When we write the vectors $(1,1,0)$ and $(0,1,1)$ in matrix form and obtain its row reduced echelon form we get $\left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\0 & 0\\ \end{array} } \right]$; which has two pivots and a zero row. Does it mean it is linearly dependent in $\mathbb{R}^2$ because of the zero row or it linearly independent in $\mathbb{R}^2$ because of two pivots?

According to Axler's Linear Algebra Done Right :


Thus, can we conclude that $(1,1,0)$ and $(0,1,1)$ are linearly independent in $\mathbb{R}^3$? Can we show it in echelon form with three pivots if they are linearly independent?

Thanks in advance for clearing my confusion..Cheers!


$(1,1,0)$ and $(0,1,0)$ are not vectors in $\mathbb R^2$, since they have three components, so you cannot discuss their linear independence in $\mathbb R^2$. Now, in $\mathbb R^3$, take $a\cdot (1,1,0)+b\cdot (0,1,0)=(0,0,0)\Rightarrow \begin{cases} a\cdot 1+b\cdot 0=0\\ a\cdot 1+b\cdot 1=0\\ a\cdot 0+b\cdot 0=0\end{cases}\Rightarrow a=0,\ b=0$. Therefore your vectors are linearly independent.

Working with matrices, your vectors are linearly independent if the matrix constructed with their components has rank equal to the number of vectors. The rank of your matrix is $2$, therefore the vectors are linearly independent.


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