# Prove these matrix-vector products are linearly dependent/independent.

I have two statements that I wanto to prove or disprove.

1. 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.

2. 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?

• Your two statement says either $b_1,...b_n$ are linearly independent, or $Ab_1, ..., Ab_n$ are linearly independent. They are sets of vectors, not a single vector or a matrix. – KittyL Jun 19 '15 at 17:30
• This is not true by the way for any matrix $A$. It is only true with invertible matrices – marwalix Jun 19 '15 at 17:32
• @ KittyL Thanks. Just a follow up question. Does $A \in \Bbb R^{n \times n}$ mean that the matrix $A$ has the dimension $n \times n$ and has real valued entries? Are all of these entries scalar values? How can the matrix be linearly independent? – qmd Jun 19 '15 at 17:38
• @marwalix Thanks. But why is that? Is that something that I need to prove in this proof? – qmd Jun 19 '15 at 17:41
• Your first statement is true if $A$ is a nonsingular matrix, and most matrices are nonsingular. It is false if $A$ is singular. – Michael Hardy Jun 19 '15 at 17:44

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?

• Thanks for your answer. For (1): In that case the resulting vectors $$Ab_1=Ab_2=\begin{pmatrix}0 \\0 \end{pmatrix}$$ Are not linearly independent because they are the same vector. So anytime the matrix $A$ has $0$ as all its entries, the first statement is false – qmd Jun 19 '15 at 17:54

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.

• Took me a while to digest but it helped me a lot. Thanks! Is there a way to accept two answers? Both your and brian's answer helped me. – qmd Jun 19 '15 at 18:18