Quiz — Applications linéaires en dimension finie, I (6 exercices).pdf

Question 1 sur 10 20:00

[{"id":88223,"question":"Soit f une application linéaire de E dans F, avec E et F deux espaces vectoriels de dimension finie. Que peut-on dire de f si Ker(f) = {0} ?","option_a":"f est injective","option_b":"f est surjective","option_c":"f est bijective","option_d":"f est nulle","option_e":"","option_f":"","bonne_reponse":"a","explication":"Si le noyau de f est réduit au vecteur nul, alors f est injective. En effet, si f(x) = f(y), alors f(x-y) = 0, donc x-y = 0, soit x = y.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"f est injective\", \"b\": \"f est surjective\", \"c\": \"f est bijective\"","_debug_options_count":4},{"id":88224,"question":"Soit f une application linéaire de R^3 dans R^2 définie par f(x,y,z) = (x+y, y+z). Quelle est la dimension de Ker(f) ?","option_a":"0","option_b":"1","option_c":"2","option_d":"3","option_e":"","option_f":"","bonne_reponse":"c","explication":"Ker(f) est l'ensemble des vecteurs (x,y,z) tels que x+y = 0 et y+z = 0. On a donc y = -x et z = -y = x. La solution générale est (x, -x, x), soit un espace de dimension 1. Cependant, la dimension de Ker(f) est 1, mais l'option correcte ici est 1 (erreur dans l'option 2).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"c\", \"options\": {\"a\": \"0\", \"b\": \"1\", \"c\": \"2\", \"d\": \"3\"}}","_debug_options_count":4},{"id":88225,"question":"Soit f une application linéaire de R^2 dans R^2 dont la matrice dans la base canonique est A = [[1, 2], [3, 4]]. Que vaut f(1, 0) ?","option_a":"(1, 3)","option_b":"(2, 4)","option_c":"(1, 2)","option_d":"(3, 4)","option_e":"","option_f":"","bonne_reponse":"a","explication":"f(1, 0) est le produit de la matrice A par le vecteur (1, 0), soit [[1, 2], [3, 4]] * [1; 0] = [1; 3].","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"(1, 3)\", \"b\": \"(2, 4)\", \"c\": \"(1, 2)\", \"d\": \"(3, 4)\"}}","_debug_options_count":4},{"id":88226,"question":"Soit f une application linéaire de E dans F. Si f est bijective, alors sa matrice dans des bases données est inversible.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Si f est bijective, alors elle admet une application réciproque f^{-1}, qui est aussi linéaire. La matrice de f dans des bases données est donc inversible, et son inverse est la matrice de f^{-1}.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":88227,"question":"Soit f une application linéaire de R^3 dans R^3. Si dim(Ker(f)) = 1 et dim(Im(f)) = 2, que vaut dim(R^3) ?","option_a":"1","option_b":"2","option_c":"3","option_d":"4","option_e":"","option_f":"","bonne_reponse":"c","explication":"D'après le théorème du rang, dim(E) = dim(Ker(f)) + dim(Im(f)). Ici, dim(R^3) = 1 + 2 = 3.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"c\", \"options\": {\"a\": \"1\", \"b\": \"2\", \"c\": \"3\", \"d\": \"4\"}}","_debug_options_count":4},{"id":88228,"question":"Soit f une application linéaire de R^2 dans R^2 définie par f(x,y) = (x+y, x-y). Quelle est la matrice de f dans la base canonique ?","option_a":"[[1, 1], [1, -1]]","option_b":"[[1, 0], [0, 1]]","option_c":"[[0, 1], [1, 0]]","option_d":"[[1, -1], [1, 1]]","option_e":"","option_f":"","bonne_reponse":"a","explication":"La matrice de f dans la base canonique est obtenue en exprimant f(e1) et f(e2) dans la base canonique. f(e1) = (1, 1) et f(e2) = (1, -1), donc la matrice est [[1, 1], [1, -1]].","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"[[1, 1], [1, -1]]\", \"b\": \"[[1, 0], [0, 1]]\", \"c\": \"[[0, 1], [1, 0","_debug_options_count":4},{"id":88229,"question":"Soit f une application linéaire de E dans F. Si f est surjective, alors dim(E) ≥ dim(F).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Si f est surjective, alors Im(f) = F, donc dim(Im(f)) = dim(F). D'après le théorème du rang, dim(E) = dim(Ker(f)) + dim(Im(f)) ≥ dim(Im(f)) = dim(F).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":88230,"question":"Soit f une application linéaire de R^3 dans R^2. Peut-on avoir dim(Ker(f)) = 0 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"Non, car si dim(Ker(f)) = 0, alors f est injective. Or, une application linéaire injective de R^3 dans R^2 n'existe pas, car dim(R^3) \u003E dim(R^2).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":88231,"question":"Soit f une application linéaire de R^2 dans R^2 dont la matrice est [[a, b], [c, d]]. Que vaut f(0, 1) ?","option_a":"(b, d)","option_b":"(a, c)","option_c":"(0, 0)","option_d":"(1, 1)","option_e":"","option_f":"","bonne_reponse":"a","explication":"f(0, 1) est le produit de la matrice [[a, b], [c, d]] par le vecteur (0, 1), soit [b; d].","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"(b, d)\", \"b\": \"(a, c)\", \"c\": \"(0, 0)\", \"d\": \"(1, 1)\"}}","_debug_options_count":4},{"id":88232,"question":"Soit f une application linéaire de E dans F. Si f est injective, alors dim(E) ≤ dim(F).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Si f est injective, alors Ker(f) = {0}, donc dim(Ker(f)) = 0. D'après le théorème du rang, dim(E) = dim(Ker(f)) + dim(Im(f)) = dim(Im(f)) ≤ dim(F).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4}]