3

The matrix in question is similar via a permutation matrix to $$ A=\begin{cases} \pmatrix{0&a_1\\ a_n&0}\oplus\pmatrix{0&a_2\\ a_{n-1}&0}\oplus\pmatrix{0&a_{(n-1)/2}\\ a_{(n+3)/2}&0}\oplus a_{(n+1)/2}&\text{when $n$ is odd},\\ \pmatrix{0&a_1\\ a_n&0}\oplus\pmatrix{0&a_2\\ a_{n-1}&0}\oplus\pmatrix{0&a_{n/2}\\ a_{...


3

So you want to know the Jordan canonical form of the $i \times i$ matrix $$ A = \sum_{r=0}^{i-1} \left( n \atop r \right) k^{n-r} t^r .$$ Since $A$ has $k^n$ as an $i$-fold repeated eigenvalue, it is sufficient to find the Jordan form for $$ A - k^n I = \sum_{r=1}^{i-1} \left( n \atop r \right) k^{n-r} t^r .$$ First consider the case $k \ne 0$. Then $$ (A- ...


2

A Jordon block of size $i$ and eigenvalue $k$ has the form $(kI+t)$ where $t^i=0$. Thus $(kI+t)^n$ will just be the truncated binomial expansion:$$\sum_{r=0}^{i-1} {n \choose r} k^{n-r}t^r$$


2

In general we have $\sigma(T|_W) \subseteq \sigma(T)$ so every eigenvalue $\lambda$ of $T|_W$ is an eigenvalue of $T$. Recall that $$\dim \ker (T-\lambda I) = \text{number of $\lambda$-blocks}$$ $$\dim \ker (T-\lambda I)^2 - \dim \ker (T-\lambda I) = \text{number of $\lambda$-blocks of size $\ge 2$}$$ $$\dim \ker (T-\lambda I)^3 - \dim \ker (T-\lambda I)^2 = ...


2

Let $n$ denote the size of the matrices $A,B$. Note that the Jordan normal form of a size-$n$ matrix $M$ can be completely recovered if one knows the rank of $(M - t I)^k$ for all eigenvalues $t$ of $M$ and $k = 1,2,\dots,n$ (via the "Weyr Characteristic"). In our case, $A + \lambda B$ has $0$ as its lone eigenvalue, so it suffices to consider the ...


1

Think of $A$ as a block diagonal matrix with blocks $\begin{bmatrix} J_0^2 & 0 \\ 0 & J_0^2 \end{bmatrix}$ and $\begin{bmatrix} J_{1/4}^3 \end{bmatrix}$. If you find matrices that square to each of these blocks then form a block diagonal matrix out of them and the result will square to $A$. You've already figured out a matrix for the second block, ...


1

We can avoid computing generalized eigenvectors. Separate this into $3$ cases. Case 1: $a = b = 0$. As you said, the matrix is already in Jordan form Case 2: $a=0$, $b \neq 0$. The matrix is upper triangular, so we quickly see that its only eigenvalues are $0$ and $b$. The block associated with $b$ has size $1$, and the because the matrix $M$ has rank $2$, ...


1

The first two basis vectors corresponding to any Jordan block of size greater than $1$ span an invariant subspace on which $T$ is given by a single Jordan block of size $2$. In particular, then, $T$ is not diagonalizable on that $2$-dimensional subspace. Or, without referring to matrices at all, if $(x-\lambda)^2$ divides the minimal polynomial of $T$, this ...


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