Exercise 3.C.6

Question

Suppose $V$ and $W$ are finite-dimensional and $T \in \mathcal{L}(V,W)$. Prove that $	ext{dim range }T = 1$ if and only if there exist a basis of $V$ and a basis of $W$
such that with respect to these bases, all entries of $\mathcal{M}(T)$ equal $1$.

Arden Shi · Accepted Answer

Step 1. Prove that if $	ext{dim range }T = 1$, it follows that all entries of $\mathcal{M}(T)$ are $1$. \
Let $t$ be a vector in $	ext{range } T$ such that 
$$
	ext{span}(t) = 	ext{range } T.
$$
Thus, it follows that there must also exist a list of vectors, $v_1, ..., v_m$ that are a basis of $V$, and satisfy
$$
Tv_1 = \dotsb = Tv_m = t
$$
Therefore, all the entries of $\mathcal{M}(T)$ are indeed $1$. \
Step 2. Prove that if all entries of $\mathcal{M}(T)$ are $1$,  then $	ext{dim range }T = 1$.
Because all entries of $\mathcal{M}(T)$ are 1, 
\begin{equation}
\begin{split}
Tv_1 &= w_1 + \dotsb + w_n \
...\
Tv_m &= w_1 + \dotsb + w_n
\end{split}
\end{equation}

Therefore, 
$$
Tv_1= \dotsb = Tv_m
$$
Now, since $Tv_1, ..., Tv_m$ is a basis of $	ext{range } T$, yet they are all equal, so $m = 1$, proving the statement.
\qed

Exercise 3.C.6

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Exercise 3.C.6

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