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.

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

9 Comments »

  1. [...] and Forms I Yesterday, we defined a Hamiltonian matrix to be the matrix-theoretic analogue of a self-adjoint transformation. So why should we separate out [...]

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  2. [...] particular, this shows that if we have a symmetric form, it’s described by a self-adjoint transformation . Hermitian forms are also described by self-adjoint transformations . And [...]

    Pingback by Symmetric, Antisymmetric, and Hermitian Forms « The Unapologetic Mathematician | July 10, 2009 | Reply

  3. [...] this all sort of makes sense. Self-adjoint transformations (symmetric or Hermitian) are analogous to the real numbers sitting inside the complex numbers. Within these, positive-definite matrices are sort of like the positive real numbers. It [...]

    Pingback by Positive-Definite Transformations « The Unapologetic Mathematician | July 13, 2009 | Reply

  4. [...] off, note that no matter what we use, the transformation in the middle is self-adjoint and positive-definite, and so the new form is symmetric and positive-definite, and thus defines [...]

    Pingback by Orthogonal transformations « The Unapologetic Mathematician | July 27, 2009 | Reply

  5. [...] of the original transformation (or just the same, for a real transformation). So what about self-adjoint transformations? We’ve said that these are analogous to real numbers, and indeed their [...]

    Pingback by The Determinant of a Positive-Definite Transformation « The Unapologetic Mathematician | August 3, 2009 | Reply

  6. [...] All the transformations in our analogy — self-adjoint and unitary (or orthogonal), and even anti-self-adjoint (antisymmetric and [...]

    Pingback by Normal Transformations « The Unapologetic Mathematician | August 5, 2009 | Reply

  7. [...] to denote transformations). We also have its adjoint . Then is positive-semidefinite (and thus self-adjoint and normal), and so the spectral theorem applies. There must be a unitary transformation [...]

    Pingback by The Singular Value Decomposition « The Unapologetic Mathematician | August 17, 2009 | Reply

  8. [...] the underlying space forms a vector space itself. Indeed, such forms correspond to correspond to Hermitian matrices, which form a vector space. Anyway, rather than write the usual angle-brackets, we will write one [...]

    Pingback by Projecting Onto Invariants « The Unapologetic Mathematician | November 13, 2010 | Reply

  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

    Luysii

    Comment by luysii | January 31, 2013 | Reply


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