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Imagine the $n \times n$ matrix A: $$\begin{bmatrix} 0 & 0 &... & 0 & a_1\\ 0 & 0 & ...& 0 & a_2\\ . & .&&.&.\\ a_1 & a_2 & ... & 0 &a_n \end{bmatrix}$$ Where $a_i$ are complex numbers. The problem is to find, the values of $a_i$, for which the matrix is diagonalizable. At first, what I tried is to split the matrix into $B+ i C$, where $B$ and $C$ are real symmetric matrices (and therefore diagonalizable) and tried finding a condition so that $B$ and $C$ commute. What I got was that $$B_{in} C_{nj} = B_{jn} C_{ni}$$ The problem I'm having is that there are matrices that don't meet this condition and are still diagonalizable. I tried finding the eigenvalues and I got $$\frac{a_n \pm \sqrt{a_n ^2 + 4 \sum_{i=1}^{n-1} a_i^2}}{2}$$ The rest of eigenvalues are 0 I believe. I don't know how to proceed from here. I saw a matrix that meets this form with $a_n=0$ that was not diagonalizable, so I don't think it's diagonalizable for all complex numbers. Thanks for reading.

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  • $\begingroup$ I'm thinking maybe the condition is that it's diagonalizable if a_n is nonzero but I'm not sure if that's true or how to prove it. $\endgroup$ Nov 25 at 7:59
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    $\begingroup$ It would be possible just to hack out the solution using the criterion that $A$ is digonalisable iff $m_A(X)$ is a product of distinct linear factors. You've got the characteristic polynomial $\chi_A(X)=X^{n-2}(X^2- a_n X+ (a_1,\dots,a_{n-1})^T (a_1,\dots,a_{n-1}))$ so you need only check (in the general case) whether $A(A^2- a_n A+ (a_1,\dots,a_{n-1})^T (a_1,\dots,a_{n-1}))=0$; and then chase the special cases when the quadratic has a root $0$. But I hope someone will suggest a clever way of getting the result. $\endgroup$ Nov 25 at 8:30
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    $\begingroup$ In the above comment I have got the sign of the $a^Ta$ term wrong, sorry. $\endgroup$ Nov 25 at 10:12
  • $\begingroup$ @ancientmathematician thank you, the case where the quadratic has two equal solutions also non diagonalizable, right? Meaning if $a_n^2 = -4 \sum a_i^2$, the matrix is non diagonalizable? $\endgroup$ Nov 25 at 10:37
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As you have reckoned, the only two possibly nonzero eigenvalues of $A$ are $\frac12\left(a_n\pm\sqrt{a_n^2+4v^\top v}\right)$, where $v=(a_1,a_2,\ldots,a_{n-1})^\top$. There are three possibilities:

  1. All eigenvalues of $A$ are zero, i.e., $a_n=v^\top v=0$. Since $A$ is nilpotent in this case, it is diagonalisable if and only if it is the zero matrix, i.e., if and only if $v=0$.
  2. $A$ has exactly one nonzero eigenvalue, i.e., $a_n\ne0=v^\top v$. Then $A$ is diagonalisable if and only if $\operatorname{rank}(A)=1$, i.e., if and only if $v=0$.
  3. $A$ has two distinct nonzero eigenvalues $\lambda_1$ and $\lambda_2$, i.e., both $v^\top v$ and $a_n^2+4v^\top v$ are nonzero. The Jordan form of $A$ is therefore $N\oplus\lambda_1\oplus\lambda_2$, where $N$ is some nilpotent matrix. However, by looking at the column space of $A$, we see that the rank of $A$ is always $\le2$. Hence $N=0$ and $A$ is diagonalisable.

In summary, $A$ is diagonalisable if and only if $v=0$ or both $v^\top v$ and $a_n^2+4v^\top v$ are nonzero.

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  • $\begingroup$ I see, thank you very much, it makes sense now. +1 for clarity. $\endgroup$ Nov 25 at 11:45

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