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If there is a Jordan block matrix with $A(i,i) = a$ for all $i=1$ to $n$, $A(i,i+1)=b$ for all $i=1$ to $n-1$ and $A(j,k)=0$ otherwise. What will be eigenvalue and eigenvector of $J$?

To calculate eigenvalues I split the matrix in a diagonal matrix $= diag(a)$ and another matrix $B$ with $A(i,i+1)=b$. As $B$ is singular matrix, one of the eigenvalues will be zero. So for Jordan block matrix eigenvalue become $=a+0 =a$. How to find rest of the eigenvalues and eigenvectors?

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what you have described is an upper triangular matrix, hence the only eighenvalue is a – kaharas Mar 12 '13 at 17:15

Note that there is only one eigenvalue in this case: that's $a$.

To find the eigenvectors, solve the linear system $$ AX=aX\quad\Leftrightarrow\quad (A-aI_n)X=0. $$ If $b\neq 0$, the solutions are easily seen to be $$ \mbox{span}(1,0,\ldots,0). $$ These are the eigenvectors.

If $b=0$, you the solutions/eigenvectors are obviously the whole vector space since $A-aI_n=0$.

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The only candidates for eigenvalues are the roots of the characteristic polynomial. So, if $A$ is an upper/lower triangular matric (as Jordan block matrices are) of dimension $n$ with $a ∈ k$ on the diagonal, $χ_A = (X-a)^n$, so $a$ is the only possible eigenvalue.

Then, the eigenvectors of $a$ are obtained by looking at the kernel of $A - aI$, since $A$ is a Jordan block, the eigenvectors are $k^×$-multiples of $e_n$ (or $e_1$ depending on whether your Jordan Block matrix is in upper or lower triangular shape).

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