Linear combinations of vectors of a basis

Question

Suppose $\dim V=n$ and $\beta=\{\beta_{1},\beta_{2}, ..., \beta_{n}\}$ is a basis of $V$.

Then, $\alpha_{1},\alpha_{2},...,\alpha_{n} $ such that $ \alpha_{k} = \lambda_{k1}\beta_{1} + \lambda_{k2}\beta{2} + ... + \lambda_{kn}$ for k=$1,2,...n$ are linearly independent if and only if
$\left| \begin{array}{cccc} \lambda_{11}&\lambda_{12} &...& \lambda_{1n}\\ \lambda_{21}&\lambda_{22} &...& \lambda_{2n}\\ . &.& .& .\\ . &.& .& .\\ . &. &. &.\\ \lambda_{n1}& \lambda_{n2}& ... &\lambda_{nn} \end{array} \right| \neq 0$

I need help in proving this.

Answer 1

Suppose that the above determinant $= 0$. Then there are two rows which are linearly dependent. Without loss of generality, we can assume that there is a scalar $c \ne 0$ such that

$(\lambda_{11},\lambda_{12} ,..., \lambda_{1n})=c(\lambda_{21},\lambda_{22} ,..., \lambda_{2n})$.

But then $\alpha_1=c \alpha_2$. Hence $\alpha_{1},\alpha_{2},...,\alpha_{n}$ are linearly dependent, a contradiction.