How to Determine if a Vector Set is Linearly Dependent or Independent?

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By the definition, a vector set \((\vec{u}_1, \vec{u}_2,\cdots, \vec{u}_n)\) is linearly independent if the equality

\[k_1 \vec{u}_1+k_2\vec{u}_2+\cdots+k_n\vec{u}_n=0 \]

holds only if all \(k_i=0, 1\le i\le n\). If one of them is not zero, we say that the vector set is linearly dependent.

If we rewrite the equality as a linear system \(A\vec{x}=\vec{0}\), where \(A\) is the matrix that has the vectors as its columns, that is

\[A=\begin{pmatrix}\vec{u}_1&\vec{u}_2&\cdots&\vec{u}_n\end{pmatrix}, \qquad \vec{x}=\begin{pmatrix}k_1\\k_2\\ \vdots\\ k_n\end{pmatrix}.\]

Then the vector set is linearly independent is equivalent to the system \(A\vec{x}=\vec{0}\) has only trivial solution. And hence we have some equivalent conditions

• \(\vec{u}_1, \vec{u}_2,\cdots, \vec{u}_n \) are linearly independent
• \(A\vec{x}=0\) has only trivial solution
• \(A\vec{x}=\vec{b}\) has unique solution
• \(\text{det} A=|A|\ne 0\)
• \(\text{Rank}A=n\)

and some other conditions. From all of those conditions we know that the most efficient way to determine if the vector set is linearly dependent or independent is using the rank. Because using rank, we only need to reduce the matrix to be the row echelon form and no other computation is needed. We don’t need to solve the system, we even don’t need to reduce the matrix to be reduced echelon form. This method need the least computation so we recommend it.

Now let us look an example.

Example: Determine if the vectors are linearly dependent or independent:

\[(1) \begin{pmatrix}-1\\3\\1\end{pmatrix}, \begin{pmatrix}2\\1\\0\end{pmatrix}, \begin{pmatrix}1\\4\\1\end{pmatrix}\qquad (2) \begin{pmatrix}2\\3\\0\end{pmatrix}, \begin{pmatrix}-1\\4\\0\end{pmatrix}, \begin{pmatrix}0\\0\\2\end{pmatrix} . \]

Solution: (1) Let

\[ A=\begin{pmatrix}-1&2&1\\3&1&4\\1&0&1\end{pmatrix} \]

Using row reduction for \(A\), we have

\[ A=\begin{pmatrix}-1&2&1\\3&1&4\\1&0&1\end{pmatrix}\sim \begin{pmatrix} 1&0&1 \\3&1&4\\ -1&2&1 \end{pmatrix}\sim \begin{pmatrix} 1&0&1 \\0&1&1\\ 0&2&2 \end{pmatrix}\sim \begin{pmatrix} 1&0&1 \\0&1&1\\ 0&0&0 \end{pmatrix} \]

The row echelon form has only 2 nonzero rows, Hence \(\text{Rank}A=2<3\). So the vectors are linearly dependent.

(2) Let \[A= \begin{pmatrix} 2&-1&0 \\3&4&0\\ 0&0&2 \end{pmatrix}\sim \begin{pmatrix} 2&-1&0 \\0&\frac{11}{2}&0\\ 0&0&2 \end{pmatrix} \]

The row echelon form has 3 nonzero rows. \(\text{Rank}A=3\). So the vectors are linearly independent.