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Let the matrix $A$ be an anti-diagonal matrix with real number elements where $a_{1n} = \lambda_1$, $a_{2,n-1} = \lambda_2,\ldots, a_{n1} = \lambda_n$. The task is to find out conditions on $\lambda_1, \ldots, \lambda_n$ when the matrix can be diagonalized. I found a similar question here Conditions for diagonalizability of $n\times n$ anti-diagonal matrices but only the complex number field is considered.

Thanks in advance

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$A$ is similar (via a permutation matrix) to $A_1\oplus A_2\oplus\cdots\oplus A_m$, where $m=\lceil\frac n2\rceil$, the first $m-1$ diagonal blocks are $2\times2$ and the last one is $2\times2$ when $n$ is even, or $1\times 1$ when $n$ is odd.

As a block diagonal matrix is diagonalisable if and only if its diagonal subblocks are all diagonalisable, it suffices to consider the problem clockwise and we may ignore the last $1\times1$ block (which is always diagonalisable) if any. So, consider $A_k=\pmatrix{0&\lambda_k\\ \lambda_{n+1-k}&0}$. It is annihilated by $p(x)=x^2-\lambda_k\lambda_{n+1-k}$.

When $\lambda_k\lambda_{n+1-k}\ne0$, $p$ is always splittable into two distinct linear factors over $\mathbb C$, and it splits over $\mathbb R$ if and only if $\lambda_k\lambda_{n+1-k}>0$. Therefore $A_k$ is diagonalisable over $\mathbb R$ if and only if $\lambda_k\lambda_{n+1-k}>0$.

If $\lambda_k\lambda_{n+1-k}=0$, $A_k$ is nilpotent. Hence $A_k$ is diagonalisable over $\mathbb C$ or $\mathbb R$ if and only if it is the zero matrix, i.e. iff $\lambda_k=\lambda_{n+1-k}=0$. In short:

A real anti-diagonal matrix $A$ is diagonalisable over $\mathbb R$ if and only if for every symmetric pair of entries $\{\lambda_k,\lambda_{n+1-k}\}$ on the anti-diagonal, either $\lambda_k\lambda_{n+1-k}>0$ or $\lambda_k=\lambda_{n+1-k}=0$.

Similarly:

A complex anti-diagonal matrix $A$ is diagonalisable over $\mathbb C$ if and only if for every symmetric pair of entries $\{\lambda_k,\lambda_{n+1-k}\}$ on the anti-diagonal, either $\lambda_k\lambda_{n+1-k}\ne0$ or $\lambda_k=\lambda_{n+1-k}=0$.

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