# When this matrix is diagonalizable?

When this matrix is diagonalizable? ($a_i \in \mathbb{R}$) $$\begin{pmatrix} &&&a_1\\ &&a_2&\\ &\ddots&&\\ a_n&&&\\ \end{pmatrix}$$ I think I should probably consider characteristic polynomial of this matrix, and if all roots are simple, then the matrix is diagonalizable.

UPD: Also I suppose that if $\forall a_i \neq 0$ then the matrix is diagonalizable, but I can't prove that.

The matrix is anti-diagonal, of course.

• Is $\;a_{ij}=0\;$ for $\;i+j\neq n\;$ in your matrix? I mean, is this an antidiagonal matrix? – DonAntonio Apr 28 '16 at 19:29
• It will help to treat the case $n=2$ first. – hmakholm left over Monica Apr 28 '16 at 19:29
• You can extract some information about the eigenvalues using the fact that the square of such a matrix is diagonal (so that the eigenvalues of the square are just the diagonal entries) and that $\det (A^2) = (\det A)^2$. – Travis Willse Apr 28 '16 at 19:30
• – thebooort Apr 28 '16 at 19:34
• Conjugate by the appropriate permutation matrix. – Adam Hughes Apr 28 '16 at 19:36

Let $e_i$ be the standard unit vectors, $i=1\ldots n$. The two-dimensional subspaces spanned by $e_i$ and $e_{n+1-i}$, $i = 1 \ldots \lfloor n/2 \rfloor$ and (if $n$ is odd) the one-dimensional subspace spanned by $e_{(n+1)/2}$, are invariant under your matrix, so everything reduces to the two-dimensional case.

If $a_1, a_2 \ne 0$, $\pmatrix{0 & a_1\cr a_2 & 0\cr}$ has two distinct eigenvalues $\pm \sqrt{a_1 a_2}$, therefore is diagonalizable. Of course if $a_1 = a_2 = 0$, you have the $0$ matrix which is diagonalizable. However, if one of $a_1$ is $0$ and the other is not, the matrix is not diagonalizable (the only eigenvalue is $0$, but the null space is one-dimensional).

Back to the general case: the matrix is diagonalizable unless for some $i$, one of $a_i$ and $a_{n+1-i}$ is $0$ and the other is not.

• I can't understand the beginning of your solution, could you please clearify me? – AnatoliySultanov Apr 28 '16 at 19:54

Lemma 1 : $A$ is diagonalizable over $\mathbb C$ if and only if $A^2$ is diagonalizable and $\ker A =\ker A^2$

Here, $A^2$ is luckily a diagonal matrix, so $A$ diagonalizable over $\mathbb C$ if and only if $\ker A =\ker A^2$, that is to say, if and only if $\forall i, a_i=0\iff a_{n-i+1}=0$

Lemma 2: $A$ is diagonalizable over $\mathbb R$ if and only if $\ker A =\ker A^2$ and all the eigenvalues of $A^2$ are nonnegative.

Here, this translates as $\forall i, (a_i=0\iff a_{n-i+1}=0) \;\text{and } a_ia_{n-i+1}\geq 0$

These results are similar to those found here Conditions of diagonalizability of $n \times n$ anti-diagonal matrix

• Could you explain me how to prove Lemma 2, please? – AnatoliySultanov Apr 28 '16 at 20:05
• @AnatoliySultanov if the eigenvalues of $A^2$ are negative then the eigenvalues of $A$ are complex and it is clearly not real-diagonalizable. but for the converse you have to check also if the $P$ in $P D P^{-1}$ is a real matrix – reuns Apr 28 '16 at 20:09