Some feed back would be appreciated. I'm am not completely confident in my proof.

Let $$X = (X_1, X_2, \ldots, X_n)$$ where $X_1, X_2, \ldots, X_n \in \mathbb{F}^n.$ Prove that these vectors form a basis of $F^n$ if and only if $$\det{(X_1, X_2, \ldots, X_n)} \neq 0.$$

Assume first that $$\det{(X_1, X_2, \ldots, X_n)} \neq 0.$$ It is given that the column vectors $X_1, X_2, \ldots, X_n$ of $X$ are linearly independent if $$\det{X} \neq 0.$$ Moreover, as there are $n$ such vectors, it follows immediately that these column vectors span $\mathbb{F}^n.$ Said differently, if $$\det{X} \neq 0,$$ the matrix $X$ is invertible and corresponds with an isomorphism $L_X$ whose nullspace has zero dimension (that is, $L_X$ contains only the zero-vector in its null space) and whose image is $\mathbb{F}^n$ (that is, the rank, or the dimension of the image, of $L_X$ is $n$). Hence, (1) a linear combination of the column vectors of $X,$ namely $X_1, X_2, \ldots, X_n,$ is zero if and only if the coefficients corresponding to that linear combination are all zero, and (2) these vectors span $\mathbb{F}^n.$

Now, assume that the vectors $X_1, X_2, \ldots, X_n$ form a basis of $\mathbb{F}^n,$ that is, that they are linearly independent and span $\mathbb{F}^n.$ It is sufficient to show that if the vectors are linearly independent, then $\det{X} \neq 0.$ So, if $X_1, X_2, \ldots, X_n$ are linearly independent, then $$a_1X_1 + X_2 + \cdots + a_nX_n = 0$$ if and only if $a_1 = a_2 = \cdots = a_n = 0.$ That is, the linear map $L_X$ corresponding to $X$ has a nullspace with zero dimension. By the rank-nullity theorem, then, the rank of, or the dimension of the image of, $L_X$ is $n.$ Indeed, it has been assumed $X_1, X_2, \ldots, X_n$ spans $\mathbb{F}^n.$ Therefore, $L_X$ is invertible, $X$ is invertible, and $$\det{(X_1, X_2, \ldots, X_n)} \neq 0.$$


If $X\in M_n(\mathbb F)$, then applying Gaussian Elimination algorithm, you can transform your matrix $X$ to a diagonal matrix $D$ with all coefficients $\ne 0$. Since Gaussian Elimination doesn't affect the determinant, $\det(X) = \det(D) \ne 0$.

Reciprocally, if $\det(X) \ne 0$ suppose $(X_1, \ldots, X_n)$ are linearly dependent, then there exists $(\alpha_1, \ldots, \alpha_{k-1}, \alpha_{k+1}, \ldots, \alpha_n)$ so that $$\alpha_1 \cdot X_1 + \ldots + \alpha_{k-1} \cdot X_{k-1} + X_k + \alpha_{k+1} \cdot X_{k+1}+\ldots + \alpha_n \cdot X_n = 0.$$ Therefore, there will be a zero row in the reduced row echelon form, thus $\det(X) = 0$, which is absurd.

  • $\begingroup$ Gaussian Elimination does usually change the determinant (even if you don't scale any rows, you often can't reduce to a diagonal matrix without row interchanges). It just can't change the determinant between 0/nonzero. $\endgroup$ – Morgan Rodgers Jan 30 at 20:09

You can simply say this:

How $det(A)=0 \iff $the row echelon form of $A$ has no null lines. I.e., the matrix has rank $n$. And this will imply that the subspace generated by those vectors has dimension $n$, and because there are only $n$ vectors... its a basis of $\mathbb{F}^n$.

(this is just a sketch, just showing another possible way)


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