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Give an example of operator $T:\mathbb{R}^2\to\mathbb{R}^2$ with $T^2=T$, but $T^{*}\neq T$.

What could I consider ?

thanks :)

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Edit: the following is a result of comments by Harald:

Let

$$S\binom{x}{y}:=\binom{x+y}{-y}\,\,,\,\,\text{or in matrix form:}\;\;[S]=\begin{pmatrix}1&1\\0&\!\!\!-1\end{pmatrix}$$

Since $\,S^2=I\,$ we get that

$$T:=\frac{1}{2}\left(S+I\right)\;\;\;\text{fulfills}\;\;\;T\binom{x}{y}=\binom{x+\frac{1}{2}y}{0}\,\,,\,\,\text{or as matrix:}\;\;[T]=\begin{pmatrix}1&\frac{1}{2}\\{}\\0&0\end{pmatrix}$$

yet

$$[T^*]=\begin{pmatrix}1&0\\{}\\\frac{1}{2}&0\end{pmatrix}\neq[T]$$

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  • $\begingroup$ That would be an oblique symmetry, not a projection. $\endgroup$ – Harald Hanche-Olsen Feb 5 '13 at 10:03
  • $\begingroup$ Perhaps (I really don't know what "oblique symmetry" is), but it fulfills the OP's conditions. $\endgroup$ – DonAntonio Feb 5 '13 at 10:04
  • $\begingroup$ But $T^2\neq T$. $T^2=I$. $\endgroup$ – Iuli Feb 5 '13 at 10:07
  • $\begingroup$ A symmetry is an operator satisfying $T^2=I$. It is a reflection through a subspace. It is oblique if the two subspaces corresponding to the two eigenvalues $\pm1$ are not orthogonal. A symmetry has the form $T=2P-1$ where $P$ is a projection; so a correct answer to the problem at hand is $\frac12(T+I)$, where $T$ is your answer. $\endgroup$ – Harald Hanche-Olsen Feb 5 '13 at 10:10
  • $\begingroup$ Oh, I see now what you (and, in fact, the OP meant). Thanks, I shall edit my answer. $\endgroup$ – DonAntonio Feb 5 '13 at 10:19
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An oblique projection would do. The equation $T^2=T$ is a defining property of projections, and $T^*=T$ (for a projection) is connected with orthogonality.

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  • $\begingroup$ can you explicity, please ? $\endgroup$ – Iuli Feb 5 '13 at 9:58
  • $\begingroup$ I added to my post before seeing your comment. Hope it provides the needed detail. $\endgroup$ – Harald Hanche-Olsen Feb 5 '13 at 10:04
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  • Restrict yourself to linear operators (matrices) and make the general ansatz

$$T(a,b,c,x\ne c^*-c)=\begin{pmatrix}a&c\\{}\\c+x&b\end{pmatrix}.$$

  • Then square it (if you're lazy plug in {{a,c},{c+x,b}}^2//FullSimplify in Wolfram Alpha) and solve one linear and one quadratic equation which stem from the condition $T^2-T=0.$

You find the family of solutions

$$T(c,x\ne c^*-c)=\begin{pmatrix}\tfrac{1}{2}\pm \tfrac{1}{2}\sqrt{1-4\ c\ (c+x)}&c\\{}\\c+x&\tfrac{1}{2}\mp \tfrac{1}{2}\sqrt{1-4\ c\ (c+x)})\end{pmatrix}.$$

For $c=0$ and $c=-x$ you'd get the most simple examples

$$\begin{pmatrix}1&0\\{}\\x&0\end{pmatrix},\ \ \ \begin{pmatrix}1&-x\\{}\\0&0\end{pmatrix},$$

with $x\ne 0$.

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    $\begingroup$ This is the solution that I had in mind. Harald’s solution gives a geometric characterization of all such operators, while yours explicitly computes all such operators. $\endgroup$ – Haskell Curry Feb 5 '13 at 10:55
  • $\begingroup$ Up to unitary equivalence, $\begin{pmatrix}1&x\\0&0\end{pmatrix}$ with $x\neq 0$ also describes all examples. $\endgroup$ – Jonas Meyer Feb 6 '13 at 17:22
  • $\begingroup$ @JonasMeyer: Sounds good. $\endgroup$ – Nikolaj-K Feb 6 '13 at 17:36
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You can take $$T = \begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix}$$

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  • $\begingroup$ One more question please, how can I calculate $T^{*}$? Is any formula I should use Riesz's Theorem ? Thanks :) $\endgroup$ – Iuli Feb 5 '13 at 10:22
  • $\begingroup$ @Iuli For operators on $\mathbb{R}^n$ the adjoint is just the transpose. $\endgroup$ – mrf Feb 5 '13 at 10:31
  • $\begingroup$ I assume you mean orthonormal basis, but that's right @DonAntonio $\endgroup$ – mrf Feb 5 '13 at 11:32
  • $\begingroup$ Oh, dear: of course, @mrf $\endgroup$ – DonAntonio Feb 5 '13 at 11:59

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