6.1 Linear dependence and independence
How do we characterise a vector space \(V\)? One possibility is to choose a set of vectors \(v_1,v_2,\cdots,v_k\in V\) which span \(V\), i.e., such that \[V=\operatorname{span}\{ v_1,v_2,\cdots,v_k\} .\] In order do this in an efficient way, we want to choose the minimum number of vectors necessary. If one vector from the set can be written as a linear combination of the others, it is redundant. This leads to the following definition.
If we have a set of three vectors that is linearly dependent, what can we say about the relationship between these vectors?
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A similar analysis to the above shows that \(3\) cases can occur:
at least one of them is \(\mathbf{0}\),
two of them are proportional to each other, or
one of them is a linear combination of the other two.
As the examples illustrate, when \(v_1, \cdots ,v_k\) are linearly dependent, then we can write one of the vectors as a linear combination of the others.
If a set of vectors is not linearly dependent they are called linearly independent.
So if the vectors are linearly independent the only way to get \(\mathbf{0}\) as a linear combination is to choose all the coefficients to be \(0\).
What these examples showed is that questions about linear dependence or independence lead to linear systems of equations. So the question of whether a set of vectors is linearly independent is the same as asking whether the corresponding system of equations has a unique solution or not.
Let \(v_1,v_2, \cdots ,v_k\in\mathbb{F}^n\) and let \(A\in M_{n,k}(\mathbb{F})\) be the matrix which has \(v_1,v_2, \cdots ,v_k\) as its columns. Then the vectors \(v_1,v_2, \cdots ,v_k\) are linearly independent if \[S(A,\mathbf{0})=\{\mathbf{0}\}\] and linearly dependent otherwise.
By the definition of \(A\) we have for \(x=(\lambda_1,\lambda_2,\cdots, \lambda_k)\) \[Ax=\lambda_1v_1+\lambda_2v_2+\cdots+\lambda_kv_k\] (this follows from (3.3)). So if \(S(A,\mathbf{0})=\{\mathbf{0}\}\) then \(v_1,v_2, \cdots ,v_k\) are linearly independent, and otherwise are not.
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As a consequence of this result we can use Gaussian elimination to determine if a set of vectors is linearly dependent or linearly independent. We consider the matrix \(A\) whose column vectors are the set of vectors under investigation and apply elementary row operations until it is in row-echelon form. If every column has a leading one the vectors are linearly independent, otherwise they are linearly dependent.
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Over \(\mathbb{C}\), we have that \(v_1=-iv_2\), so they are linearly dependent. However, over \(\mathbb{R}\) they are linearly independent as there are no non-zero real solutions to \(\lambda_1v_1+\lambda_2v_2=\mathbf{0}\).
So this example shows that it is important to know exactly which vector space we are working in when considering linear dependence.
As a consequence of this relation to systems of linear equations we have the following fundamental result.
If \(v_1,v_2, \cdots ,v_k\in \mathbb{F}^n\) over \(\mathbb{F}\) are linearly independent, then \(k\leq n\). So any set of linearly independent vectors in \(\mathbb{F}^n\) over \(\mathbb{F}\) can contain at most \(n\) elements.
Let \(A\) be the \(n\times k\) matrix consisting of the columns \(v_1,v_2, \cdots ,v_k\), then by Theorem 6.8 the vectors are linearly independent if \(S(A,\mathbf{0})=\{\mathbf{0}\}\), but by Corollary 3.42 this gives \(k\leq n\).
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Although we stated Theorem 6.8 and Corollary 3.42 in the real case, both theorems also hold if we are looking at complex matrices and complex systems of equations, meaning this result holds for vectors in \(\mathbb{C}^n\) over \(\mathbb{C}\) (but not for \(\mathbb{C}^n\) over \(\mathbb{R}\) - we will explore the differences between these further in the next section).
In the case where we have \(n\) vectors in \(\mathbb{R}^n\) we can also use our determinant to see whether they are linearly independent, thanks to a theorem that connects the determinant, the invertibility of a matrix, and the linear independence of its columns (or rows).
Let \(A\in M_{n}(\mathbb{F})\), then the following properties are equivalent:
\(\det A\neq 0\),
\(A\) is invertible,
the column vectors of \(A\) are linearly independent over \(\mathbb{F}\).
the row vectors of \(A\) are linearly independent over \(\mathbb{F}\).
Let us first show that (1) implies (3). Let \(a_1,a_2,\cdots ,a_n\) be the column vectors of \(A\). If \(a_1,a_2,\cdots ,a_n\) are linearly dependent, then there exists an \(1\leq i \leq n\) and some \(\lambda_j\in \mathbb{F}\) such that \[a_i=\sum_{j\neq i} \lambda_j a_j ,\] and using linearity in the \(i\)th component we get \[\begin{split} d_n(a_1,\cdots,a_i,\cdots ,a_n)&=d_n\bigg(a_1,\cdots,\sum_{j\neq i} \lambda_j a_j,\cdots ,a_n\bigg)\\ &=\sum_{j\neq i} \lambda_j d_n(a_1, \cdots ,a_j,\cdots ,a_n)=0,\\ \end{split}\] where in the last step we used that there are always at least two equal vectors in the argument of \(d_n(a_1, \cdots ,a_j,\cdots ,a_n)\), so by part (ii) of Proposition 4.4 we get \(0\).
Now we show that (3) implies (2). If the column vectors are linearly independent, then the corresponding homogeneous equation has a unique solution. This means that the reduced row echelon form of \(A\) is the identity matrix, and hence \(A\) is invertible.
Then (2) implies (1) since if \(A^{-1}A=I\) we get by the product formula for determinants that \(\det A^{-1}\det A=1\), hence \(\det A\neq 0\), as we have previously seen in 4.
Finally, we have that (3) and (4) are equivalent since \(\det A = \det A^t\) so if \(\det A \neq 0\) then \(\det A^t\neq 0\), and the columns of \(A^t\) are the rows of \(A\).□
This is one of the most important results about determinants and it is often used when one needs a criterion for invertibility or linear independence.
In particular, note that we can use this for checking linear independence of vectors in \(\mathbb{R}^n\) over \(\mathbb{R}\) and \(\mathbb{C}^n\) over \(\mathbb{C}\).