Let $A$ be an $n\times n$ nonsingular matrix. Let $\mathbf{v}, \mathbf{w}$ be linearly independent vectors in $\R^n$. Prove that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.

Suppose that we have a linear combination
\[c_1(A\mathbf{v})+c_2(A\mathbf{w})=\mathbf{0},\]
where $c_1, c_2$ are scalars.
Out goal is to show that $c_1=c_2=0$.
Factoring out $A$, we have
\[A(c_1\mathbf{v}+c_2\mathbf{w})=\mathbf{0}.\]

Note that since $A$ is a nonsingular matrix, the equation $A\mathbf{x}=\mathbf{0}$ has only the zero solution $\mathbf{x}=\mathbf{0}$.
The above equality yields that $c_1\mathbf{v}+c_2\mathbf{w}$ is a solution to $A\mathbf{x}=\mathbf{0}$.
Hence, we have
\[c_1\mathbf{v}+c_2\mathbf{w}=\mathbf{0}.\]

By assumption, the vectors $\mathbf{v}$ and $\mathbf{w}$ are linearly independent, and this implies that $c_1=c_2=0$.

We have shown that whenever we have $c_1(A\mathbf{v})+c_2(A\mathbf{w})=\mathbf{0}$, we must have $c_1=c_2=0$. This yields that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is to ignore the logic and write down whatever you know.
For example, some students started with $c_1\mathbf{v}+c_2\mathbf{w}=\mathbf{0}$ and since $\mathbf{v}$ and $\mathbf{w}$ are linearly independent, we have $c_1=c_2=0$.
Multiplying by $A$, we have $c_1A\mathbf{v}+c_2A\mathbf{w}=\mathbf{0}$ and $c_1=c_2=0$.

This argument is totally wrong.

The above argument is wrong because it started with different vectors than we want to prove to be linearly independent.
There is nothing wrong about the mathematical operations in the above arguments. However, that argument does not prove that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.

We should first assume that $c_1A\mathbf{v}+c_2A\mathbf{w}=\mathbf{0}$ and prove that $c_1=c_2=0$.

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