Let $A \in \mathbb{R}^{n \times n}$ where $k_i \in \mathbb{R}^n, i = 1,2, \cdots , n$ are column vectors of $A$ and satisfy the following condition
$k_i = (i+2)k_{i+2}, i= 1,2, \cdots, n-2$
for $n>3$ choose one eigenvalue of $A$ then determine what is the smallest possible dimension for the eigenspace of the chosen eigenvalue.
What I've been trying so far:
$A = [3k_3 \quad 4k_4 \quad 5k_5 \quad \cdots \quad (n-1)k_{n-1} \quad nk_n \quad k_{n-1} \quad k_n ]$
what I can see here is for $n>3$, there are $2$ zero column vectors ($2$ linearly dependent vector). And determinant of $A$ has to be $0$.
Do you have any hint? I would really appreciate it.
In this matrix are odd columns multiples of each other, and it is so in the family of even columns as well. Therefore is $0$ an eigenvalue of multiplicity at least $n-2.$ This is the smallest possible dimension of the eigenspace related to $0,$ as it has at least $(n-2)$ independent eigenvectors.
Since for any column of the matrix $$k_{i+2}=\frac{1}{i+2}k_i,$$ the generating eigenvectors can be easily described as the vectors ${v}_i$ with all coordinates $0$ except $i-$th which is equal to $(i+2)$ and $(i+2)-$th equal to $-1.$ Or, if you prefer, $$v_i=(\underbrace{0,\dots,0},\underbrace{i+2}_{i-th},0,\underbrace{-1}_{(i+2)-nd},\underbrace{0,\dots,0})^T,$$ where the number of underbraced $0$s is appropriate.