Very nice video. You can also prove the theorem on invariant subspaces for diagonalizable operators by induction, by showing that whenever U is an invariant subspace and u in U is written as u = v1 + ... + vk with each vj an eigenvector in V_j with eigenvalue lambda_j, then each v_j belongs to U. The induction is on the number of nonzero vectors among the vk. If there's only 1, it's already in U. If not, note that Tu - lambda_1*u = (lambda_2 - lambda_1)v_2 + ... + (lambda_k - lambda_1) v_k is another vector in U written as a combination of the eigenvectors, then by induction all the nonzero vectors v_2, ..., v_k are in U (none of them are "zeroed out" because lambda_j - lambda_1 is nonzero if j > 1). Then v_1 = u - (v_2 + ... + v_k) is a difference of vectors in U, and so v_1 is in U as well. Thus all the eigenvectors v_1, ..., v_k in the combination u = v1 + ... + vk are actually in U. This allows you to skip the use of the Vandermonde determinant, but of course I wouldn't claim it's a "simpler" approach at all lol. Apologies if some typos above, it's been a minute since module theory.
You are absolutely right, it is possible to avoid the Vandermonde determinant argument by applying your ideas. Nonetheless, the Vandermonde determinant method is a common trick in algebra, and it is useful to have it in the toolbox. Here is another related situation, where the same Vandermonde argument may be applied: Suppose we have a vector-valued polynomial: P(t) = v_n t^n + v_{n-1} t^{n-1} + ... v_1 t + v_0, where all v_i belong to vector space V. Lemma. If we know that for some n+1 of distinct points t_0, ..., t_n, the values P(t_k) belong to a subspace U, then all coefficients v_n, ..., v_0 belong to U.
@@yulybillig2814Ah, had never seen that one. Another nice application of VdM determinant. Can also be done inductively by looking at difference quotient, mean value theorem, etc. Thanks for the reply 👍
Also! You can reduce that one to the case of a scalar polynomial of degree n having n+1 zeroes, just by composing with any linear functional on V that is 0 on the subspace U.
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
Спасибо, вы очень хорошо объясняете как жаль что я нашел ваш канал только в конце семестра
Very nice video. You can also prove the theorem on invariant subspaces for diagonalizable operators by induction, by showing that whenever U is an invariant subspace and u in U is written as u = v1 + ... + vk with each vj an eigenvector in V_j with eigenvalue lambda_j, then each v_j belongs to U. The induction is on the number of nonzero vectors among the vk. If there's only 1, it's already in U. If not, note that Tu - lambda_1*u = (lambda_2 - lambda_1)v_2 + ... + (lambda_k - lambda_1) v_k is another vector in U written as a combination of the eigenvectors, then by induction all the nonzero vectors v_2, ..., v_k are in U (none of them are "zeroed out" because lambda_j - lambda_1 is nonzero if j > 1). Then v_1 = u - (v_2 + ... + v_k) is a difference of vectors in U, and so v_1 is in U as well. Thus all the eigenvectors v_1, ..., v_k in the combination u = v1 + ... + vk are actually in U. This allows you to skip the use of the Vandermonde determinant, but of course I wouldn't claim it's a "simpler" approach at all lol. Apologies if some typos above, it's been a minute since module theory.
You are absolutely right, it is possible to avoid the Vandermonde determinant argument by applying your ideas. Nonetheless, the Vandermonde determinant method is a common trick in algebra, and it is useful to have it in the toolbox.
Here is another related situation, where the same Vandermonde argument may be applied: Suppose we have a vector-valued polynomial: P(t) = v_n t^n + v_{n-1} t^{n-1} + ... v_1 t + v_0, where all v_i belong to vector space V.
Lemma. If we know that for some n+1 of distinct points t_0, ..., t_n, the values P(t_k) belong to a subspace U, then all coefficients v_n, ..., v_0 belong to U.
@@yulybillig2814Ah, had never seen that one. Another nice application of VdM determinant. Can also be done inductively by looking at difference quotient, mean value theorem, etc. Thanks for the reply 👍
Also! You can reduce that one to the case of a scalar polynomial of degree n having n+1 zeroes, just by composing with any linear functional on V that is 0 on the subspace U.