When is a complex symmetric matrix with only the last row and column being non zero diagonalizable?

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.

• 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. Nov 25 '21 at 7:59
• 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. Nov 25 '21 at 8:30
• In the above comment I have got the sign of the $a^Ta$ term wrong, sorry. Nov 25 '21 at 10:12
• @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? Nov 25 '21 at 10:37

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.

• I see, thank you very much, it makes sense now. +1 for clarity. Nov 25 '21 at 11:45