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

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
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 $W$.
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.