# Linear transformations and constant vectors

Let's call a column vector in $R^n$ constant if all the entries of it are the same. Let $A:R^n\rightarrow R^n$ be a linear transformation and let $B$ and $C$ be two bases in $R^n$. Suppose that $b_1 + b_2 + ... + b_n$ and $c_1 + c_2 + ... + c_n$ are both constant vectors and each column of the matrix $[A]_B$ (i.e the matrix of $A$ in the basis $B$) is also a constant vector. Show that in this case each column of the matrix $[A]_C$ is also a constant vector.

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Have you tried to solve the problem yourself? Where are you stuck? –  Ben Millwood Dec 10 '12 at 18:59
I actually have, but I didn't get far. Tomorrow is my final exam and just trying to solve the previous year's exam paper. –  Nima Dec 10 '12 at 19:01

If $h=(1,1,\ldots,1)\in\mathbb R^n$ then a vector $v\in\mathbb R^n$ is constant iff $v\in\text{span}(h)$. Thus $b_1+b_2+\ldots+b_n\in\text{span}(h)$ and $c_1+c_2+\ldots+c_n\in\text{span}(h)$.
The columns of $[A]_B$ are constant $\Longrightarrow Ab_i=\lambda_i(b_1+b_2+\ldots+b_n)\in\text{span}(h)$ for all $i \in \{1,2,\ldots,n\}$ for some $\lambda_i \in \mathbb R$.
Since $B$ is a basis of $\mathbb R^n$ this means that $\text{Im}A\subseteq\text{span}(h)$.
Therefore $Ac_i\in \text{span}(h)$ for all $i \in \{1,2,\ldots,n\} \Longrightarrow Ac_i=\mu_ih$ for some $\mu_i\in \mathbb R$. Since $c_1+c_2+\ldots+c_n\in\text{span}(h)$ and $C$ is a basis this means that $c_1+c_2+\ldots+c_n\neq 0 \Longrightarrow \text{span}(c_1+c_2+\ldots+c_n)=\text{span}(h)\Rightarrow \\h=\mu (c_1+c_2+\ldots+c_n) \Rightarrow Ac_i=\mu\mu_i(c_1+c_2+\ldots+c_n), \ \ \forall i \in \{1,2,\ldots,n\}.$
Therefore the columns of $[A]_C$ are constant.

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@copper.hat: I corrected that. Thanks! –  P.. Dec 10 '12 at 20:09
Let $S = \text{sp} \{e \}$, where $e = (1,...,1)^T$. You are given that $Be \in S$ and $Ce \in S$. So both $B,C$ (and hence their inverses) act as multiplication by a constant on $S$.
You are also given that $[A]_B e_i \in S$ for all $i$ ($e_i$ is the $i$th unit vector). It follows that since $x = \sum x_i e_i$ that $[A]_B x \in S$ for all $x$.
You have $[A]_C = C^{-1} B [A]_B B^{-1} C$. Since $[A]_B B^{-1} Cx \in S$, it follows from the above remarks that $[A]_C x = C^{-1} B [A]_B B^{-1} Cx \in S$ for all $x$. Setting $x = e_i$ shows that each column of $[A]_C$ is in $S$.