神经网络-配套-Ch6pres课件.ppt

1、61线性变换62Hopfield 网络的问题 网络的输出重复地乘以权矩阵 W。这个重复操作的结果是什么？输出是收敛,趋向无穷大，振荡？这一章将研究矩阵乘法，它表示了一般的线性变换。63线性变换一个变换包含三个部分：1.一个被称为定义域的元素集合 X=xi，2.一个被称为定义域的元素集合 Y=yi，和3.一个将每个 x i X 和一个元素 yi Y 相联系的规则。一个变换是线性的，如果：1.对所有的 x 1，x 2 X，A(x 1+x 2)=A(x 1)+A(x 2)，2.对所有的 x X，a ，A(a x )=a A(x)。64例子 旋转变换旋转变换是否是线性变换?1.2.65矩阵表示-(1)

2、两个有限维向量空间之间的任意线性变换都可由矩阵的乘法来表示。设 v1,v2,.,vn 是向量空间 X 的一个基，u1,u2,.,um 是向量空间 Y 的一个基。即对任意两个向量 x X 和 y Y，有xxivii1=n=yyiuii1=m=设 A:XYA xy=Axjvjj1=nyiuii1=m=66矩阵表示-(2)因为 A 是一个线性运算，xjA vjj1=nyiuii1=m=A vjaijuii1=m=因为 ui 是 Y 的基集,xjaijuii1=mj1=nyiuii1=m=(The coefficients aij will make up the matrix representat

3、ion of the transformation.)67矩阵表示-(3)uiaijxjj1=ni1=myiuii1=m=uiaijxjj1=nyii1=m0=由于 ui 是线性无关的,aijxjj1=nyi=a11a12 a1na21a22 a2nam1am2 amnx1x2xny1y2ym=这等价于矩阵乘法68小结 A linear transformation can be represented by matrix multiplication.To find the matrix which represents the transformation we must transfor

4、m each basis vector for the domain and then expand the result in terms of the basis vectors of the range.A vjaijuii1=m=Each of these equations gives us one column of the matrix.69例子-(1)Stand a deck of playing cards on edge so that you are looking at the deck sideways.Draw a vector x on the edge of t

5、he deck.Now“skew”the deck by an angle q,as shown below,and note the new vector y=A(x).What is the matrix of this transforma-tion in terms of the standard basis set?610例子-(2)A vjaijuii1=m=To find the matrix we need to transform each of the basis vectors.We will use the standard basis vectors for both

6、 the domain and the range.A sjaijsii1=2a1 js1a2 js2+=611例子-(3)We begin with s1:A s11s10s2+ai1sii1=2a11s1a21s2+=This gives us the first column of the matrix.If we draw a line on the bottom card and then skew thedeck,the line will not change.612例子-(4)Next,we skew s2:A s2q tans11s2+ai2sii1=2a12s1a22s2+

7、=This gives us the second column of the matrix.613例子-(5)变换矩阵是：A1q tan01=614基变换Consider the linear transformation A:XY.Let v1,v2,.,vn be a basis for X,and let u1,u2,.,um be a basis for Y.xxivii1=n=yyiuii1=m=A xy=Axy=The matrix representation is:a11a12 a1na21a22 a2nam1am2 amnx1x2xny1y2ym=615新基集Now let

8、s consider different basis sets.Let t1,t2,.,tn be a basis for X,and let w1,w2,.,wm be a basis for Y.yyiwii1=m=xxitii1=n=The new matrix representation is:a11a12 a1na21a22 a2nam1am2 amnx1x2xny1y2ym=Axy=616A 和 A 之间的关系?Expand ti in terms of the original basis vectors for X.titjivjj1=n=Expand wi in terms

9、 of the original basis vectors for Y.wiwjiujj1=m=wiw1iw2iwmi=tit1it2itni=617A 和 A 之间的关系?(续)Btt1t2tn=xx1t1x2t2xntn+Btx=Bww1w2wm=yBwy=Bw1ABtxy=ABw1ABt=Axy=ABtxBwy=Axy=相似变换618例子-(1)Take the skewing problem described previously,and find thenew matrix representation using the basis set s1,s2.t10.5s1s2+=t

10、2s1s2+=Btt1t20.5111=BwBt0.5111=t10.51=t211=(Same basis fordomain and range.)619例子-(2)ABw1ABt2 32 3231 31qtan010.5111=A2 3qtan1+2 3qtan23qtan23qtan1+=A5 32 3231 3=For q=45:A1 10 1=620例子-(3)Try a test vector:x0.51=x10=yAx5 32 3231 3105 323=yAx1 10 10.511.51=yB1y0.511111.512 32 3231 31.515 32 3=Check u

11、sing reciprocal basis vectors:621特征值和特征向量Let A:XX be a linear transformation.Those vectorsz X,which are not equal to zero,and those scalarsl which satisfy A(z)=l zare called eigenvectors and eigenvalues,respectively.Can you find an eigenvectorfor this transformation?622计算特征值Azlz=AlIz0=AlI0=A1 10 1=S

12、kewing example(45):1l101l0=1l20=l11=l21=1l101lz00=z110=对于这个变换只有一个特征向量。210=z0 10 0z10 10 0z11z2100=623对角化Perform a change of basis(similarity transformation)usingthe eigenvectors as the basis vectors.If the eigenvalues aredistinct,the new matrix will be diagonal.Bz1z2zn=z1z2 zn,Eigenvectorsl1l2 ln,EigenvaluesnB1ABl10 00 l2 000 l=624例子A1 11 1=1l111l0=l22ll l20=l10=l22=1l111lz00=1 11 1z11 11 1z11z2100=z21z11=z111=l10=1111z11111z12z2200=l22=z211=z22z12=AB1AB1 21 21 21 21 11 11 1110 00 2=对角形式：