The Unapologetic Mathematician

Mathematics for the interested outsider

Self-Adjoint Transformations

Let’s now consider a single inner-product space V and a linear transformation T:V\rightarrow V. Its adjoint is another linear transformation T^*:V\rightarrow V. This opens up the possibility that T^* might be the same transformation as T. If this happens, we say that T is “self-adjoint”. It then satisfies the adjoint relation

\displaystyle\langle v,T(w)\rangle=\langle T(v),w\rangle

What does this look like in terms of matrices? Since we only have one vector space we only need to pick one orthonormal basis \left\{e_i\right\}. Then we get a matrix

\displaystyle\begin{aligned}t_i^j&=\langle e_j,T(e_i)\rangle\\&=\langle T(e_j),e_i\rangle\\&=\overline{\langle e_i,T(e_j)\rangle}\\&=\overline{t_j^i}\end{aligned}

That is, the matrix of a self-adjoint transformation is its own conjugate transpose. We have a special name for this sort of matrix — “Hermitian” — even though it’s exactly equivalent to self-adjointness as a linear transformation. If we’re just working over a real vector space we don’t have to bother with conjugation. In that case we just say that the matrix is symmetric.

Over a one-dimensional complex vector space, the matrix of a linear transformation T is simply a single complex number t. If T is to be self-adjoint, we must have t=\bar{t}, and so t must be a real number. In this sense, the operation of taking the conjugate transpose of a complex matrix (or the simple transpose of a real matrix) extends the idea of conjugating a complex number. Self-adjoint matrices, then, are analogous to real numbers.

June 23, 2009 - Posted by | Algebra, Linear Algebra


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  9. This was exactly what I needed. Plowing through Thorpe: Elementary Aspects of Differential Geometry on my own. On p. 58 he notes that the Weingarten map Lp is self-adjoint

    Lp(v) dot v = Lp(w) dot v; v,w vectors in a real finite dimension vector space. Elsewhere he notes that the map is linear. So a matrix is definitely involved. This post saved me from plowing through Axler, Strang again (it’s been years) to find what such a matrix must look like.

    Thanks and thanks to Google


    Comment by luysii | January 31, 2013 | Reply

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