Normal Transformations

All the transformations in our analogy — self-adjoint and unitary (or orthogonal), and even anti-self-adjoint (antisymmetric and “skew-Hermitian”) transformations satisfying $T^*=-T$ — all satisfy one slightly subtle but very interesting property: they all commute with their adjoints. Self-adjoint and anti-self-adjoint transformations do because any transformation commutes with itself and also with its negative, since negation is just scalar multiplication. Orthogonal and unitary transformations do because every transformation commutes with its own inverse.

Now in general most pairs of transformations do not commute, so there’s no reason to expect this to happen commonly. Still, if we have a transformation $N$ so that $N^*N=NN^*$ , we call it a “normal” transformation.

Let’s bang out an equivalent characterization of normal operators while we’re at it, so we can get an idea of what they look like geometrically. Take any vector $\lvert v\rangle$ , hit it with $N$ , and calculate its squared-length (I’m not specifying real or complex, since the notation is the same either way). We get

$\displaystyle\lVert\lvert N(v)\rangle\rVert^2=\langle N(v)\vert N(v)\rangle=\langle v\rvert N^*N\lvert v\rangle$

On the other hand, we could do the same thing but using $N^*$ instead of $N$ .

$\displaystyle\lVert\lvert N^*(v)\rangle\rVert^2=\langle N^*(v)\vert N^*(v)\rangle=\langle v\rvert NN^*\lvert v\rangle$

But if $N$ is normal, then $N^*N$ and $NN^*$ are the same, and thus $\lVert\lvert N(v)\rangle\rVert^2=\lVert\lvert N^*(v)\rangle\rVert^2$ for all vectors $\lvert v\rangle$

Conversely, if $\lVert\lvert N(v)\rangle\rVert^2=\lVert\lvert N^*(v)\rangle\rVert^2$ for all vectors $\lvert v\rangle$ , then we can use the polarization identities to conclude that $N^*N=NN^*$ .

So normal transformations are exactly those that the length of a vector is the same whether we use the transformation or its adjoint. For self-adjoint and anti-self-adjoint transformations this is pretty obvious since they’re (almost) the same thing anyway. For orthogonal and unitary transformations, they don’t change the lengths of vectors at all, so this makes sense.

Just to be clear, though, there are matrices that are normal, but which aren’t any of the special kinds we’ve talked about so far. For example, the transformation represented by the matrix

$\displaystyle\begin{pmatrix}1&1&0\\{0}&1&1\\1&0&1\end{pmatrix}$

has its adjoint represented by

$\displaystyle\begin{pmatrix}1&0&1\\1&1&0\\{0}&1&1\end{pmatrix}$

which is neither the original transformation nor its negative, so it’s neither self-adjoint nor anti-self-adjoint. We can calculate their product in either order to get

$\displaystyle\begin{pmatrix}2&1&1\\1&2&1\\1&1&2\end{pmatrix}$

since we get the same answer, the transformation is normal, but it’s clearly not unitary because if it were we’d get the identity matrix here.

August 5, 2009 - Posted by John Armstrong | Algebra, Linear Algebra

3 Comments »

[…] and Eigenvectors of Normal Transformations Let’s say we have a normal transformation . It turns out we can say some interesting things about its eigenvalues and […]

Pingback by Eigenvalues and Eigenvectors of Normal Transformations « The Unapologetic Mathematician | August 6, 2009 | Reply
[…] have a diagonal matrix with respect to some basis. First of all, such a transformation must be normal. If we have a diagonal matrix we can find the matrix of the adjoint by taking its conjugate […]

Pingback by The Complex Spectral Theorem « The Unapologetic Mathematician | August 10, 2009 | Reply
[…] 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 (orthogonal, if […]

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

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