Prove these matrix-vector products are linearly dependent/independent. I have two statements that I wanto to prove or disprove.


*

*Let $A \in \Bbb R^{n \times n}$ and $b_1,\ldots,b_n \in \Bbb R^n$ be linearly independant. Then, $Ab_1,\ldots,Ab_n$ are also linearly independent.

*Let $A \in \Bbb R^{n \times n}$ and $b_1,\ldots,b_n \in \Bbb R^n$ such that $Ab_1,\ldots,Ab_n$ are linearly independent. Then $b_1,\ldots,b_n$ are linearly independent.
If I was dealing with vectors $\vec{x},\vec{y},\vec{z}$ then I would know how to show they are lin. independent by showing that the only solution to 
$$\lambda_1 \vec{x}+\lambda_2\vec{y}+ \lambda_3 \vec{z}=0$$
is $\lambda_1=\lambda_2=\lambda_3=0$.
But how does one do this with matrices and single vectors? Can you even say a single vector $b \in \Bbb R^n$ is linearly independent. I always thought you need at least two vectors? How can a matrix be linearly independent?
 A: For (1) consider the case $n=2$ and 
\begin{align*}
A&=\begin{bmatrix}0&0\\0&0\end{bmatrix} &
\vec b_1 &= \begin{bmatrix}1\\0\end{bmatrix} &
\vec b_2 &= \begin{bmatrix}0\\1\end{bmatrix}
\end{align*}
What is the answer to your question in this case? 
To generalize this example, suppose $A$ is nonsingular. Then $\DeclareMathOperator{rank}{rank}\rank(A)=m<n$. It follows that each of $A\vec b_1,\dotsc,A\vec b_n$ lives in an $m$-dimensional subspace of $\Bbb R^n$. Since $m<n$ this implies that the vectors $A\vec b_1,\dotsc,A\vec b_n$ are linearly dependent.
For (2) suppose that
$$
\lambda_1 \vec b_1+\dotsb+\lambda_n \vec b_n=\vec 0
$$
Applying $A$ to the equation gives
$$
\lambda_1 A\vec b_1+\dotsb+\lambda_n A\vec b_n=A\vec 0=\vec 0
$$
But we are given that $\{A\vec b_1,\dotsc,A\vec b_n\}$ are linearly independent. What does this allow us to conclude about $\lambda_1,\dotsc,\lambda_n$? How does this prove/disprove the statement?
A: If $Ab_1,\ldots,Ab_n$ are linearly independent, then there are no scalars $\lambda_1,\ldots,\lambda_n$ satisfying $\lambda_1 Ab_1+\cdots+\lambda_n Ab_n=0$ except $\lambda_1=\cdots=\lambda_n=0$.  Now observe that
$$
\lambda_1 Ab_1+\cdots+\lambda_n Ab_n=A(\lambda_1 b_1 + \cdots+\lambda_n b_n).
$$
Therefore the latter expression cannot be $0$ unless $\lambda_1 = \cdots = \lambda_n = 0$.  If the $A(\lambda_1 b_1 + \cdots+\lambda_n b_n)=0$, one cannot infer that $\lambda_1 b_1 + \cdots+\lambda_n b_n=0$ without more information, but one can say that if $\lambda_1 b_1 + \cdots+\lambda_n b_n=0$ then $A(\lambda_1 b_1 + \cdots+\lambda_n b_n)=0$, and we've just shown that cannot happen unless $\lambda_1 b_1 + \cdots+\lambda_n b_n=0$.
Thus $b_1,\ldots,b_n$ must be linearly independent in that case.
If $b_1,\ldots,b_n$ are linearly independent, that doesn't mean $Ab_1,\ldots,Ab_n$ are linearly independent unless $A$ has linearly independent columns.  Notice that no matter which vector $b$ is, the vector $Ab$ is a linear combination of the columns of $A$.  If those are linearly dependent, then and $n$ linear combinations of them, i.e. any $Ab_1,\ldots,Ab_n$, will be linearly dependent.
