T/F Exercise in Linear Algebra 
*

*A set containing a single vector is linearly independent.

*The set of vectors $\{v , kv\}$ is linearly dependent for every scalar $k$.

*If $A$ has size $n\times n$ and $I$ (identity matrix), $A, A^2,\dots,A^{n^2}$  are distinct matrices, then $\{ I, A, A^2,\dots,A^{n^2}\}$ is linearly dependent.


My answers... 


*

*For $k*v=0$ , $k$ must be  zero. So i think this is true(does the set contain the zero vect0r??? Is the set with only zero vect0r dependent?? Then it will be false)

*With $k=0$, the set is linearly independent so that is false.

*Since $n\times n$ matrix has $n^2$ dimension, and the set $\{ I, A, A^2,\dots,A^{n^2}\}$ has $n^2+1$ elements, it must be linearly dependent. So that is true.


But with textbook's answer, I had wrong answers for all of this. Which one is my mistake?
 A: *

*By definition, a set of vectors $\{v_1,\ldots,v_n\}$ in a vector space over a field $F$ is linearly dependent if there exist $\alpha_1,\ldots,\alpha_n \in F$ such that $\alpha_i \neq 0$ for at least one $i$ and $\alpha_1 v_1 + \ldots + \alpha_n v_n = 0$. So if you begin with the set $\{0\}$ containing only the zero vector, is it linearly dependent?

*If $k = 0$, we can write $(-k) \cdot v + 1 \cdot (kv) = 0$, where the second coefficient ($1$) is always nonzero. So with the exception of the case where $k = 1$ (where the set $\{v,kv\}$ is just $\{v\}$), we will have linear dependence. However, as Arnauld pointed out in the comments, in the case where $k = 1$ the set is just $\{v\}$, so in general the statement made in 2. is false.

*If the question did, in fact, say that $I,A,A^2,\ldots,A^{n^2}$ are all distinct and $A$ is an $n\times n$ matrix, then you are correct: the set $\{I,A,\ldots,A^{n^2}\}$ lies in an $n^2$-dimensional vector space but has order $n^2+1$. Therefore it must be linearly dependent.


Edit: So, to clarify, I agree with you on #3, and I believe on #2 that the book author didn't take into account $k = 1$, in which $\{kv,v\} = \{v\}$, and the statement is then false (unless $v = 0$). For $k \neq 1$, the set $\{kv,v\}$ will always be linearly dependent.
