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?

  • $\begingroup$ Okay so I saw this on the related items just now - math.stackexchange.com/questions/28061/… - so that confirms you can use it to determine if it's a basis - but what about Span and Linear Independence? $\endgroup$
    – eWizardII
    Nov 5, 2011 at 23:44

1 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.

  • $\begingroup$ Thanks, a few sections later they give a similar explanation like you said - it just wasn't in the one where they are expecting you to solve these types of problems. $\endgroup$
    – eWizardII
    Nov 6, 2011 at 18:37
  • 3
    $\begingroup$ Does the converse apply for all of those? For example, if det=0, does that always imply that the rows are linearly independent? $\endgroup$ Oct 7, 2013 at 16:28
  • 6
    $\begingroup$ @Asad equivalent means they are all true or all false. If $\det(A)=0$ means the second one is false, which means ALL are false. So yes, $\det(A)=0$ implies the rows are linearly DEPENDENT. $\endgroup$
    – N. S.
    Oct 7, 2013 at 16:33
  • 1
    $\begingroup$ @N.S. I see, thanks. $\endgroup$ Oct 7, 2013 at 17:06

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