General question in linear algebra:

Let $T:V\to W$ be a linear transformation and $\{v_1\,\dots,v_n \}$ a basis for B. If $\{v_1\,\dots,v_k \}$ is a basis for $kerT$.

Can I conclude that the vectors $\{T(v_{k+1}), \dots , T(v_n)\}$ are linearly independent in W?

A clear explanation would be appreciated.

  • $\begingroup$ No, let $T$ be the zero map. $\endgroup$ – Randall Jan 23 '18 at 16:59
  • $\begingroup$ Assuming T is not the zero map? $\endgroup$ – Moshe King Jan 23 '18 at 17:00
  • $\begingroup$ @Randall if $T$ is the zero map then $k = n$ $\endgroup$ – Omnomnomnom Jan 23 '18 at 17:00
  • $\begingroup$ Still no, for the same reasons. $T$ is allowed to collapse stuff. $\endgroup$ – Randall Jan 23 '18 at 17:00
  • $\begingroup$ Hmmm, need to re-read. $\endgroup$ – Randall Jan 23 '18 at 17:01

Yes, you can indeed conclude that those vectors are linearly independent.

In particular, let $c_{k+1},\dots,c_n$ be such that $$ c_{k+1}T(v_{k+1}) + \cdots + c_n T(v_n) = 0 $$ we can then write $$ T(c_{k+1}v_{k+1} + \cdots + c_n v_n) = 0 $$ which is to say that $c_{k+1}v_{k+1} + \cdots + c_n v_n \in \ker T$, which is the span of $\{v_1,\dots,v_k\}$. However, since the set $\{v_1,\dots,v_n\}$ is linearly independent, the vector $c_{k+1}v_{k+1} + \cdots + c_n v_n$ can only be in the span of $\{v_1,\dots,v_k\}$ if $c_{k+1} = \cdots = c_n = 0$.

  • $\begingroup$ Wow thank you for this explanation. A few more questions: If the dimension of $W \lt n-k$ then how is that possible? Also, can I conclude that $\{T(v_{k+1}),…,T(v_n)\}$ is a basis for $ImT$? $\endgroup$ – Moshe King Jan 23 '18 at 17:10
  • $\begingroup$ It is in fact impossible to have $\dim W < n - k$. In particular, the kernel will necessarily satisfy $\dim \ker T \geq \dim V - \dim W$. And yes, you can indeed conclude that $\{T(v_{k+1}),\dots,T(v_n)\}$ is a basis for the image of $T$. $\endgroup$ – Omnomnomnom Jan 23 '18 at 17:13
  • $\begingroup$ Nice and clear. $\endgroup$ – Randall Jan 23 '18 at 17:14
  • $\begingroup$ Thank you. Great answer! $\endgroup$ – Moshe King Jan 23 '18 at 17:15
  • $\begingroup$ @Randall glad you think so $\endgroup$ – Omnomnomnom Jan 23 '18 at 17:22

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