Question-86

Consider the set \(\displaystyle S=\{( 1,1,1) ,( -2,1,1) ,( 1,-2,-2)\}\). Which of the following are true?

• The cardinality of \(\displaystyle S\) is the dimension of \(\displaystyle \mathbb{R}^{3}\)
• \(\displaystyle S\) is a linearly dependent set
• \(\displaystyle S\) spans \(\displaystyle \mathbb{R}^{3}\)
• \(\displaystyle S\) spans a plane in \(\displaystyle \mathbb{R}^{3}\)
• \(\displaystyle S\) is a basis of \(\displaystyle \mathbb{R}^{3}\)
• Vectors in \(\displaystyle S\) are pairwise orthogonal

Answer

• The cardinality of \(\displaystyle S\) is the dimension of \(\displaystyle \mathbb{R}^{3}\)
• \(\displaystyle S\) is a linearly dependent set
• \(\displaystyle S\) spans \(\displaystyle \mathbb{R}^{3}\)
• \(\displaystyle S\) spans a plane in \(\displaystyle \mathbb{R}^{3}\)
• \(\displaystyle S\) is a basis of \(\displaystyle \mathbb{R}^{3}\)
• Vectors in \(\displaystyle S\) are pairwise orthogonal

Solution

Let us add the vectors in \(\displaystyle S\) and perform row reduction:

\[ \begin{equation*} \begin{bmatrix} 1 & 1 & 1\\ -2 & 1 & 1\\ 1 & -2 & -2 \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 1 & 1\\ 0 & 3 & 3\\ 0 & -3 & -3 \end{bmatrix} \end{equation*} \]

\[ \begin{equation*} \begin{bmatrix} 1 & 1 & 1\\ 0 & 3 & 3\\ 0 & -3 & -3 \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix} \end{equation*} \]

\[ \begin{equation*} \begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix} \end{equation*} \]

The matrix has rank \(\displaystyle 2\). It follows that \(\displaystyle S\) is linearly dependent and \(\displaystyle \text{span}( S)\) is a plane passing through the origin. \(\displaystyle S\) is not a basis. We can also see that vectors in \(\displaystyle S\) cannot be pair-wise orthogonal (\(\displaystyle S\) is linearly dependent).