Let $T$ be diagonalizable linear operator on a finite dimensional vector space $V$. Show that $V=\ker T\oplus \text{range}T$
Since $\dim V<\infty \implies$ so if we can prove that $\ker T\cap \text{Image} T=\{0\}$ then we are done.
Let $x\in \ker T\cap \text{Image}$ then $Tx=0$ and $x=Ty\implies T^2y =0$
Now since $T$ is diagonalizable so $V$ has a basis consisiting of eigen vectors of $T$ say $\{v_1,v_2\ldots ,v_n\}$
Hence $x=\sum_{i=1}^n c_iv_i\implies Tx=\sum_{i=1}^n c_iTv_i=\sum c_i\lambda_iv_i$
where $\lambda_i$ is an eigen value of $T$ corresponding to eigen vector $v_i$.
Now $Tx=0\implies \sum c_i\lambda_iv_i=0\implies c_i\lambda_i=0\forall i$ since $\{v_1,v_2\ldots ,v_n\}$ is linearly independent.
Claim : $c_i=0\forall i$.If not then $c_i\neq 0$ for some $i\implies \lambda_i=0\implies v_i\in \ker T$
But I can't show that $c_i=0$ from here.
How can I show it?
Please help.
Best Answer
Some of the $\lambda_i$'s are equal to $0$; otherwise, $\ker T=\{0\}$ and $V=T(V)$. Suppose that $\lambda_1=\cdots=\lambda_k=0$ and that $\lambda_i\neq0$ if $i>k$. Then, if $x=\sum_{j=1}^mc_jv_j$, $0=T(x)=\sum_{j=k+1}^nc_j\lambda_jv_j$. Therefore, each $c_j\lambda_j$ is differente from $0$ and, since $\lambda_j\neq0$, $c_j=0$.
So, $\ker T=\langle v_1,\ldots,v_k\rangle$ and $\operatorname{range}T=\langle v_{k+1},\ldots,v_n\rangle$.