Solutions to Homework 8 - Math 3410

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1 Soluions o Homework 8 - Mah 34. (Page 9: #.49(b)) Le a, b, c, d R be fixed. Show ha F : R R where F ((x, y)) = (ax + by, cx + dy). Soluion. Le (x, y ), (x, y ) R and k R. Then and F ((x, y ) + (x, y )) = F ((x + x, y + y )) = (a(x + x ) + b(y + y ), c(x + x ) + d(y + y )) = (ax + by, cx + dy ) + (ax + by, cx + dy ) = F ((x, y )) + F ((x, y )). F (k(x, y )) = F ((kx, ky )) = (a(kx ) + b(ky ), c(kx ) + d(ky )) Therefore F is linear. = k(ax + by, cx + dy ) = kf ((x, y )).. (Page 9: #.(b),(c)) Show he following maps are no linear: (b) F : R 3 R where F ((x, y, z)) = (x +, y + z). (c) F : R R where F ((x, y)) = (xy, y). Soluion. (b) F is no linear since F ((,, )) = ( +, + ) = (, ). Therefore F is no linear since i doesn map he zero vecor o he zero vecor. (c) Le x = y = and c =. Noe ha bu F ((, )) = (, ) = (, ) F ((, )) = F (, ) = (, ) = (4, ). Thus F ((, )) F ((, )) and F is no linear. 3. (Page 9: #.) Find F (a, b) where F : R R and F (, ) = (3, ) and F (, ) = (, ). Soluion. Le (a, b) R. Since {(, ), (, )} is a basis of R we deermine c, c such ha Tha is (a, b) = c (, ) + c (, ). a = c b = c + c. Solving his sysem, we see ha c = a and c = b c = b a. Therefore (a, b) = a(, ) + (b a)(, ). I follows ha F (a, b) = af (, ) + (b a)f (, ) = a(3, ) + (b a)(, ) = (3a, a) + (b 4a, b a) = (b a, b 3a).

2 4. (Page 9: #.6(b)) For each linear map F find a basis and dimension of he kernel and image of F : R 4 R 3 defined by F (x, y, z, ) = (x + y + 3z +, x + 4y + 7z +, x + y + 6z + ). Soluion. Noe ha x F ( y z ) = x + y + 3z + x + 4y + 7z + x + y + 6z + or F ( x) = A x where x = (x y z ) T and 3 A = By Gauss-Jordan eliminaion i may be shown ha A U = Therefore he pivos are in he firs and hird columns. Tha is, x and z are he pivo variables and y and are he free variables. Solving for he pivo variables in erms of he free variables we ge Therefore x + y = or x = y + z + = or z = ker(f ) = { x R 4 A x = } = { x R 4 U x = } = { x R 4 x = y +, z = } Thus an elemen x R 4 if x y + y z = y Hence ker(f ) = span( As explained in class, im(f ) = colsp(a). = y, ). + Since A U and he pivos are in he firs and hird columns of U. I follows ha he firs and hird columns of A are a basis of colsp(a). Tha is, 3 im(f ) = span(, 7 ). 6 Therefore dim(ker(f )) = and dim(im(f )) =.

3 . (Page 9: #.63(a)) Consider he linear mapping F : R 4 R 3 given by F ( x) = A x where A = 3 Soluion. Recall ha ker(f ) equals he soluion space (or nullspace) of A and ha im(f ) = colsp(a). By Gauss-Jordan eliminaion i may be shown ha 4 A U = 3. Therefore he pivos are in he firs and second columns. Tha is, x and y are he pivo variables and z and are he free variables. Solving for he pivo variables in erms of he free variables we ge Therefore x + 4 z = or x = 4 z + y z + 3 = or y = z 3 ker(f ) = { x R 4 A x = } = { x R 4 U x = } = { x R 4 x = 4 z +, y = z 3 } Thus an elemen x R 4 if x 4 z + y z = z 3 z Hence ker(f ) = span( 4 = y, 3 4 ). I remark here ha an alernaive answer is 4 ker(f ) = span(, 3 ). + (This can be obained jus by muliplying each of he vecors in he previous spanning se by. - Why does his sill work?). Recall ha im(f ) = colsp(a). Since A U and he pivos are in he firs and second columns of U. I follows ha he firs and second columns of A are a basis of colsp(a). Tha is, im(f ) = span(, ). 3 Therefore dim(ker(f )) = and dim(im(f )) =. 3 3

4 6. (Page 9: #.64) Find a linear mapping F : R 3 R 3 whose image is spanned by (,, 3) and (4,, 6). Soluion. Consider he marix A = Define F : R 3 R 3 by x 4 F ( y ) = z 3 6 x y z As explained in class his defines a linear mapping. Moreover, im(f ) = colsp(a). Noe ha A U = As he pivos occurs in columns one and wo of U i follows ha columns one and wo of A span colsp(a) = im(f ). 7. (Page 9: #.67) Suppose F : V U is linear. Show ha (a) he image of any subspace of V is a subspace of U; (b) he preimage of any subspace of U is a subspace of V. Soluion. (a) Le W be a subspace of V. Then he image of W is F (W ) = {F ( w) w W }. I claim ha F (W ) is a subspace of U. Noe ha F ( ) = and hus F (W ). Le w, w W. Then F ( w ) and F ( w ) are elemens of F (W ). Noe ha F ( w ) + F ( w ) = F ( w + w ) since F is linear. However, as W is a subspace of V, w + w W. Thus F ( w ) + F ( w ) F (W ). Le c K and F ( w) F (W ) for some w W. Then cf ( w) = F (c w) since F is linear. However, as W is a subspace of V, c w W. (b) Le W be a subspace of U. The preimage of W is he se F (W ) = { v V F ( v) W }. Since F ( ) = i follows ha F (W ). Le v, v F (W ). Then F ( v ), F ( v ) W. As W is a subspace of U i follows ha F ( v ) + F ( v ) W and as F is linear we have F ( v + v ) = F ( v ) + F ( v ) and hus F ( v + v ) W. Therefore v + v F (W ). 4

5 8. (Page 9: #.68) Show ha if F : V U is ono, hen dim(u) dim(v ). Deermine all linear maps F : R 3 R 4 ha are ono. Soluion. I will jus assume here ha U and V are finie dimensional. However, he resul is rue in general. Noe ha im(f ) is a subspace of V. As explained in class F is ono if and only if im(f ) = U. Now we have he ideniy dim(ker(f )) + dim(im(f )) = dim(v ). Bu since F is ono his is dim(ker(f )) + dim(u) = dim(v ). As dim(ker(f )) i follows ha dim(u) dim(v ). Suppose ha F : R 3 R 4 is an ono map. Then by he firs par of he problem 4 = dim(r 4 ) dim(r 3 ) = 3 which is a conradicion. Thus here are no ono linear mappings from R 3 o R 4.