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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 eigen value and eigen vector of J.

To calculate evalues 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 evalues will be zero. So for Jordan block matrix evalue become =a+0 =a. How to find rest of the evalues and evectors.

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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

2 Answers 2

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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