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I recently came across this question: suppose we have an $n$ by $n$ matrix whose diagonal values ($a_{i,i}$) are 1 and 2 of the secondary diagonal values ($a_{i,n+1-i}$) are $r$, and all other values are 0. what are the eigenvalues.

This question looks similar to some other questions such as this one Suppose A is an n-by-n matrix with its diagonal entries are n and other entries are one. Find determinant of A..

I've tried several approaches. I am thinking whether we can find the eigenvalues recursively. For example, when $n=$2, the eigenvalues are $1-r$ and $1+r$. However, I found it hard for general cases when $n>2$. To fix the idea, what I was thinking is that we just assume the two $r$'s occur on the first value and the third value. However, how do we use $n-1$ by $n-1$ such matrix to express such $n$ by $n$ matrix? And if there is any better way to approach this question?

DA_PA
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  • Isn’t your matrix upper-triangular? – Ted Shifrin Dec 05 '21 at 03:07
  • I guess no? But generally when n > 3, yes – DA_PA Dec 05 '21 at 04:51
  • I’ve never encountered “secondary diagonal” before — I guess this is the anti-diagonal. – Ted Shifrin Dec 05 '21 at 04:56
  • Sure, let's call it anti-diagonal. Any thoughts on this problem then? – DA_PA Dec 05 '21 at 05:41
  • If the two off diagonal entries are on the same side of the diagonal, then you have either an upper or lower triangular matrix. The diagonal entries of any triangular matrix are its eigenvalues. So in this particular case, all the eigenvalues must be $1$. On the other hand, when $n = 2$, the eigenvalues are $\lambda = 1 \pm r$. So the answer is going to depend on $n$ and which two rows have $r$ in the anti-diagonal. – Paul Sinclair Dec 05 '21 at 16:56

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