 1Www2023/5/22407.4 Diagonalization and the Pseudoinverse :Similar Matrices:A and S-1AS andW-1AWn7E For any basis w1,w2,wn,find the matrix B in three steps.Change the input basis to the standard basis with W.

45、The matrix in the standard basis is A.Then change the output basis back to the ws with W-1.The product B=W-1AW represents ITI:B ws to ws=W-1 standard to ws A standard W ws to standardq I:V V (改變input basis I(v)=v),T:V V(標準基底 T(v)=Av)I:V V (改變output basis I(v)=v)nA change of basis produces a similari

46、ty transformation in the matrix.2023/5/22417.4 Diagonalization and the Pseudoinverse :The Singular Value Decomposition(SVD)n當Input V=Rn與 output W=Rm，用不相同的基底v1,v2,vn與u1,u2,um且T(v)=Av：qSVD:U-1AV=，其中vj在A的nullspace當j=r+1,n。qA與代表相同的線性轉換T。nA 是對應於標準基底，而對應於SVD其中的vs與us基底n正交矩陣V與U分別代表在Rn與Rm的identity線性轉換In與Imn矩

47、陣U-1AV=對應於線性轉換ImTIn U=u1 u2 um ,V=v1 v2 vn forfor jjjj rj ru0Avr=rank(A)2023/5/22427.4 Diagonalization and the Pseudoinverse :The Singular Value Decomposition(SVD)n7F The matrix in the new bases comes from A in the standard bases by U-1AV:vs to us=U-1 standard to us A standard V vs to standard The S

48、VD chooses orthonormal bases(U-1=UT and V-1=VT)that diagonalize A.2023/5/22437.4 Diagonalization and the Pseudoinverse :The Pseudoinverse A+nSVD 使得Avj=juj,A=UVT,反過來說滿足A-1 uj=vj/j 的矩陣”A-1”是存在的，因為A不必然是方陣其反矩陣A-1不一定存在，因此將A-1 改寫為A+，稱為虛擬反矩陣(Pseudoinverse)。即:quj ,j=1,r 屬於A的行空間(column space)，經由A+對應回到A的列空間(r

49、ow space)。quj ,j=r+1,m 屬於A的left nullspace，A+uj=0.1 for and for jjjjjrjrA uvA u011TT12121 .0nnrAV Uv vvu uu2023/5/22447.4 Diagonalization and the Pseudoinverse :The Pseudoinverse A+n當m=n=r,A-1存在，A-1=A+n+幾近是單位矩陣I.nIf 1=2,2=3 then 1111 by by by 1 1 0 0 0rrnnmnnm rth row1213 0 02 0 01 0 00 0 0 3 00 1 00

50、 0 00 0 00 0 0 rth column2023/5/22457.4 Diagonalization and the Pseudoinverse :The Pseudoinverse A+n7H The pseudoinverse A+is the n by m matrix with these 2 properties:qAA+=projection matrix onto the column space of A.qA+A=projection matrix onto the row space of A.n例題：找到矩陣 的虛擬反矩陣A+q解：先考慮SVD將A分解為UTAV