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I have matrix that represent $T:V \to V$ (linear map over $\mathbb{R}$) according to basis $E$.

$E$ is not an orthonormal set.

how can I know if this $T$ is self-adjoint.

I know that if $E$ was orthonormal basis we would take the transpose matrix.

but what about here ?

the matrix is : $$\begin{bmatrix} 1 & 2 \\ 2 & 1 \\ \end{bmatrix}$$

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  • $\begingroup$ Have you studied inner products? $\endgroup$ – Mhenni Benghorbal May 15 '13 at 11:16
  • $\begingroup$ If you don't specify $E$ or, rather, its Gramian matrix, we can't say anything. I know you made your question more precise so that it could be answered in another thread. You should have edited this one instead. There is not a real reason to leave this one now. $\endgroup$ – Julien May 18 '13 at 22:36
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Without an orthonormal matrix I'm afraid you'll have to do it the long way: we're looking for a matrix $\,T^*\,$ s.t. (I'm assuming the usual, euclidean inner product):

$$\overline x:=\binom xy\,,\,\,\overline y=\binom ab\implies\langle T\, \overline x\,,\,\overline y\rangle=\langle \overline x\,,\,T^*\,\overline y\rangle\;\iff $$

$$\left\langle\,\begin{pmatrix}1&2\\2&1\end{pmatrix}\binom xy\,,\,\binom ab\,\right\rangle =\left\langle\,\binom xy\,,\,\begin{pmatrix}p&q\\r&s\end{pmatrix}\binom ab\,\right\rangle\iff$$

$$\left\langle\,\binom{x+2y}{2x+y}\,,\,\binom ab\,\right\rangle=\left\langle\,\binom xy\,,\,\binom{pa+qb}{ra+sb}\,\right\rangle\iff$$

$$ax+2ay+2bx+by=pax+qbx+ary+bsy\iff$$

$$(1-p)ax+(1-s)by+(2-r)ay+(2-q)bx=0$$

As the above has to be true for any two vectors $\,\overline x\,,\,\overline y\,$ , you can now choose particular cases and deduce the possible values of $\,p,q,r,s\,$ (which, of course, you then need to check in the general case), for example: with $\,x=a=1\,,\,y=b=0\,,\,$ :

$$(1-p)=0\implies p =1\;\;,\;\;etc.\ldots$$

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  • $\begingroup$ Thank you very much !! you helped a lot $\endgroup$ – wantToLearn May 15 '13 at 12:05
  • $\begingroup$ Unless I misunderstood, didn't this computation assume the basis is orthonormal? The original matrix the OP started with is symmetric, so in that case is automatically representing a self-adjoint transformation. You need to insert the inner products of the basis elements. That is, if $\langle v_1,v_1\rangle = A$, $\langle v_1,v_2\rangle=B$, and $\langle v_2,v_2\rangle = C$, then expand ... If this matrix of inner products is $P$ and the original matrix for the transformation is $M$, then self-adjointness should be given by $M^TP = PM$, or something like that. $\endgroup$ – Ted Shifrin May 15 '13 at 14:40
  • $\begingroup$ No, this computation doesn't assume anything: that's what the OP wanted. If we knew we were wroking with an orthonormal basis the adjoint has a very simple form. $\endgroup$ – DonAntonio May 15 '13 at 14:51
  • $\begingroup$ In the original problem there is also given that $(v_1,v_1)=2,(v_1,v_2)=-1,(v_2,v_1)=-1, (v_2,v_2)=1$ $\endgroup$ – Sandra West May 16 '13 at 0:21
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    $\begingroup$ @DonAntonio, forgive me for insisting here. You're assuming the usual, Euclidean inner product, but you're also assuming an orthonormal basis. How else would $\langle (x+2y)v_1,av_1\rangle$ turn out to be $a(x+2y)$? $\endgroup$ – Ted Shifrin May 18 '13 at 22:03
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@SandraWest, so we need to compute. (Or use the matrix short-cut in my comment.) Remember that $T(v_1)=v_1+2v_2$ and $T(v_2)=2v_1+v_2$. It suffices (by linearity) to check whether $\langle Tx,y\rangle=\langle x,Ty\rangle$ when $x$ and $y$ are basis vectors. You can see the only thing that really requires checking is whether $\langle Tv_1,v_2\rangle = \langle v_1,Tv_2\rangle$.

Well, \begin{align*} \langle Tv_1,v_2\rangle & =\langle v_1+2v_2,v_2\rangle=−1+2=1, \quad\text{and} \\ \langle v_1,Tv_2\rangle &= \langle v_1,2v_1+v_2\rangle = 4−1=3 \,, \end{align*}

so T is not self-adjoint.

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