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We have $A=(a_{ij})\in\mathbb M_n(\mathbb R)$ such that $a_{ij}= \begin{cases} i, & \quad \text{if } i+j=n+1\\ 0 ,& \quad \text{otherwise. } \\ \end{cases}$

Then what will be the Eigen value of $A?$

MY TRY:I am unable to proceed a bit.

2 Answers 2

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Try squaring the matrix, and you will have that the diagonal entries are $a_{i\times i}=i\cdot(n-i)$. This gives a recipe for a polynomial equation in $A$. For $n$ even, that is \begin{equation*} \prod_{i=1}^{\frac{n}{2}}(A^2-i\cdot(n-i+1))=0 \end{equation*} For $n$ odd, that is \begin{equation*} \left(A^2-\left(\frac{n+1}{2}\right)^2\right)\prod_{i=1}^{\frac{n-1}{2}}(A^2-i\cdot(n-i+1))=0 \end{equation*} Thus in all cases we have eigenvalues $\pm\sqrt{i\cdot(n-i+1)}$ for $i=1,2,..., \lfloor\frac{n}{2}\rfloor$. That is all the eigenvalues in the even case. In the odd case there is one more: $\frac{n+1}{2}$ (and it is positive because there is an obvious eigenvector).

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    What will be the case if $A_{n\times n} = \begin {bmatrix} 0 & \cdots & 0 & 0 & 1\\ 0 & \cdots & 0 & 1 & 0\\ 0 & \cdots & 1 & 0 & 0\\ \vdots & \vdots & \dots & \vdots & \vdots\\ 1 & \cdots & 0 & 0 & 0\\ \end {bmatrix}?$ – 2017-01-22
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    Have you tried squaring that matrix? What is the result when you square it? – 2017-01-22
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    Yes ,I think $A^2=I $ for any order. – 2017-01-22
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    Exactly! That means $A^2-I=0$, so all eigenvalues must be either $1$ or $-1$. Can you find an eigenvector for each value? – 2017-01-22
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    So,Minimal polynomial of this matrix is $x^2-1=0$.but can we conclude characteristics polynomial? – 2017-01-22
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    It's going to be hard to spell this out in the comments. Perhaps ask another question. The result is probably dependent on even or odd; however, $(x^2-1)^{\lfloor\frac{n}{2}\rfloor}$ is probably a large factor of the characteristic polynomial. – 2017-01-22
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    Ok,Thanks a lot!!! – 2017-01-22
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For the given situation the matrix $A_{n\times n} $ matrix can be described as $$A_{n\times n} = \begin {bmatrix} 0 & \cdots & 0 & 0 & 1\\ 0 & \cdots & 0 & 2 & 0\\ 0 & \cdots & 3 & 0 & 0\\ \vdots & \vdots & \dots & \vdots & \vdots\\ n & \cdots & 0 & 0 & 0\\ \end {bmatrix} $$

We can check that a pair of eigenvalues will be definitely $\pm \sqrt{n}$. Can you check that out? Hope it helps.