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.
Best Answer
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 $\mathbf{C}^n$.