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I understand that when solving for a linear dep/independent matrix, you can take the determinant of the matrix and if it is zero, then it is linearly dependent. However, how can I go about doing this for something like two $3\times 1$ vectors?

Example:

$$\begin{bmatrix} -6 \\ -1 \\ -7 \end{bmatrix}, \begin{bmatrix} -1 \\ -5 \\ 4\end{bmatrix} $$

The RREF yield infinitely many solutions, because R3 0=0. Does this tell me that the system is linearly independent if it has infinitely many solutions?

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  • $\begingroup$ Hint: Two vectors are dependent if and only if they are collinear. $\endgroup$ – AnonSubmitter85 Feb 20 '14 at 9:35
  • $\begingroup$ So, to clarify. Collinear vectors just mean they are scalar multiples of each other, right? $\endgroup$ – KnowledgeGeek Feb 20 '14 at 9:39
  • $\begingroup$ You are correct. $\endgroup$ – AnonSubmitter85 Feb 20 '14 at 9:50
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You should verify the rank of the matrix with the two vector as columns. Take for instance the $2$x$2$ submatrix and verify that has a non null determinant. Thus the rank is two, and it is maximum.

Hence, the vectors you showed are linearly independent.

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Just set the equation $Ax=y$, and look at when there are solutions for $y$. Infinite solutions does not imply dependence, infinite solutions only gives a criteria of where the solution space exists and what the nullity is. You must have full rank in order for the matrix to be non singular. Further if you have more vectors than the dimension of the space, then there exist linear dependent vectors, so look at the dimension. What is the dimension of your space? What is the maximum allowable independent vectors that can exist?

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  • $\begingroup$ so in order to have ndependence each vector of y that corresponds to a pivot MUST have only the zero solution. $\endgroup$ – sophie-germain Feb 20 '14 at 10:08
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Define a matrix $A\in\mathbb{R}^{3\times 2}$ by setting $A:=\begin{bmatrix}-6 & -1 \\ -1 & -5 \\ -7 & 4 \end{bmatrix} $.

By performing the row operations on the matrix $A$, we can show that the null space $\mathcal{N}(A)=\left\{\begin{bmatrix}0 \\ 0\end{bmatrix}\right\}$.

Since the null space of $A$ contains only the zero vector, we have $Ax=0$ if and only if $x=\begin{bmatrix}0 \\ 0 \end{bmatrix}$.

But $Ax$ is the combinations of the column vectors of the matrix $A$, and thus the only combination of the vectors $v_1:=\begin{bmatrix}-6 \\ -1\\-7\end{bmatrix}$, $v_2:=\begin{bmatrix}-1 \\ -5\\4\end{bmatrix}$ which produces the zero vector $\begin{bmatrix}0 \\ 0\\0\end{bmatrix}$ is the trivial combination $(0,0)$.

In other words, we have $\displaystyle\sum_{1\le k\le 2}c_kv_k=0$ if and only if $c_k=0$ for all $1\le k\le 2$. Therefore by the definition of linear independence, the vectors $v_1,v_2$ are linearly independent.

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