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A) Let $V$ a vector space over $\mathbb{R}$ and $T: V \rightarrow V$ a linear mapping, such that $T^n=0$ for a natural $n \geq 2$. If $x \in V$, then the set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly independent if and only if $T^{n-1}x \neq 0$.

B) Let $V$ a vector space over $\mathbb{R}$ and $n \in \mathbb{N}$ odd. If the vectors $x_1, \dots,x_n$are linearly independent, then the same stands also for the vectors $x_1+x_2, x_2+x_3, \dots, x_{n-1}+x_n, x_n+x_1$.

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I have done the following:

For $(A)$:

for the direction $\Rightarrow$

The set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly independent:

$c_0x +c_1 Tx +c_2 T^2x+ \dots+c_{n-1} T^{n-1}x=0 \Rightarrow c_0=c_1=c_2= \dots=c_{n-1}=0$

$c_0x +c_1 Tx +c_2 T^2x+ \dots+c_{n-1} T^{n-1}x=0 \overset{\cdot T}{\Rightarrow} \\ c_0Tx +c_1 T^2x +c_2 T^3x+ \dots+c_{n-1} T^{n}x=0 \Rightarrow \\ c_0Tx +c_1 T^2x +c_2 T^3x+ \dots+c_{n-2}T^{n-1} =0 \overset{\cdot T}{\Rightarrow} \\ c_0T^2x +c_1 T^3x +c_2 T^4x+ \dots+c_{n-2}T^{n} =0 \Rightarrow \\ c_0T^2x +c_1 T^3x +c_2 T^4x+ \dots+c_{n-3}T_{n-1} =0 \overset{\cdot T}{\Rightarrow} \\ c_0T^3x +c_1 T^4x +c_2 T^5x+ \dots+c_{n-3}T_{n} =0 \Rightarrow \\ c_0T^3x +c_1 T^4x +c_2 T^5x+ \dots+c_{n-4}T^{n-1}=0 \overset{\cdot T}{\Rightarrow} \dots \Rightarrow c_0T^{n-2}x+c_1T^{n-1}=0 \overset{\cdot T}{\Rightarrow} \\ c_0T^{n-1}x+c_1T^{n}=0 \Rightarrow \\ c_0T^{n-1}x=0 \overset{c_0=0}{\Rightarrow} T^{n-1}x \neq 0$

Is this correct??

for the direction $\Leftarrow$

Knowing that $T^{n-1}x \neq 0$ we have to show that $c_0x +c_1 Tx +c_2 T^2x+ \dots+c_{n-1} T^{n-1}x=0 \Rightarrow c_1=c_2=\dots c_{n-1}=0$, right? But how could I do that??

For $(B)$:

$x_1, \dots,x_n$are linearly independent:

$c_1 x_1+c_2x_2+ \dots c_nx_n=0 \Rightarrow c_1=c_2= \dots =c_n=0 \ \ \ (*)$

$a_1(x_1+x_2)+a_2(x_2+x_3)+\dots+a_{n-1}(x_{n-1}+x_n)+a_n(x_n+x_1)=0 \Rightarrow \\ (a_1+a_n)x_1+(a_1+a_2)x_2+\dots(a_{n-1}+a_n)x_n=0 \overset{(*)}{\Rightarrow} \\ a_1+a_n=a_1+a_2= \dots =a_{n-1}+a_n=0$

Can we conclude from here that $a_1=a_2= \dots =a_{n-1}=a_n=0$ ??

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  • $\begingroup$ How do you conclude that $T^{n-1}x \neq 0$? $$c_0T^{n-1}x=0 \overset{c_0=0}{\Rightarrow} T^{n-1}x \neq 0$$ $\endgroup$ – Fermat Jun 28 '14 at 20:02
  • $\begingroup$ @Fermat I thought that since $c_0 =0$, it can be that $T^{n-1}x \neq 0$.. $\endgroup$ – Mary Star Jun 28 '14 at 20:03
  • $\begingroup$ One direction is clear. $T^{n-1}x$ can not be zero otherwise the given set will be a linearly dependent set. $\endgroup$ – Fermat Jun 28 '14 at 20:04
  • $\begingroup$ @MaryStar Being $c_0=0$, does not imply that $T^{n-1}x \neq 0$ $\endgroup$ – Fermat Jun 28 '14 at 20:09
  • $\begingroup$ @Fermat Ahaa... Do you have any idea how to show this?? $\endgroup$ – Mary Star Jun 28 '14 at 20:12
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To prove (A), it is clear that if $T^{n-1} x = 0$, then the set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly dependent.

Conversely, suppose the set is linearly dependent, and let $$ c_{0} x + c_{1} Tx + \dots + c_{n-1} T^{n-1} x = 0\tag{eq} $$ be a relation where not all coefficients are zero.

If $c_{0} = c_{1} = \dots = c_{k-1} = 0$, and $c_{k} \ne 0$, apply $T^{n-1-k}$ to (eq) to get $$ c_{k} T^{n-1} x = 0, $$ so that $T^{n-1} x = 0$ as $c_{k} \ne 0$.


As to (B), note that if $$ a_{1} (x_{1} + x_{2}) + a_{2} (x_{2} + x_{3}) + \dots + a_{n} (x_{n} + x_{1}) = 0, $$ then $$ (a_{n} + a_{1}) x_{1} + (a_{1} + a_{2} ) x_{2} + \dots + (a_{n-1} + a_{n}) x_{n} = 0, $$ so that $$ a_{n} + a_{1} = a_{1} + a_{2} = \dots = a_{n-1} + a_{n} = 0.\tag{eq2} $$ Now note that since $n$ is odd $$ 0 = (a_{1} + a_{2}) - (a_{2} + a_{3}) + \dots - (a_{n-1} + a_{n}) + (a_{n} + a_{1}) = 2 a_{1}, $$ so $a_{1} = 0$ and thus, by (eq2), all $a_{i}$ are zero.

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  • $\begingroup$ As for $(A)$: Is the formulation of the exercise of my post wrong?? Can we only say that $$\{x,Tx,T^2x, \dots, T^{n-1}x \} \text{ linearly dependent } \Leftrightarrow T^{n-1}x=0$$ ?? Could you explain me further why it stands that if $T^{n-1}x=0$ then the set $\{x, Tx, T^2x, \dots, T^{n-1}x\}$ is linearly dependent?? What do you mean by saying "...apply $T^{n-1-k}$ to (eq)" ?? $$$$ As for $(B)$: I haven't understood how the fact that $n$ is odd implies that $$0=(a_1+a_2)-(a_2+a_3)+ \dots -(a_{n-1}+a_n)+(a_n+a_1)=2a_1$$ Could you explain it to me?? $\endgroup$ – Mary Star Jun 28 '14 at 22:37
  • $\begingroup$ @MaryStar, your formulation is fine, and it is exactly the same thing to say $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly independent if and only if $T^{n-1}x \neq 0$ or $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly dependent if and only if $T^{n-1}x = 0$. It is an elementary fact, that you should master, that any set of vectors $v_{1}, \dots, v_{n}$ containing zero is linearly dependent: if $v_{1} = 0$, say, then $1 \cdot v_{1} + 0 \cdot v_{2} + \dots + 0 \cdot v_{n} = 0$. (continues) $\endgroup$ – Andreas Caranti Jun 29 '14 at 9:43
  • $\begingroup$ Then $T$ is a linear operator, and so it is its power $T^{n-1-k}$, so you can calculate $T^{n-1-k} (c_{0} x + c_{1} Tx + \dots + c_{n-1} T^{n-1} x) = T^{n-1-k} (0) = 0$. Finally, for $n = 2$ we have $(a_{1} + a_{2}) - (a_{2} + a_{1}) = 0$, for $n = 3$ we have $(a_{1} + a_{2}) - (a_{2} + a_{3}) + (a_{3} + a_{1}) = 2 a_{1}$, etc. Can you see the pattern? $\endgroup$ – Andreas Caranti Jun 29 '14 at 9:46

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