---
title: "Axler 7.A"
source: https://www.jemoka.com/posts/kbhaxler_7_a/
---

This is not actually like a proper review of a chapter, instead, it is an opinionated review of what I think Jana thinks Axler thinks is important about 7.A.
Note that all of the &ldquo;proofy things&rdquo; in this section are poofy because problems with putting trips prior to the end of the year.
Here&rsquo;s an outline:
We defined the adjoint We learned some properties of the adjoint; importantly, that \((A+B)^{*} = A^{*} + B^{*}\), \((AB)^{*} = B^{*} A^{*}\), \((\lambda T)^{*} = \bar{\lambda}T^{*}\); a correlary is that \(M^{*}M\) is self-adjoint We defined normal, self-adjoint, and unitary With those definitions, we showed that eigenvalues of self-adjoint matricies are real Then, we created two mildly interesting intermediate results Over \(\mathbb{C}\), \(Tv\) is orthogonal to all \(v\) IFF \(T\) is the zero matrix Over \(\mathbb{R}\), \(Tv\) is orthogonal to all \(v\) and \(T\) is self-adjoint, then \(T\) is the zero matrix The latter of which shows that Eigenvectors of \(T\) corresponding to distinct eigenvalues are orthogonal if \(T \in \mathcal{L}(V)\) is normal This, all, builds up to the result of the Complex Spectral Theorem, which you should know adjoint Suppose \(T \in \mathcal{L}(V,W)\), we define the adjoint as a \(T^{*} \in \mathcal{L}(W,V)\) that:
\begin{equation} \langle Tv,w \rangle = \langle v, T^{*}w \rangle \end{equation}
the usual verifications could be made (setting \(w = w_1+w_2\), then applying inner product additivity, etc.) to show that \(T^{*}\) is a linear map.
properties of the adjoint \((S+T)^{*} = S^{*} + T^{*}\) \((\lambda T)^{*} = \bar{\lambda}T^{*}\) \((T^{*})^{*} = T\) \(I^{*} = I\) &ldquo;identity is self adjoint&rdquo; and, importantly, \((ST)^{*} = T^{*} S^{*}\) All of these results are findable by chonking through the expressions for inner product properties.
adjoint is the conjugate transpose For complex-valued matricies, and the Euclidean Inner Product, the adjoint is the conjugate transpose of the matrix.
To show this ham-fistedly, first convince yourself that the property of \((ST)^{*} = T^{*} S^{*}\) holds for the act of conjugate transpose of matricies as well. Now, we will show how we can get to the adjoint definition from that result:
Consider:
\begin{equation} \langle v,w \rangle \end{equation}
we can represent this as the product of two &ldquo;matricies&rdquo;: an \(n \times 1\) &ldquo;matrix&rdquo; named \(v\), and a \(n \times 1\) &ldquo;matrix&rdquo; named \(w\); specifically:
\begin{equation} \langle v,w \rangle = v^{*} w \end{equation}
where \(v^{*}\) is the conjugate-transpose of \(v\) (the dagger is nonstandard notation, to distinguish from \(*\) the adjoint defined above which we haven&rsquo;t shown yet). This is by definition of how matricies multiply and how the Euclidean Inner Product works.
So then consider the same for the above:
\begin{equation} \langle Tv,w \rangle = (Tv)^{*}w = v^{*} T^{*}w = \langle v, T^{*}w \rangle \end{equation}
Axler gives an arguably better proof involving representing the matricies w.r.t. the orthonomal bases, and then showing that the inner products just swapped slots:
Buncha matrix adjectives And given we now know what the adjoint is, we can make some definitions:
self-adjoint \begin{equation} A = A^{*} \end{equation}
wow.
Namely:
\begin{equation} \langle Tv, w \rangle = \langle v, T^{*}w \rangle = \langle v, Tw \rangle \end{equation}
normal \begin{equation} A A^{*} = A^{*} A \end{equation}
As in, if the operator commutes with its own adjoint.
unitary \begin{equation} A^{*} = A^{-1} \end{equation}
or, that \(A\) has orthonormal columns: an unitary operator is invertible, and the inverse of its atrix representation is its transpose
Eigenvalues of self-adjoint matricies are real So, if we have:
\begin{equation} Tv = \lambda v \end{equation}
we want to show that \(\lambda\) is real. To do this, we can show that \(\lambda = \bar{\lambda}\) which would mean the \(b\) component is \(0\).
Now, recall that self-adjoint means \(\langle Tv,w \rangle = \langle v, Tw \rangle\).
Construct now: \(\lambda \|v\|^{2}\) &mdash;
\begin{equation} \lambda \|v\|^{2} = \lambda \langle v,v \rangle = \langle \lambda v,v \rangle = \langle Tv, v \rangle = \langle v, Tv \rangle = \langle v, \lambda v \rangle = \bar{\lambda} \langle v,v \rangle = \bar{\lambda} \|v\|^{2} \end{equation}
So we have \(\lambda \|v\|^{2} = \bar{\lambda} \|v\|^{2}\), which means \(\lambda = \bar{\lambda}\), as desired.
Two less important intermediate results &hellip;that we just trust Axler&rsquo;s word + our intuition for:
7.14: Over \(\mathbb{C}\), \(Tv\) is orthogonal to all \(v\) IFF \(T\) is the zero matrix 7.16: Over \(\mathbb{R}\), \(Tv\) is orthogonal to all \(v\) and \(T\) is self-adjoint, then \(T\) is the zero matrix Why is it different? ASK JANA IDK
Eigenvectors of \(T\) corresponding to distinct eigenvalues are orthogonal if \(T \in \mathcal{L}(V)\) is normal Prove depended on the minor results from before
and then voodo whitchcraft.
Complex Spectral Theorem On \(\mathbb{C}\), and with \(T \in \mathcal{L}(V)\), the following statements are equivalent:
\(T\) is normal \(T\) has an orthonormal basis of eigenvectors and so \(T\) is diagonalizable w.r.t. that orthonormal basis of eigenvectors This proof depends on Schur&rsquo;s Theorem.
The real number version requires that \(T\) is self-adjoint.
Things to ask jana why is 7.14 and 7.16 different
