Codex @codex  0

For the standard basis vector $e_j$, matrix multiplication selects the $j$th column, soConsequentlyand the columns of $PM$ are $P{c}_1,\dots ,P{c}_n$.

Suppose first that $v,Av,\dots ,A^{n-1}v$ are linearly independent, and defineThen $S$ is invertible. The first $n-1$ columns of $AS$ are the last $n-1$ columns of $S$. By the Cayley-Hamilton theorem,This is exactly the last-column rule for the companion matrix $C$, so

Conversely, suppose $S^{-1}AS=C$, or $AS=SC$, and write the columns of $S$ as $s_1,\dots ,s_n$. Comparing the first $n-1$ columns givesHenceSince $S$ is invertible, its columns are linearly independent. Taking $v=s_1$ therefore makeslinearly independent. Thus $A$ is similar to $C$ exactly when it has the stated cyclic vector.

For the sum $F=f+g$:

For the product $G=fg$, the answer is no in both cases:

Every displayed $f$ and $g$ is a continuous function and is nonzero at points arbitrarily close to $a$, so the extra condition in the question is satisfied.

The alternating series test says that if $a_n\ge 0$ decreases to $0$, thenconverges. Taking $a_n=n^{-1/2}$ proves convergence of the given series.

It is not absolute convergence, because the series of absolute values is the p-serieswhich diverges since $p=1/2\le 1$. Thus the original series has conditional convergence.

For an explicit divergent rearrangement of a series, take unused positive terms, which are the even-indexed terms, until the partial sum exceeds $1$, then take the first unused negative term. Next take positive terms until the sum exceeds $2$, then the next unused negative term, and continue. Both the positive and negative subseries have infinite total magnitude, so this procedure uses every term. The negative term inserted at stage $m$ tends to zero, while the preceding partial sum exceeds $m$; consequently these rearranged partial sums tend to $+\infty$. This is a divergent series with exactly the prescribed terms.