# Determinant of a matrix and linear independence (explanation needed)

It is written on Wikipedia that:

$n$ vectors in $\mathbb R^n$ are linearly independent if and only if the determinant of the matrix formed by taking the vectors as its columns is non-zero

Can someone explain this to me? You do not have to give a complete proof, just in simple terms explain what the determinant of that matrix has to do with linear independence? And why it has to be non-zero? And are vectors allowed to be rows instead of columns in that matrix?

• It is because the determinant is linear on the columns and alternating (switching two columns changes the sign). If a column is a linear combination of the others, then by linearity the determinant is equal to a linear combination of determinans of matrices that have a repeated column. When you have a repeated column and switch them, the sign is supposed to change, but the resulting matrix is the same. the only number that stays the same after changing sign, is $0$.
– user545963
Apr 3, 2018 at 15:27
• And, yes you can put them as rows, since the determinant of a matrix is equal to the determinant of its transpose.
– user545963
Apr 3, 2018 at 15:31
• The determinant can be seen as a function that is specifically designed to test for the linear dependency of vectors. From this point of view there is no case to answer.
– Karl
Apr 3, 2018 at 20:52
• To illustrate my point with a perhaps poor example. If you define $\sin$ in terms of a power series you might ask the question what has $\sin$ got to do with circles? This question is made redundant if you start with the idea of $\sin$ as a coordinate on a unit circle. Hopefully you see what I'm driving at as I know the analogy isn't great.
– Karl
Apr 3, 2018 at 21:17

Here a simple geometric explanation:

• $2$ vectors in the plane are linearly independent if and only if they span a parallelogram with a non-zero area
• $3$ vectors in 3D-space are linearly independent if and only if they span a parallelepiped with a non-zero volume
• $n$ vectors in $\mathbb{R}^n$ are linearly independent if and only if they span an $n$-dimensional parallelepiped with a non-zero volume

The determinant is a so-called "volume form" that gives for $n$ vectors in $\mathbb{R}^n$ the $n$-dimensional volume of the parallelepiped that is spanned by those vectors (up to a sign which gives rise to the so called orientation).

So, if $n$ vectors in $\mathbb{R}^n$ are linearly dependent, they cannot span an $n$-dimensional parallelepiped and hence produce a volume of zero.

• You should perhaps also include the basis-independent definition (if you have vectors spanning some n-dimensional volume, then transforming the vectors according to a linear map $\mathbf{A}$ changes the volume by the factor $\det\mathbf{A}$). And the OP can look up the operation called “exterior product” for more on the relationship between determinants and measures. Apr 3, 2018 at 18:19
• @Roman Odalsky: I had rather the Weierstrass approach to determinants in mind. After understanding the determinant as a volume, it is no surprise anymore when they appear in $n$-dimensional substitution (as a special application of transformed measures). I was rather tempted to include a shorte note on Cavalieri's principle which produces immediately the invariance of the determinant when a multiple of a column/row is added to another column/row. Apr 3, 2018 at 18:57
• And, of course, one vector in the line ($\mathbb R$) is linearly independent if and only if it spans a line with non-zero length. :-) Apr 4, 2018 at 0:02

The determinant relates to the invertibility of the matrix. The statement is equivalent to saying that no two columns are linearly dependent. If they were, then when you turn it into a reduced form (like RREF) you get a row or column of zeros. This would mean that the determinant is zero, and therefore the columns are linearly dependent.

The determinant will take a matrix, and give back a real number, such that multiplication is preserved. That is:

If $AB = N$ then $det(A)det(B) = det(N)$

This implies that since $AA^{-1} = I$, then $det(A)det(A^{-1}) = 1$. Or, rearranging, $det(A^{-1}) = \frac{1}{det(A)}$

Right away, a neat property. The determinant of an inverse matrix is the reciprocal of the determinant of the original matrix. However, if the determinant of the original matrix is $0$, we run into a clear problem. The issue was the assumption of a well defined inverse, which clearly can't exist when $det(A) = 0$.

1. The n vectors are linearly dependent iff the zero vector is a nontrivial linear combination of the vectors (definition of linearly independent).

2. The zero vector is a nontrivial linear combination of the vectors iff the matrix times some nonzero vector is zero (definition of matrix multiplication)

3. The matrix times some nonzero vector is zero iff the augmented matrix of the vectors and the zero vector has a nontrivial solution (definition of solving an augmented matrix)

4. The augmented matrix has a nontrivial solution iff the augmented matrix can be row reduced to be a matrix augmented with the zero vector where the matrix has a row without a pivot column (that is, a zero row). These row reduction operations won't affect the zero vector, so that part will still be the zero vector, and if a row has a pivot column, then the corresponding variable must be zero to match the zero vector. So to have a nontrivial solution, there must be some free variable that can be nonzero.

5. Row reduction operations do not change whether a matrix has a determinant of zero (subtracting two rows leaves the determinant unchanged, switching rows multiplies by -1, multiplying a row by a scalar multiplies the determinant by that factor).

6. The determinant of a matrix with a zero row is zero (this can be verified by expanding the determinant about that row).

7. By 4, the vectors are linearly dependent iff the reduced form has a zero row. By 5 and 6, the reduced form has a zero row iff the original matrix has determinant zero. Therefore, the vectors are linearly dependent iff the matrix has determinant zero.

The determinant is an n-linear (multilinear) alternating form.

(Let us assume that our determinants are with respect to an arbitrary base of the space considered, it's just a technical aspect, for rigor, but you should consider the determinant of a family with respect to a certain base.)

## Alternating characteristic

What is relevant here is the alternating characteristic, let's take $(x_1, \ldots, x_n)$ a family of vectors and $f : \mathbb{K}^n \to \mathbb{K}$ a $n$-linear alternating form from the $\mathbb{K}$-vector space of dimension $n$ to $\mathbb{K}$.

If there is $(i, j) \in \{ 1, 2, \ldots, n \}^2$ such that $i \neq j$ and $x_i = x_j$, then $f(x_1, \ldots, x_n) = 0$.

Use the $n$-linear characteristic, and you get that:

If $(x_1, \ldots, x_n)$ is not linearly independent, then $f(x_1, \ldots, x_n) = 0$.

## The case of the determinant

Well, this applies to $\det$ also, so that, if $\det (x_1, \ldots, x_n) \neq 0$, then $(x_1, \ldots, x_n)$ is linearly independent.

Finally, we define the determinant of a matrix as the determinant of columns (or lines, because $\det$ is invariant with respect to the transposed of a matrix).

Consider the wedge product $\det V = \Lambda^n V$ for a vector space $V$ of dimension $n$ over a field $k$. Any $f\in \operatorname{End}(V)$ induces a corresponding action on the space $\det V$ given by $\overline{f}(v_1 \wedge \cdots \wedge v_n) = f(v_1) \wedge \cdots \wedge f(v_n)$, giving a homomorphism $\operatorname{End}(V) \to \operatorname{End}(\det V)$. But the $\det V$ is $1$-dimensional, so $\operatorname{End}(\det V)$ is canoncially isomorphic to $k$. Thus we have a multiplicative map $\det:\operatorname{End}(V) \to k$.

If $g$ is invertible, then $(\det g)(\det g^{-1}) = \det 1\not = 0$; in particular, $\det g\not = 0$. Conversely, if $f$ is not invertible, then it is not surjective, and $f(v_1), \dots, f(v_n)$ are therefore not linearly independent for any basis $\{v_i\}$ of $V$. . Thus assume without loss of generality that $f(v_1) = \lambda_2 f(v_2) + \cdots + \lambda_n f(v_n)$ for some constants $\lambda_i$. But then $$\overline{f}(v_1 \wedge \cdots \wedge v_n) = f(v_1) \wedge \cdots \wedge f(v_n) = \sum_{i=2}^n \lambda_i f(v_i) \wedge f(v_2) \wedge \cdots \wedge f(v_n) =0$$ by linearity, forcing $\det f = 0$.

Let me put it in simple terms.

Consider a $2 \times 2$ matrix $$\left( \begin{array}{c c} a & b \\ c & d \end{array} \right).$$

It's inverse can be shown to be $$\frac{1}{a d - b c} \left( \begin{array}{c c} d & -b \\ -c & a \end{array} \right).$$ What you notice is that the matrix is invertible if and only if $$a d - b c \neq 0$$. Hence the relationship between the invertibility of a matrix and the determinant (for a $2 \times 2$ matrix). Invertibility is equivalent to the matrix having linearly independent columns. (Why that is the case is another discussion.)

Now, one can come up with a similar formula for the inverse of a $3 \times 3$ matrix. And, again, the determinant comes into the picture in that formula. Ditto for any $n \times n$ matrix (although no one in their right mind would write down that formula).

Hence the relationship between linearly independent columns and the determinant.

• "no one in their right mind would write down that formula" hmmm.... Apr 3, 2018 at 17:49
• I probably should have said "no one in their right might would use that formula in practice" Apr 3, 2018 at 18:57

First, note that the determinant of a matrix $A$ and its transpose $A^T$ are equal. The determinant of $A^T$ is $0$ iff there exists a sequence of row operations which produces a row of all zeros. Equivalently, there exists a linear combination of the columns of $A$ that is equal to $\bf{0}$.

To answer the part of the question no one else did: Yes, they can be rows instead. The underlying reasoning doesn't really change, it just 'transposes'.