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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.

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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... – 2017-01-29
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    I am weak in english, i suppose :p – 2017-01-29

2 Answers 2

2

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 $a_3=0$.

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    Nice. Got it. Thank u. – 2017-01-29
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    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? – 2017-01-29
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    @Shobhit, $T^2=0$, $T\ne0$ imply that the rank of $T$ is at least $1$, bu that is trivial in this case. – 2017-01-29
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    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? – 2017-01-29
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    answer please. i have doubts – 2017-01-29
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    @Shobhit, ask a separate question. – 2017-01-29
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    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. – 2017-01-30
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    @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? – 2017-12-05
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    @ManeeshNarayanan, but $u$ is not necessarily in the image of $T$. – 2017-12-05
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    @lhf you are right!!!Rank of $T$ is at least $3$. – 2017-12-05
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    @lhf Thank you very much. – 2017-12-05
1

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?

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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 :) – 2017-01-29
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    Try checking the [rank-nullity theorem](https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem) – 2017-01-29
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    I know the rank nullity theorom. I dont follow. – 2017-01-29
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    @Shobhit what didn't you follow? Do you already understand why $\Bbb R^5 = N(T) \oplus R(T)$? – 2017-01-29
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    $\Bbb R^5 = N(T) \oplus R(T)$ is false in general. – 2017-01-29
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    @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. – 2017-01-29
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    @lhf what do you mean by "in general"? In our context, $T$ a linear transformation $\Bbb R^5 \to \Bbb R^5$ and so $\Bbb R^5 = N(T) \oplus R(T)$ – 2017-01-29
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    Try $T(x,y)=(y,0)$ on $\mathbb R^2$. – 2017-01-29
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    Answer my questions please – 2017-01-29
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    @Shobhit I will; I am just seeking clarification regarding the notation. I may have been using the wrong names for the wrong things – 2017-01-29
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    @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 $\Bbb R^2 = N(T) \oplus R(T) = \text{span}\{(1,0)\} \oplus \text{span}\{(0,1)\}$ – 2017-01-29
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    $R(T)$ is the image space of $T$, the *column* space for a matrix. – 2017-01-29