Using the Determinant to verify Linear Independence, Span and Basis

Question

Can the determinant (assuming it's non-zero) be used to determine that the vectors given are linearly independent, span the subspace and are a basis of that subspace? (In other words assuming I have a set which I can make into a square matrix, can I use the determinant to determine these three properties?)

Here are two examples:

• Span Does the following set of vectors span $\mathbb R^4$: $[1,1,0,0],[1,2,-1,1],[0,0,1,1],[2,1,2,-1]$? Now the determinant here is $1$, so the set of vectors span $\mathbb R^4$.
• Linear Independence Given the following augmented matrix:

$$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 2 & 0 \end{array}\right], $$ where again the determinant is non-zero ($-2$) so this set S is linearly independent.

Of course I am in trouble if you can't make a square matrix - I figure for spans you can just rref it, and I suppose so for linear independence and basis?

Answer 1

Most introductory books on Linear Algebra have a Theorem which says something like

Let $A$ be a square $n \times n$ matrix. Then the following are equivalent:

• $A$ is invertible.
• $\det(A) \neq 0$.
• The columns of $A$ are linearly independent.
• The columns of $A$ span $R^n$.
• The columns of $A$ are a basis in $R^n$.
• The rows of $A$ are linearly independent.
• The rows of $A$ span $R^n$.
• The rows of $A$ are a basis in $R^n$.

• The reduced row echelon form of $A$ has a leading 1 in each row.

  and many other conditions.....

What does this mean, it simply means that if you want to check if any of these conditions is true or false, you can simply pick whichever other condition from the list and check it instead..

Your question is: Can instead of third or fourth condition, check the second? That's exactly what the Theorem says: YES.