Let $V$ be a finite-dimensional vector space and let $T$ be a linear operator on $V$. Suppose that $\text{rank}(T^{2}) = \text{rank}(T)$. Prove that the range and null space of $T$ have only the zero vector in common.
MY ATTEMPT
Let $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ be a basis for $V$, where $\alpha_{1},\alpha_{2},\ldots,\alpha_{k}$ is a basis for the null space $N$ of $T$ and $T\alpha_{k+1},\ldots, T\alpha_{n}$ is a basis for the range $R$ of $T$. We must prove that $\alpha_{1},\ldots,\alpha_{k},T\alpha_{k+1},\ldots,T\alpha_{n}$ are linear independent, which implies that $N\cap T = \{0\}$. Let us assume that $\alpha\in N\cap R$. Then we can express it in both basis: $\alpha = a_{1}\alpha_{1} + \ldots + a_{k}\alpha_{k}$ and $\alpha = a_{k+1}T\alpha_{k+1} + \ldots + a_{n}T\alpha_{n}$. Consequently, one has
\begin{align*}
& a_{1}\alpha_{1} + \ldots + a_{k}\alpha_{k} – a_{k+1}T\alpha_{k+1} – \ldots – a_{n}T\alpha_{n} = 0 \Longrightarrow\\\\
& a_{k+1}T(T\alpha_{k+1}) + \ldots + a_{n}T(T\alpha_{n}) = a_{k+1}T^{2}\alpha_{k+1} + \ldots + a_{n}T^{2}\alpha_{n} = 0
\end{align*}
Since $\text{rank}(T^{2}) = \text{rank}(T)$, let $\beta_{k+1},\beta_{k+2},\ldots,\beta_{n}$ be a basis for $R(T^{2})$, from whence we get
\begin{align*}
& a_{k+1}\sum_{j=k+1}^{n}b^{k+1}_{j}\beta_{j} + \ldots + a_{n}\sum_{j=k+1}^{n}b^{n}_{j}\beta_{j} = 0 \Longrightarrow\\\\
& (a_{k+1}b^{k+1}_{k+1} + \ldots + a_{n}b^{n}_{k+1})\beta_{k+1} + \ldots + (a_{k+1}b^{k+1}_{n} + \ldots + a_{n}b^{n}_{n})\beta_{n} = 0 \Longrightarrow\\\\
& a_{k+1}b^{k+1}_{k+1} + \ldots + a_{n}b^{n}_{k+1} = \ldots = a_{k+1}b^{k+1}_{n} + \ldots + a_{n}b^{n}_{n} = 0
\end{align*}
Then I get stuck. If we prove that $a_{k+1} = a_{k+2} = \ldots = a_{n} = 0$, then $a_{1} = a_{2} = \ldots = a_{k}$, and $N\cap T = \{0\}$. Could someone help me to complete this demonstration or exhibit any theoretical flaw?
Best Answer
I appreciate the enthusiasm with which you dove into this, applying definitions, and mucking in with bases. However, there's a key fact that I think you're missing, which simplifies things: $$R(T^2) \subseteq R(T).$$ This is true for any operator $T$ (even nonlinear operators!), and can be proven in a single line: $$v \in R(T^2) \implies (\exists w \in V)(v = T(Tw)) \implies v \in R(T).$$ Therefore, if we know that $\dim R(T^2) = \dim R(T)$, then these spaces must be the same, and $T|_{R(T)}$ (i.e. $T$ restricted to $R(T)$), is actually a surjective linear operator! As such, $T|_{R(T)}$ is injective (recall, surjective linear operators on finite-dimensional spaces are invertible), hence $$T(a_{k+1} T v_{k+1} + \ldots + a_n T v_n) = 0 \implies a_{k+1} T v_{k+1} + \ldots + a_n T v_n = 0,$$ as required.
As I originally hinted in the comments, there's a slightly slicker way to do this using the rank-nullity theorem, and the dual observation $$N(T) \subseteq N(T^2).$$ By the rank-nullity theorem, $\dim N(T^2) = \dim N(T)$, hence $$N(T) = N(T^2).$$ If $x \in N(T) \cap R(T)$, then there exists some $v \in V$ such that $x = Tv$, but at the same time, $Tx = 0$. Thus $T^2 v = Tx = 0$, hence $v \in N(T^2) = N(T)$. Thus $x = Tv = 0$, as required.