I have a pretty simple question that confused me:

V is a vector space of a finite dimension. $T: V \to V$ is a linear transformation. The information that's been given in question: $\operatorname{Im} T = \ker T$

I want to know if what I'm doing is right: I took only for exmaple: $\dim \ker T = \dim \operatorname{Im} T = 2$

I take $D = \{v_1,v_2\}$ as some basis for $\ker T$ and $\operatorname{Im} T$

then I add vectors and let's say $B = \{v_1,v_2,v_3,v_4\}$ is basis of $V$, and I demand that $v_3$ and $v_4$ follow these requirements: $T(v_3) = v_1$, $T(v_4) = v_2$. (I did this for a reason, I didn't write here the whole question) But I have to prove that this is basis of $V$, linearly independent and spanning $V$.

So I tried to prove that it is linearly independent: I assume : $a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 = 0$ Then I applied $T$ on it : $T(a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4) = 0$

$a1 * T(v1) + a2* T(v2) + a3*T(v3) + a4* T (v4) = 0$.

$v_1$ and $v_2$ are from $\ker T$ so $T(v_1) = 0$ and $T(v_2) = 0$,

so I get that there is a linear dependence, but I don't know where is the dependence: is the action that I did means that $\{v_1,v_2,v_3,v_4\}$ are linearly dependent or it means ONLY that $T(v_1)$, $T(v_2)$, $T(v_3)$, $T(v_4)$ are dependent?

Thank you

  • $\begingroup$ The information was: Im T = Ker T , so I can understand that they are the same subspace as I see it $\endgroup$ – CnR Jul 5 '13 at 14:13
  • $\begingroup$ Ah, apologies, I should have read closer. It does say that $\ker(T)=\mathrm{im}(T)$. $\endgroup$ – Zev Chonoles Jul 5 '13 at 14:19

Since $\dim\mathop{\rm Ker} T > 0$, the operator $T$ is singular, hence it reduces the dimension of the space (because $\dim\mathop{\rm Im} T = n - \dim\mathop{\rm Ker} T < n$), meaning it maps any base to a linearly dependent set (of vectors that span its image).

In other words, $T(B)$ is a set of $4$ linearly dependent vectors, because $T(B) \subseteq \mathop{\rm Im} T$, which is a 2-dimensional space (so it cannot have more than 2 linearly independant vectors).

  • $\begingroup$ Okay I understand your explanation, thank you. Can you answer me on the last question that I wrote about the dependence? I want to be sure that the way I did it is correct generally.. if I assume : a1v1+a2v2+a3v3+a4v4=0 and apply T on it and get that the ai-s are all 0 it means that only the T(v) are dependent or we can get by it that the vi -s are dependent? $\endgroup$ – CnR Jul 5 '13 at 15:38
  • $\begingroup$ No, this is not correct. If you have an expression of a form $\sum_{k=1}^n \alpha_k v_k = 0$ that is possible only when all $\alpha_k = 0$, that means that $v_k$ are independent. Same goes for vectors $T(v_k)$, i.e., if $\sum_{k=1}^n \alpha_k T(v_k) = 0$ is possible only when all $\alpha_k = 0$, then all $T(v_k)$ are independent. But, as I explained, in your case this is impossible (due to the dimension of $\mathop{\rm Im}T$). On the other hand, if $\sum_{k=1}^n \alpha_k v_k = 0$ for some $\alpha_k$ that are not all equal to zero, vectors are dependent (same for $T(v_l)$ instead of $v_l$). $\endgroup$ – Vedran Šego Jul 5 '13 at 15:48
  • $\begingroup$ In your example, $a_1T(v_1)+a_2T(v_2)+a_3T(v_3)+a_4T(v_4)=0$ doesn't yield $a_k = 0$. It is easy to show that $T(v_1) = T(v_2) = 0$ (because $v_1$ and $v_2$ span image ($T(v_3) = v_1$ and $T(v_4) = v_2$) and the image is same as kernel). So, your equation will be true for any $a_1$ and $a_2$. $\endgroup$ – Vedran Šego Jul 5 '13 at 15:59
  • $\begingroup$ I'm sorry, I had a mistake. I meant to ask: if I assume : a1v1+a2v2+a3v3+a4v4=0 and apply T on it and find out that there are ai-s that are not 0 so the whole expression is 0 [dependent], does it mean that only the T(v) are dependent or we can get by it that the vi -s are dependent? $\endgroup$ – CnR Jul 5 '13 at 19:04
  • $\begingroup$ Only $T(v_k)$ are dependent. To make it easier to see this, take $T \equiv 0$, then any set of vectors $\{v_k\}$, be it dependent or independent, will be mapped to zero (which is dependent by definition). A bit more complex example, where ${\rm Ker}\ T = {\rm Im}\ T$, you can take (switch to matrix notation; it's easier):$$T = \begin{bmatrix} 0 & {\rm I} \\ 0 & 0 \end{bmatrix},$$where $\rm I$ denotes identity matrix of order $n/2$. Now, see what happens with the vectors of the canonical base $e_k = \begin{bmatrix} 0 & \dots & 0 & 1 & 0 & \dots & 0 \end{bmatrix}^T$ ($1$ is in $k$-th position). $\endgroup$ – Vedran Šego Jul 5 '13 at 20:21

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