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This is a problem from Linear Algebra Done Right 4th edition Chapter 7B (This is an open access book available from Professor Sheldor Axler's website). I am self-studying and need some help.

The problem goes like this:

T is a operator on a finite-dimensional vector space over F (field).

(1) Suppose F = R. Prove that 𝑇 is diagonalizable if and only if there is a basis of 𝑈 such that the matrix of 𝑇 with respect to this basis equals its transpose.

(2) Suppose F = C. Prove that 𝑇 is diagonalizable if and only if there is a basis of 𝑈 such that the matrix of 𝑇 with respect to this basis commutes with its conjugate transpose.

For context, the chapter is about real and complex spectral theorem. The book has covered dual space, eigenvalues, inner product space, self-adjoint and normal operator.

I have spent hours stuck at (1). I am trying to show that T = T* (adjoint of T), through the correspondence between T* and T' (dual map of T), but stuck at showing that if transpose of M(T) = M(T), then 𝑈 has a basis consisting of eigenvectors of T

As additional information, here is my current limited progress. I am stuck at finishing the <= part. I know M(T) = M(T') with the corresponding basis, but I don't know how to proceed to infer T* = T from this.

Or maybe there are other simpler way to prove the statement.

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  • $\begingroup$ Hint: For (1), if $V$ is a finite-dimensional real vector space, you can associate to any basis $B$ of $V$ the positive definite symmetric bilinear form $b$ with respect to which $B$ is orthonormal, that is if $B = \{e_1,\ldots,e_n\}$ then $b(e_i,e_j) = \delta_{ij}$. $\endgroup$
    – krm2233
    Apr 19 at 4:08
  • $\begingroup$ Is there other more elementary ways? The book hasn't covered positive definite or bilinear form yet. In fact they follow closely after this chapter. There has to be a way without using these knowledge? $\endgroup$
    – Tim
    Apr 19 at 4:12
  • $\begingroup$ $T=T^*$ certainly need not hold, but note that a diagonal matrix equals its transpose. This is perhaps sneaky. The converse is just the real spectral theorem. $\endgroup$ Apr 19 at 4:39
  • $\begingroup$ @TedShifrin but according to the book, the real spectral theorem ask the basis to be orthonormal, which doesn't neccessarily hold from the assumption? This is the main obstacle for me from proving the converse I believe. $\endgroup$
    – Tim
    Apr 19 at 4:44
  • $\begingroup$ @Tim a positive definite bilinear form is an inner product, and chapter 7 is about inner product spaces, so I'm pretty sure what I suggest is exactly what Axler is asking for, but perhaps I should have phrased what I said as follows: given a vector space $V$ equipped with a basis $B$ there is a unique inner product on $V$ with respect to which $B$ is orthonormal. $\endgroup$
    – krm2233
    Apr 19 at 7:42

2 Answers 2

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If $T$ is diagonalizable then certainly there is a basis of $U$ such that the matrix of $T$ with respect to this basis equals its transpose, since a diagonal matrix always equals its transpose.

Conversely, suppose there is a basis $u_1, \ldots, u_n$ such that $A = A^{\text{t}}$, where $A$ is the matrix of $T$ with respect to $u_1, \ldots, u_n$. Thinking of $A$ as a linear operator on $\mathbf{R}^n$, notice that $A$ is self-adjoint. Thus we can apply the real spectral theorem to obtain an orthonormal basis $x_1, \ldots, x_n$ of $\mathbf{R}^n$ consisting of eigenvectors of $A$. For each $k \in \{ 1, \ldots, n \}$, suppose that $x_k = (x_{k,1}, \ldots, x_{k,n})$ and let $$ v_k = x_{k,1} u_1 + \cdots + x_{k,n} u_n. $$ You can now show that $v_1, \ldots, v_n$ is a basis of $U$ consisting of eigenvectors of $T$, i.e. $v_1, \ldots, v_n$ diagonalizes $T$.

A similar approach will work for the second part of the exercise. Think of the matrix of $T$ as an operator on $\mathbf{C}^n$.

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    $\begingroup$ The shift of perspective on A is something I didn't think of. Thank you. $\endgroup$
    – Tim
    Apr 19 at 5:13
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The point of Axler's question seems to be being missed here. If $V$ is a finite-dimensional vector space over a field $F$ (where $F=\mathbb R$ or $\mathbb C$) then it is meaningless to speak of the adjoint $T^*$ of a linear endomorphism $T$ of $V$ unless $V$ is equipped with an inner product.

One can, however, consider all possible inner products on $V$ and ask whether there exists one with respect to which $T$ and its adjoint have a particular property. Axler's question is the assertion that a linear map $T \colon V \to V$ is diagonalisable in the real case if and only if there exists an inner product on $V$ with respect to which $T$ is self-adjoint, and in the complex case if and only if $T$ commutes with its adjoint with respect to some inner products on $V$.

To see that this is equivalent to the formulation Axler's question actually uses note that, given any basis $B=\{e_1,\ldots,e_n\}$ of $V$, the prescription that $\langle e_i,e_j \rangle_B = \delta_{ij}$ completely determines an inner products $\langle -,- \rangle_B$ on $V$. Moreover, by Gram-Schmidt, one knows that any inner products space possesses an orthonormal basis, thus the map $B \mapsto \langle -,-\rangle_B$ is a surjection from the set of all bases of $V$ to the set of all possible inner products on $V$.

Using this, it is straight-forward to prove the claims about diagonalisability: If $T$ is diagonalisable, then if $B$ is a basis of eigenvectors for $T$, then when $F= \mathbb R$, $T$ is certainly self-adjoint with respect to the inner products $\langle -,-\rangle_B$, while if there is some inner product with respect to which $T$ is self-adjoint, then the spectral theorem shows that $T$ has an orthonormal basis of eigenvectors.

Similarly in the case $F=\mathbb C$, $T$ will commute with its adjoint $T^*$ with respect to the Hermitian inner product $\langle -,-\rangle_B$, while again the spectral theorem guarantees that if $T$ commutes with its adjoint with respect to some inner product, then it has an orthonormal basis of eigenvectors.

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  • $\begingroup$ Thank you for this answer, I found it very enlightening. I must admit I assumed $U$ came equipped with an inner product, but upon rereading the exercise it seems clear that Axler did not intend this, and in fact defining our own inner product on $U$ provides a cleaner proof than the one I provided in my answer. I will only make minor changes to my answer, given that OP has already accepted it. $\endgroup$
    – terran
    Apr 19 at 10:23
  • $\begingroup$ @krm2233 Am I understanding correctly that you are implying that any basis of U can be orthonormal by some definition of inner product on U? (I believe, reading the book from the start to this point, Axler has not explicitly mentioned this anywhere before though. But there certainly may be some implicit connections I am missing) $\endgroup$
    – Tim
    Apr 19 at 11:48
  • $\begingroup$ Personally I would still be inclined to accept terran's solution as the author intention. My understanding of the solution is that it does not assume an inner product on U, rather it defined another linear operator S on Rn that along with the standard basis of Rn corresponds to the same matrix. And if we choose to define the Euclidean inner product on Rn, which is conventional, then the standard basis is an orthonormal basis. These all seem reasonable given the content of the book before this exercises. Of course, I may be missing something you are implying $\endgroup$
    – Tim
    Apr 19 at 13:04
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    $\begingroup$ Yes that's correct. It may help to think of it this way: if $V$ and $W$ are (say) $\mathbb R$-vector spaces and $W$ is an inner product space, then any linear isomorphism $\phi\colon V \to W$, allows us to equip $V$ with an inner product: for $v_1,v_2 \in V$, we set $\langle v_1,v_2\rangle = \langle \phi(v_1), \phi(v_2) \rangle$. Now a basis $\{e_1,\ldots,e_n\}$ of $V$, gives an isomorphism $\phi\colon V \to \mathbb R^n$ where if $v = \sum_{i=1}^n a_i e_i$ then $\phi(v) = (a_i)_{i=1}^n$. The inner product you get by pulling back the dot product on $\mathbb R^n$ makes the basis orthonormal. $\endgroup$
    – krm2233
    Apr 19 at 13:08
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    $\begingroup$ Hopefully the previous comment makes it clear that terran's answer is completely equivalent to mine. My main point was to emphasize that, if you have a linear map $T\colon V \to V$ and $V$ is just a vector space, then you can't talk about $T^*$, the only thing you have is the transpose $T'\colon V' \to V'$, the linear map induced on the dual space $V' = \text{Hom}(V,F)$, but you can't relate the two without an identification of $V$ with $V'$ (which an inner product gives you). $\endgroup$
    – krm2233
    Apr 19 at 13:15

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