User:IssaRice/Linear algebra/Properties of a list of vectors and their images

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Let V and W be two finite-dimensional vector spaces, let T : V \to W be a linear transformation, and let v_1, \ldots, v_n \in V be a list of vectors (n is not necessarily equal to \dim V or \dim W).

Given the above background setting, we can now place various conditions on the list v_1, \ldots, v_n and the transformation T to conclude various things about the list Tv_1, \ldots, Tv_n. For instance, the first row below says that if v_1, \ldots, v_n is a linearly independent list and T is an injective map, then Tv_1, \ldots, Tv_n is also linearly independent.

Condition on v_1, \ldots, v_n Condition on T Conclusion about Tv_1, \ldots, Tv_n Notes
Linearly independent Injective Linearly independent
Spans V Surjective Spans W
Basis for V Bijective Basis for W Combine previous two rows
Basis for V No condition No conclusion Even though we have placed a strong condition on v_1, \ldots, v_n, if we don't place any conditions on T, then there is not much we can conclude about Tv_1, \ldots, Tv_n. If T sends everything to zero, then we just get 0,\ldots,0, which neither spans W nor is it a linearly independent list.
No condition Bijective No conclusion Even though we have placed a strong condition on T, if we don't place conditions on the list of vectors, then there is not much we can say about the list of the images. For instance, if V = W = \mathbf R^2, if T is the identity map, and if our list is (1,0),(1,0), then the image is also (1,0),(1,0), which is linearly dependent and does not span W.
Orthonormal Is an isometry Orthonormal
Eigenvectors of T corresponding to distinct non-zero eigenvalues No condition Linearly independent; eigenvectors of T corresponding to distinct eigenvalues Eigenvectors corresponding to distinct eigenvalues are linearly independent, so v_1, \ldots, v_n is a linearly independent list. Since each v_j is an eigenvector, we see that Tv_1, \ldots, Tv_n is the list \lambda_1 v_1, \ldots, \lambda_n v_n, where each \lambda_j is the corresponding eigenvalue. Since by assumption none of the eigenvalues are zero, the new list is also linearly independent.