I am studying linear algebra by Hoffman Kunze. In the introductory part of chapter 6, I have the following:
Here is an illustration of what we have in mind. Perhaps the simplest matrices to work with, beyond the scalar multiples of the identity, are the diagonal matrices:
$$ D=\left[\begin{array}{ccccc} c_{1} & 0 & 0 & \cdots & 0 \\ 0 & c_{2} & 0 & \cdots & 0 \\ 0 & 0 & c_{3} & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & c_{n} \end{array}\right]\tag{6-1} $$ Let, $T$ be a linear operator on an $n$-dimensional space $V$. If we could find an ordered basis $B=\left\{\alpha_{1}, \ldots, \alpha_{n}\right\}$ for $V$ in which $T$ were represented by a diagonal matrix $D$ [as in (6-1)], we would gain considerable information about $T$. For instance, simple numbers associated with $T$, such as the rank of $T$ or the determinant of $T$, could be determined with little more than a glance at the matrix $D$. We could describe explicitly the range and the null space of $T$. Since $[T]_{B}=D$ if and only if $$ T \alpha_{k}=c_{k} \alpha_{k}, \quad k=1, \ldots, n $$ the range would be the subspace spanned by those $\alpha_{k}$ 's for which $c_{k} \neq 0$ and the null space would be spanned by the remaining $\alpha_{k}$ 's. Indeed, it seems fair to say that, if we knew a basis $B$ and a diagonal matrix $D$ such that $[T]_{B}=D$, we could answer readily any question about $T$ which might arise.
My Doubt:
Here how does it conclude that the range would be the subspace spanned by those $\alpha_{k}$ 's for which $c_{k} \neq 0$ and the null space would be spanned by the remaining $\alpha_{k}$ 's ? Please explain it with proof or any theorem that they have used and why is it true for the case of daigonal matrices only.
Thanks for any help!
If $c_k\ne0$, then $\alpha_k$ belongs to the range of $T$, since it is equal to $T\left(\frac1{c_k}\alpha_k\right)$. On the other hand, if $v$ belongs to the range of $T$, then $v=T(a_1\alpha_1+a_2\alpha_2+\cdots+a_n\alpha_n)$ for some scalars $a_1,a_2,\ldots,a_n$. In other words, $v=a_1T(\alpha_1)+a_2T(\alpha_2)+\cdots+a_nT(\alpha_n)$. But this is a linear combination of those $T(\alpha_k)$'s for which $c_k\ne0$. Since $T(\alpha_k)$ is a multiple of $\alpha_k$, this proves that the range of $T$ is the span of those $\alpha_k$'s such that $c_k\ne0$.
On the other hand, if $c_k=0$, $T(\alpha_k)=c_k\alpha_k=0$. So, $\alpha_k\in\ker T$. And if $v\in\ker T$, and $v$ cannot be written as a linear combination of those $\alpha_k$'s such that $c_k\ne0$, then $v=a_1\alpha_1+a_2\alpha_2+\cdots+a_n\alpha_n$, where, for some $k$, $a_k\ne0$. But$$0=T(v)=a_1c_1\alpha_1+a_2c_2\alpha_2+\cdots+a_nc_n\alpha_n.$$Since the $\alpha_k$'s are linearly independent and some $c_k$'s are equal to $0$, you must have $a_kc_k=0$ those $k$'s such that $c_k\ne0$. So, $a_k=0$ for those $k$'s. And this proves that $v$ is a linear combination of those $\alpha_k$'s such that $c_k=0$.