Linear independence of vectors and rank of a matrix

Question

I have a basic query on the linear independence of vectors. Consider the following matrix $$\begin{bmatrix}0 & 1 & 2 \\1 & 2 & 1 \\2 & 7 & 8 \end{bmatrix}$$

I am told that this matrix has $2$ linearly independent rows - $row$ $1$ and $row$ $2$. $Row$ $3$ is a linear combination of $rows$ $1$ and$ 2$. Specifically, $row$ $3$ = $3$ $(row 1)$ + $2$ $(row 2)$. So the rank of the matrix is $2$.

What I can't understand is how can $rows$ $1$ and $2$ be independent? By the above formula for $row$ $3$, $row$ $1$ can be expressed in terms of row 3 and $row$ $2$. Similarly, $row$ $2 $can be expressed in terms of $row$ $3$ and $row$ $1$.

So, $row$ $1$ and$ row$ $2$ are linearly dependent. Can you please explain where I am wrong?

Thanks, Debashish

Answer 1

$3\cdot \mbox{row}_1 + 2\cdot \mbox{row}_2-1\cdot \mbox{row}_3=0$ says that the three rows are linearly dependent.

Now you can also check that in your case:

$\mbox{row}_1$ and $\mbox{row}_2$ are linearly independent;

$\mbox{row}_2$ and $\mbox{row}_3$ are linearly independent;

$\mbox{row}_1$ and $\mbox{row}_3$ are linearly independent.

The reason is that two rows are linearly dipendent iff the corresponding components are proportional (with the same constant).

Answer 2

Look back at the definition of linear independence. Two vectors $v$ and $w$ are linearly independent if and only if $\lambda v+\mu w= 0$ implies that $\lambda=0=\mu$. Thus $v$ and $w$ are linearly depedent if and only if there exists a couple $(\lambda,\mu)\in \mathbb{R}^2\setminus \left\{(0,0)\right\}$ such that $\lambda v+\mu w=0$. It follows that $v=\frac{\mu}{\lambda}w$ (if $\lambda\neq 0)$ or $w=\frac{\lambda}{\mu}v$ (if $\mu\neq 0)$. From this you can clearly see that the first two rows are independent. Can you spot the error in your reasoning?

Answer 3

We say a set of vectors $\{v_1,\ldots,v_n\}$ is linearly dependent if there exist scalars $a_1,\ldots,a_n$, not all zero, such that

$$a_1v_1+\cdots+a_nv_n=0.$$

A set of vectors is linearly independent if it is not linearly dependent (i.e. if the only scalars that satisfy the equation are $a_1=\cdots a_n=0$).

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Let $v_1$, $v_2$, $v_3$ be the vectors formed by the rows of your matrix.

The set of all three vectors $\{v_1,v_2,v_3\}$ is linearly dependent because

$$3v_1+2v_2-v_3=0$$

This means we can write any one of the three rows as a linear combination of the other two rows.

However the set of vectors formed by just the first two rows, $\{v_1,v_2\}$ is linearly independent because the only scalars $a_1$,$a_2$ such that

$$a_1v_1+a_2v_2=0$$

are $a_1=a_2=0$.