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In my preparations for an upcoming exam, I'm working through the past exams available from my department. One of the questions asks for an example of a linear operator on some vector space with more than one right inverse.

In another part of the same problem, I proved that right-inverse uniqueness implies invertibility, but I feel like I'm missing how that result will help me find or construct an example (assuming that it's applicable.)

I've considered various square matrices (because I'm looking for an operator) over various fields, but nothing has worked yet and I don't feel like I'm going about it the right way.

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If you want to have an operator $T \colon V \to V$ with more than one right inverse, you need an infinite-dimensional $V$. –  Daniel Fischer Sep 12 '13 at 21:21
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@wil Square matrices will never work. –  azarel Sep 12 '13 at 21:22
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2 Answers 2

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Let $V$ be the space of infinite sequences of real numbers, and let $T$ take the sequence $x_1,x_2,x_3,\dots$ to the sequence $x_1+x_2,x_3,x_4,\dots$. Then $T$ has at least two right inverses: for example, $U$, which takes $y_1,y_2,y_3,y_4\dots$ to $0,y_1,y_2,y_3\dots$ and $V$, which takes $y_1,y_2,y_3,y_4\dots$ to $y_1,0,y_2,y_3\dots$.

If you're allowed operators $T:V\to W$ where $V\neq W$, then you can use a matrix such as $T=\begin{pmatrix}1 & 0\end{pmatrix}:\mathbb R^2\to\mathbb R:\begin{pmatrix}a\\b\end{pmatrix}\mapsto a$. Then $S(x)=\begin{pmatrix}1\\x\end{pmatrix}:c\mapsto\begin{pmatrix}c\\xc\end{pmatrix}$ is linear and an inverse for $T$, for any $x$.

If $V$ is a finite-dimensional vector space, then no linear map $T:V\to V$ will have more than one right inverse. Suppose that $S_1,S_2$ are two right inverses for $T$. Then $TS_1x=TS_2x=x$, so $T(S_1-S_2)x=0$ for any $x\in V$. So, for each $x\in V$, $(S_1-S_2)x\in\ker T$. By the rank-nullity theorem, $\dim\ker T = \dim\textrm{Im} T-\dim V$. But we must have $\dim\textrm{Im} T=\dim V$, because, for any $x\in V$, $x=Ix=TS_1x=T(S_1x)\in\textrm{Im} T$. So $\dim\ker T=0$, so the kernel of $T$ is the singleton set $\{0\}$. So $(S_1-S_2)x=0$ for every $x\in V$; i.e., $S_1-S_2=0$, so $S_1=S_2$.

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In the context of vector spaces the inverse is assumed to be linear. Then the shift operator you describe only has one inverse. –  Ittay Weiss Sep 12 '13 at 22:26
    
Thanks; I'll change it. –  Donkey_2009 Sep 12 '13 at 22:29
    
This is similar to what I ended up doing after @Daniel Fischer's comment to my original post. The paragraph about why infinite dimensions are needed is much appreciated. –  wil Sep 13 '13 at 4:10
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$d/dx$ is an operator on the vector space of differentiable functions with lots of right-inverses: $\int_a^x\quad dx$ - one for each $a$.

(Not sure what the right/left convention is here, but they are one-sided inverses anyway.)

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