Linear algebra. Rank. linear independence.

[caffeinemachine]

Let $V$ be a finite dimensional vector space. Let $T$ be a linear transformation on $V$ with eigenvalue $0$. A vector $v \in V$ is

said to have rank $r > 0$ w.r.t eigenvalue $0$ if $T^rv=0$ but $T^{r-1}v\neq 0$. Let $x,y \in V$ be linearly independent and have

ranks $r_1$ and $r_2$ w.r.t eigenvalue $0$ respectively. Show that $\{x,Tx,\ldots ,T^{r_1-1}x,y,Ty,\ldots , T^{r_2-1}y \}$ is a

linearly independent set of vectors.

I can see that $\{ x, Tx, \ldots , T^{r_1-1}x \}$ are Linearly Independent, and $\{ y, Ty, \ldots, T^{r_2-1}y \}$ are linearly independent, but now I am stuck. Please help.

 

[Opalg]

Let $V$ be a finite dimensional vector space. Let $T$ be a linear transformation on $V$ with eigenvalue $0$. A vector $v \in V$ is

said to have rank $r > 0$ w.r.t eigenvalue $0$ if $T^rv=0$ but $T^{r-1}v\neq 0$. Let $x,y \in V$ be linearly independent and have

ranks $r_1$ and $r_2$ w.r.t eigenvalue $0$ respectively. Show that $\{x,Tx,\ldots ,T^{r_1-1}x,y,Ty,\ldots , T^{r_2-1}y \}$ is a

linearly independent set of vectors.

I can see that $\{ x, Tx, \ldots , T^{r_1-1}x \}$ are Linearly Independent, and $\{ y, Ty, \ldots, T^{r_2-1}y \}$ are linearly independent, but now I am stuck. Please help.

As stated, this result is clearly false. For example, given $x$ satisfying those conditions for some $r_1>1$, you could take $y=Tx$ (and $r_2 = r_1-1$).

For the result to be true, you need some extra conditions, such as requiring that neither $x$ nor $y$ is in the range of $T$, and that $r_1\ne r_2$. Then the generalised eigenspaces generated by $x$ and $y$ will correspond to different Jordan blocks of $T$ and will therefore be linearly independent.