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?
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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.
On
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.)
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 $f(x_1, \ldots, x_n) = 0$.
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).
On
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 $det(A) = 0$.
On
Here a simple geometric explanation:
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.
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The n vectors are linearly dependent iff the zero vector is a nontrivial linear combination of the vectors (definition of linearly independent).
The zero vector is a nontrivial linear combination of the vectors iff the matrix times some nonzero vector is zero (definition of matrix multiplication)
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)
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.
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).
The determinant of a matrix with a zero row is zero (this can be verified by expanding the determinant about that row).
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.
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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'.
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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 $\bf{0}$.
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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 $\det f = 0$.
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.