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Normally in trying to determine whether or not a linear transformation is self adjoint or normal I find the coordinate matrix with respect to an orthonormal basis and see whether it equals its conjugate transpose or if it commutes with its conjugate transpose.

The definition of an adjoint of a linear transformation is given to me as follows:

Let V be a finite-dimensional inner product space and let T be a linear operator on V. Then there exists a unique linear operator T* called the adjoint of T such that $\langle T(x),y\rangle=\langle x, T^{*}(y)$.

With this in mind, determine if the following linear transformation is self adjoint, normal, or neither:

$T:\mathbb{R}^3\rightarrow \mathbb{R}^3$ defined by $T(a,b,c)=(-a+b,5b,4a-2b+5c)$

A self adjoint linear transformation is such that $T^*=T$ so I believed that we only had to ensure that $\langle (-a+b,5b,4a-2b+5c),(a',b',c')\rangle = \langle (a,b,c),(-a'+b',5b',4a'-2b'+5c') \rangle$. But this reasoning would imply that this operator (and any linear operator on a vector space over $\mathbb{R}$ is self adjoint. However, The coordinate matrix for this transformation with respect to the standard basis is:

\begin{pmatrix} -1 & 1 & 0 \\ 0 & 5 & 0 \\ 4 & -2 & 5 \\ \end{pmatrix}

Which is not self adjoint. So what is wrong with the reasoning I followed to test if this is self adjoint using the definition?

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I believed that we only had to ensure that [condition for adjointness]. But this reasoning would imply that this operator (and any linear operator on a vector space) over R is self adjoint.

Why? Not every linear operator satifies that condition. The one you gave is an example. Let $x=(1,0,0)$ and $y=(0,1,0)$ and let the transformation act on the right of row vectors.

Then $\langle Tx\cdot y\rangle=\langle (-1,1,0)\cdot(0,1,0)\rangle=1$ but

$\langle x\cdot Ty\rangle=\langle (1,0,0)\cdot(0,5,0)\rangle=0$.

Thus this operator is not self-adjoint.

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  • $\begingroup$ and I just realized that the dot products weren't equal symbolically either...boy I'm silly. Thank-you for the answer! $\endgroup$ – illysial Jul 8 '14 at 16:51
  • $\begingroup$ Haven't you made some calculation errors here? T(1, 0, 0) = (-1, 0, 4), not (-1,1,0) afaict. $\endgroup$ – Björn Lindqvist Oct 20 '17 at 12:51
  • $\begingroup$ @BjörnLindqvist That's what happens if you apply the transformation on the left side... which is what I said I was going to do, apparently! I will correct the inconsistency.. $\endgroup$ – rschwieb Oct 20 '17 at 13:00

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