Question:
Prove that for a $m \times n$ matrix $A$, if $A^TA$ is invertible, then $A$ has linearly independent column vectors.
I am hitting a complete blank with this proof, I have the following jotted down so far about stuff that I know.
What I know so far:
Let $A$ be an $m \times n$ matrix and suppose $A^TA$ is invertible.
We know $A^T$ is an $n\times m$ matrix, hence $A^TA$ is an $n\times n$ square matrix with nonzero determinant.
We also know that $A^TA\bar{x}=\bar{0}$ has only the trivial solution and $A^TA = \bar{b}$ is consistent and has exactly one solution.
The column and row vectors of $A^TA$ are linearly independent.
How can I use some of the above to show that $A$ has linearly independent column vectors?
Best Answer
For each $i$, let $A_i$ be the $i$th column of $A$. Let $A_1 x_1 + \cdots + A_n x_n = 0$ be a linear dependence relation. Then $Ax = 0$, where $x$ is the column vector $(x_1\cdots x_n)^T$. So $A^TAx = 0$. Invertibility of $A^T A$ implies $x = 0$. Thus $x_1 = \cdots = x_n = 0$, showing that $\{A_1,\ldots, A_n\}$ is linearly independent, as desired.