# Can I determine if a set spans a vector space by only taking the determinant?

After reviewing chapters and questions on Span, Linear Independence and Basis, something occurred to me. Determinant is always nonzero for a set to span, or a set to be linear independent. I know taking the determinant is key to find linear independence but what about proving that set spans a vector space? Can I use the determinant alone to prove it?

If you have a set of $n$ vectors in the space $\mathbb R^n$, then the set is linearly independent if and only if it spans $\mathbb R^n$, so it spans $\mathbb R^n$ if the determinant of the matrix, produced by stacking the vectors into a matrix, is nonzero.

Of course, if you have less than $n$ vectors in $\mathbb R^n$, then you cannot possibly span the whole set (because you can only span a space of $m$ or fewer dimensions with $m$ vectors).

If you have more than $n$ vectors, then stacking them into a matrix will create a matrix of size $n\times m$ where $m>n$. You cannot calculate the determinant of this matrix, because it is not square, but it still holds that the vectors span $\mathbb R^n$ if and only if there exists a $n\times n$ square submatrix of the original matrix that does have a nonzero determinant.

• in a case in which I have more than n vectors, solving the I will need to prove a unique solution to the coefficient matrix of nxm to prove that the set spans the vector space , correct? – user2171775 Apr 15 '15 at 7:37
• @user I don't know what you mean by "solution to the coefficient matrix". – 5xum Apr 15 '15 at 7:42
• you mentioned that if I have more than n vectors, stacking them into a matrix of size (nxm ) .. this is the matrix I was referring as the coefficient matrix. – user2171775 Apr 15 '15 at 7:44
• @user2171775 Yes, but I don't know what a "solution" of a matrix is. A matrix is not an equation, and thus has no solution. – 5xum Apr 15 '15 at 7:46

Given $m$ vectors $\{v_1,\dots,v_m\}$ of an $n$-dimensional vector space $V$ equipped with an inner product $\langle . ,.\rangle$ you may calculate their Gram-Matrix $G$ defined by $g_{ij}:=\langle v_i,v_j\rangle$. Then $\det(G)=$ the squared $m$-dimensional volume of the parallelepiped spanned by $\{v_1\dots v_m\}$. From here: $\{v_1,\dots,v_m\}$ linear independent iff $\det(G)\neq0$.

Edit: In case $V=\mathbb R^n$ with the usual dot product the Gram-matrix is just $A^tA$ where $A$ is the matrix whose column vectors are $v_1,\dots,v_m$.

I will give you a long proof of your question.

Let $V$ be a finite, $n$-dimensional vector space over a field $\mathbb{F}$ with infinite elements such as $\mathbb{R}$ or $\mathbb{C}$.

Let $\varepsilon$ be a $\varepsilon:V^n\rightarrow\mathbb{F}$ type map, that maps $k$ vectors into a scalar in a way, that is linear in every argument separately. We call such a map a "multilinear functional". Let us also assume, that $\varepsilon$ is alternating, by which we mean that if any two of $\varepsilon$'s arguments are the same, then $\varepsilon$ is zero, when evaluated on them.

It is easy to prove, that over a field with infinte elements, an alternating multilinear functional is always also antisymmetric, meaning that it will change sign, when you switch two arguments, and it is trivial to prove, that an antisymmetric multilinear functional is alternating.

Let $x_1,...,x_n\in V$ be a set of linearly dependent vectors. Then $$\varepsilon(x_1,...,x_n)=0,$$since if those vectors are linearly dependent, then there exists $\alpha_1,...,\alpha_n\in\mathbb{F}$ scalars that not all $\alpha$s are zero, and $$0_V=\alpha_1x_1+...+\alpha_nx_n,$$ and then, assuming $\alpha_i\neq0$, then $$x_i=-\frac{\alpha_1}{\alpha_i}x_1-...-\frac{\alpha_n}{\alpha_i}x_n,$$ and if you plug this expression in to $\varepsilon(x_1,...,x_n)$ in lieu of $x_i$, by linearity, this expression will collapse into sums that all contain a vector twice, and thus are zero.

It is then not hard to show, that $\varepsilon(y_1,...,y_n)$, where $y_1,...,y_n\in V$ is only zero, if ${y_1,...,y_n}$ forms a linearly dependent set (unless $\varepsilon$ assigns zero to all $n$-tuples of vectors).

Let us assume, that ${y_1,...,y_n}$ is linearly independent, yet $\varepsilon(y_1,...,y_n)=0$, and let $\alpha_{1,i_1},...,\alpha_{n,i_n}$ (all $i$s range from 1 to n) be a set of arbitrary scalars. Since the set ${y_1,...,y_n}$ is linearly independent and contains $n$ elements, it is a basis of $V$, then arbitrary $z_1,...,z_n$ vectors are expressible as $$z_i=\sum_{j=1}^n\alpha_{i,j}y_j.$$ Plugging the $z_i$s into $\varepsilon$ yields $$\varepsilon(z_1,...,z_n)=\sum_{i_1,...,i_n=1}^n \alpha_{1,i_1}...\alpha_{n,i_n}\varepsilon(y_{i_1},...,y_{i_n}),$$ but since $\varepsilon$ is alternating, if $\varepsilon$ on $y_1,...,y_n$ is zero, it is zero on all permutations of $y_1,...,y_n$, so $\varepsilon(z_1,...,z_n)=0$.

The vectors $z_1,...,z_n$ were arbitrary, and since ${y_1,...,y_n}$ is a basis, any arbitrary vector is expressible as a linear combination of the $y$s, our result implies, that if $\varepsilon$ is zero on an independent system, it is zero on all $n$-tuples of vectors. Nonzero $\varepsilon$s exist however (by nonzero $\varepsilon$s we mean those that don't take zero value on all $n$-tuples of vectors), since it is provable by the alternating property that the set of $n$-linear alternating multilinear functionals over an $n$-dimensional vector space form a 1-dimensional vector space, therefore we proved that any nonzero $n$-linear alternating functional must take nonzero value on a linearly independent set.

The determinant of a $n\times n$ matrix, when viewed as a function of its columns, is, by the properties of the determinant, an alternating $n$-linear functional on the vector space of columns with $n$ elements, therefore, the determinant is nonzero if and only if the columns are independent.

If you have more than $n$ columns, then those columns span $\mathbb{R}^n$ if and only if there exists a subsystem of $n$ columns that are linearly independent, and you can use the determinant method to analyse that.

If you have less than $n$ columns, then those columns will never span $\mathbb{R}^n$.