# The Unapologetic Mathematician

## Mathematics for the interested outsider

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

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 […]

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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 […]

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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 […]

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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 […]

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

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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 […]

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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 […]

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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. Comment by luysii | January 31, 2013 | Reply