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
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.