Are vectors $e_i$ linearly independent if and only if the matrix $A$ is nonsingular?

Question

Let $V$ be an inner product space over $\mathbb{R}$ or $\mathbb{C}$, and let $e_1, e_2, \dots, e_n$ be a collection of vectors in $V$, not necessarily orthonormal. (Here, $n$ has nothing to do with the dimension of $V$.) Let $\langle\cdot,\cdot\rangle$ denote the inner product in $V$, and form the $n \times n$ matrix $A$ with entries $\langle e_i, e_j\rangle$ (where $i, j = 1, \dots, n$). Are the vectors $e_i$ linearly independent if and only if the matrix $A$ is nonsingular?

My thoughts on the problem so far. Let $F$ be $\mathbb{R}$ or $\mathbb{C}$ and $f$ the linear map $F^n \to V$ which sends the $i$th standard basis vector to $e_i$. I want to show that, if $v \in F^n$ is nonzero, then $Av$ is nonzero. I've observed that $v^t Av = \langle f(v), f(v)\rangle$ (which is positive). But I don't know what to do next. Can someone help me out?

Answer 1

Let me denote by $v_1, \ldots, v_n$ the vectors in $V$ (so that we won't confuse them with $e_i \in \mathbb{F}^n_{\mathrm{col}}$). Assume that $A$ is singular. Take $x \neq 0$ with $Ax = 0$. Then

$$ 0 = x^t(Ax) = x^tAx = \left< f(x), f(x) \right> = ||f(x)||^2 $$

and thus $f(x) = 0$. This means that $f$ is not one-to-one and so $\dim \mathrm{Im}(f) < n$ and since $\{ v_1, \ldots, v_n \} \in \mathrm{Im}(f)$, they must be linearly dependent.

If $A$ is not singular, then it is enough to show that $f$ is one-to-one to deduce that $\{ v_1, \ldots, v_n \}$ are linearly independent. Indeed, if $f(x) = 0$, then $0 = ||f(x)||^2 = x^tAx$. Since $A$ is positive, this implies that $Ax = 0$ showing that $x = 0$.

Answer 2

Do one direction at a time.

First, suppose your vectors are linearly dependent. A linear relation between them will corespond directly to a linear relation between the columns (or rows) of $A$, so $A$ is singular.

Second, suppose that $A$ is singular. Since it is real symmetric (or herimitian) it can be diagonalized; the diagonal form must have a $0$ somewhere on the diagonal. This $0$ corresponds (through the diagonalizing basis change) to a linear combination of your vectors whose inner product with itself is $0$ -- that is, a linear relation between your vectors.