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Let $M: P_{1}\rightarrow P_{1}$ be defined by $M(f)=f^{'}+f$, i.e. $M(a_{0}+a_{1}x)=a_{1}+a_{0}+a_{1}x$. Find the adjoint $M^{*}$ of $M$, i.e find $M^{*}(a_{0}+a_{1}x_{1})$, assuming that $L^{2}(0,1)$ inner product is imposed on $P_{1}$. $\langle f(x),g(x)\rangle =\int^{1}_{0}f(x)g(x)dx$.

81235
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  • This looks like a homework question; please see how to ask such questions. In any case, people here don't tend to like textbook-style problems with no work shown and phrased as commands. – Nate Eldredge Jun 10 '13 at 00:37

1 Answers1

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Let's find an orthogonal basis for the given inner product.
Luckily, $\|1\|=\sqrt{\langle 1,1\rangle}=\sqrt{\int_0^11}=1$. But $x$ is not orthogonal to $1$, so we have to modify it: $$\langle x,1\rangle =\int_0^11\cdot x\,dx=\frac12\,,$$ so that $(x-\frac12)\perp 1$. So is its double, $p:=\,2x-1$. Its norm can be obtained: $\|p\|^2=\|(2x-1)\|^2=\int_0^1(2x-1)^2\,dx=\frac43-2+1=\frac13$.

Then, use the general fact that a vector $v$ can be decomposed in the orthogonal basis $e_1,e_2,...$ as $$v=\frac{\langle v,e_1\rangle}{\|e_1\|^2} e_1+\frac{\langle v,e_2\rangle}{\|e_2\|^2} e_2+\dots$$

So that, now we can apply this to arrive at $M^*$: $$M^*(1)= \langle M^*(1),\,1\rangle\, +3\,\langle M^*(1),\,2x-1\rangle\cdot (2x-1)= \\ = \langle 1,M(1)\rangle+3\,\langle1,M(2x-1)\rangle\cdot(2x-1)\,.$$ Similarly one can obtain $M^*(x)$.

Berci
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  • can you make use of $<M(f),g>=<f,M^{*}(g)>$? – 81235 Jun 09 '13 at 23:43
  • Didn't I use it? – Berci Jun 09 '13 at 23:49
  • One can also introduce an orthonormal basis, and work with matrices. Now we can fix $(1,\sqrt3\cdot(2x-1))$, now both these has norm $1$, and are orthogonal to each other (w.r.t. the $L^2$ inner product). Then, coordinating $M$ in this basis, we get $$[M]=\pmatrix{1&2\sqrt3\0&1},$$ so its adjoint (in the same basis!) is $$[M^*]=\pmatrix{1&0\2\sqrt3&1},.$$ Looks odd, but it gives the same as my solution. – Berci Jun 10 '13 at 00:00
  • My approach: $<M(f),g>=<f,M^{*}(g)>$, $f=a_{0}+a_{1}x$,$g=b_{0}+b_{1}x$, $<M(f),g>=\int^{1}{0}(a{1}+a_{0}+a_{1}x)(b_0+b_{1}x)dx$, but I don't how to continue. Can you prove it following my idea? – 81235 Jun 10 '13 at 00:02
  • For start, you have to calculate $\int_0^1x^2$ and $\int_0^1x$ and $\int_0^11$ anyway. Please read through my answer. If something is not clear about it, ask more specifically. – Berci Jun 10 '13 at 00:05
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    I have read your answer carefully, according your answer,$M^{*}(x)=7x-3$,am I right? – 81235 Jun 10 '13 at 00:15
  • I think, for calculating adjoint one needs to work in orthonormal or at least orthogonal basis (or, either one can write up the Gram matrix of the inner product in the standard basis). Using orthonormal basis $e_1,e_2,..,e_n$ you probably know that a vector's $i$th coordinate can be obtained by $\langle v,e_i\rangle$. I have used the modification of this statement to orthogonal bases. – Berci Jun 10 '13 at 00:17
  • http://math.stackexchange.com/questions/182598/finding-the-adjoint-of-an-operator – 81235 Jun 10 '13 at 00:18
  • Yes, I got $M^(1)=12x-5$ and $M^(x)=7x-3$. From that you can write up it for general $M^*(a+bx)$.\ – Berci Jun 10 '13 at 00:18
  • Is $M^{*}$ linear operator? – 81235 Jun 10 '13 at 00:19
  • And I am thinking if the number of basis is too large, how can you calculate the basis? – 81235 Jun 10 '13 at 00:21
  • let us continue this discussion in chat – Berci Jun 10 '13 at 00:53
  • Yes, $M^*$ is linear. The basis has $2$ elements, as $P_1$ has dimension two. Its standard basis consists of $1$ and $x$. The orthogonal basis that I used consists of $1$ and $(2x-1)$. – Berci Jun 10 '13 at 00:57