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Suppose $v_1, v_2, v_3$ are (row) vectors in $\mathbb{R}^3$, and they are parallel, then what you can say about the rank of the matrix:

\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}

Note: So this is a $3 \times 3$ matrix with rows $v_1, v_2, v_3$.

The book points out $\text{rank } > 1$, but why is this true?

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    I'm assuming that 'parallel' vectors means that $v_1 = \lambda_1 v_2 = \lambda_2 v_3$. If this is correct, then rank is $\leq 1$. This is just a true thing... – 2017-02-05

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If the vectors are parallel (that is, each $v_i$ is a constant multiple of each $v_j$) then the rank of the matrix is actually $\leq 1$ because the dimension of the row space (the span of the rows) is $\leq 1$. If some $v_i$ is non-zero then the rank will be one. If $v_1 = v_2 = v_3 = 0$ then the rank will be zero.

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    Maybe you wanted to show that the dimension of the kernel is $> 1$. – 2017-02-05
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    thanks! And what is they are not parallel? – 2017-02-05
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    The dimension will be the dimension of the space they span. If they are not parallel, it will be at least $2$. – 2017-02-05