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Let $ T: \mathbb R^5 \to \mathbb R^5$ be a linear transformation such that $T^4(x)=0$. Then find the $Rank(T)$. Options given are

a) $Rank(T)=2$ , b) $Rank(T)=3$

I dont have any idea where to begin. I posed the same question in chat in MSE, but we were not able to answer it. So i am posting it here. You can assume that $T^k \ne 0$ for $k \le 3$. Help appreciated.

Harsh Kumar
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Shobhit
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  • I think you must (and not only "can") assume $;T^k\neq0;$ for $;k\le 3;$ , otherwise the options are wrong as $;T=0;$ is a possible option... – DonAntonio Jan 29 '17 at 10:20

2 Answers2

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Since $T^3\ne0$, take $u \in \mathbb R^5$ such that $T^3u\ne0$.

Then $Tu, T^2u, T^3u$ are linearly independent and so the rank of $T$ is at least $3$.

Indeed, applying $T^2$ to $a_1 Tu + a_2 T^2u + a_3 T^3u=0$ gives $a_1=0$.

Applying $T$ to $a_2 T^2u + a_3 T^3u=0$ gives $a_2=0$ and then $a_3=0$.

lhf
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  • Nice. Got it. Thank u. – Shobhit Jan 29 '17 at 11:44
  • It was given that T^4(x)=0, is that why u took T^3, T^2, T and proved them independent? How would the answer have changed if it was given T^2(x) = 0? – Shobhit Jan 29 '17 at 11:52
  • @Shobhit, $T^2=0$, $T\ne0$ imply that the rank of $T$ is at least $1$, bu that is trivial in this case. – lhf Jan 29 '17 at 11:53
  • But in another question, we have, T: R^7 -> R^7 such that T^2=0, but the options are a) <=3, b) >3 , c) =5 d) =6. And by your method, in the comment above rank should be 1. Can u explain this? – Shobhit Jan 29 '17 at 12:23
  • answer please. i have doubts – Shobhit Jan 29 '17 at 18:43
  • @Shobhit, ask a separate question. – lhf Jan 29 '17 at 19:17
  • Your answer is not clear, applying your method to a different question is giving the wrong answer. Only u cam clarify your answer, not others. – Shobhit Jan 30 '17 at 03:32
  • @lhf Why you didn't count $u$ in the linearly independent set. ${u, Tu, T^2u, T^3 u}$ is L.I. Right? Which statement of the theorem guarantees the connection between that rank and the number of elements in the above L.I sets? can you please explain? –  Dec 05 '17 at 13:37
  • @ManeeshNarayanan, but $u$ is not necessarily in the image of $T$. – lhf Dec 05 '17 at 13:38
  • @lhf you are right!!!Rank of $T$ is at least $3$. –  Dec 05 '17 at 13:50
  • @lhf Thank you very much. –  Dec 05 '17 at 13:52
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Start by assuming that $\{e_1, e_2, e_3, e_4, e_5\}$ is a basis for $\Bbb R^5$.

We start by trying to create some $T$ that satisfies what you describe, a $T$ such that $T^4(x) = 0\ \forall x$.

Try this:

$$\begin{cases}T(e_1) = 0\\T(e_{i+1}) = e_i, i > 1 \end{cases}$$

So what is the power of $T$ that takes all the vectors from the basis to $0$? Well, $T^1$ takes only the first, $e_1$, $T^2$ takes $e_1, e_2$, ..., $T^4$ takes $e_1, e_2, e_3$ and $e_4$. Thus it is only for $T^5$ that we have $T^5(x) = 0$. So what if we redefine our $T$ as

$$\begin{cases}T(e_1) = 0\\T(e_2) = 0\\T(e_{i+1}) = e_i, i > 2\end{cases}$$

can you see that $T^4(x) = 0$ and that $T^3 \neq 0$? If this is our $T$, then rank $ T = 3$. Can you see why?

So your answer could be $3$. Is it the only valid option? We can prove so. Remember that if $T$ is a linear transformation from $\Bbb R^5$ to itself, then $\Bbb R^5 = N(T) \oplus R(T)$.

Write a basis for each of the nullity of $T$ and the row space of $T$. If $T$ has rank $2$, then the row space has dimension $2$ and the nullity has space $3$. Now, if a vector $v$ belongs to the null space, $T^k(v) = 0\ \forall k>0$. However, the image of a vector belonging to the row space can still be a vector belonging to the row space, or belonging to the nullity. But we know that $T^4(x) = 0$ so it must be the case that every time you apply $T$ to every vector you have, some vectors of the row space get mapped to a vector from the nullity. In what ways can this happen? Try sketching a bit. Can you work this out?

RGS
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  • Yes, i got it. Thank u. Just one question : the part where u say $R^5$ is the direct sum of N(T) and R(T). How? Also, after that, what you wrote is beyond me. Still, thank u for your help :) – Shobhit Jan 29 '17 at 10:24
  • Try checking the rank-nullity theorem – RGS Jan 29 '17 at 10:40
  • I know the rank nullity theorom. I dont follow. – Shobhit Jan 29 '17 at 10:57
  • @Shobhit what didn't you follow? Do you already understand why $\Bbb R^5 = N(T) \oplus R(T)$? – RGS Jan 29 '17 at 11:10
  • $\Bbb R^5 = N(T) \oplus R(T)$ is false in general. – lhf Jan 29 '17 at 11:22
  • @RSerrao no, i dont understand how u wrote that R^5 is a direct sum of N(T) and R(T). I am saying i know rank nullity theorom, but i dont follow how this proves your case. Please explain. Also please explain what is direct sum. Thank u. – Shobhit Jan 29 '17 at 11:30
  • @lhf what do you mean by "in general"? In our context, $T$ a linear transformation $\Bbb R^5 \to \Bbb R^5$ and so $\Bbb R^5 = N(T) \oplus R(T)$ – RGS Jan 29 '17 at 11:34
  • Try $T(x,y)=(y,0)$ on $\mathbb R^2$. – lhf Jan 29 '17 at 11:36
  • Answer my questions please – Shobhit Jan 29 '17 at 12:36
  • @Shobhit I will; I am just seeking clarification regarding the notation. I may have been using the wrong names for the wrong things – RGS Jan 29 '17 at 12:39
  • @lhf I see that $N(T)$ is generated by the vector $(1, 0)$ and that $(1,0)$ is also the vector that generates the image of $T$, but isn't $R(T)$ the row space of $T$? And given that the matrix representation of your $T$ has $(0, 1)$ as only non-null row, we have $\Bbb R^2 = N(T) \oplus R(T) = \text{span}{(1,0)} \oplus \text{span}{(0,1)}$ – RGS Jan 29 '17 at 12:42
  • $R(T)$ is the image space of $T$, the column space for a matrix. – lhf Jan 29 '17 at 19:19