Question-25

Which of the following statements is/are correct?

• \(\mathbb{R}^{n}\) has a unique set of orthonormal basis vectors.

• \(\mathbb{R}^{n}\) does not have a unique set of orthonormal basis vectors.

• Linearly independent vectors in \(\mathbb{R}^{n}\) are orthonormal.

• Orthonormal vectors in \(\mathbb{R}^{n}\) are linearly independent.

NoteAnswer

• \(\mathbb{R}^{n}\) has a unique set of orthonormal basis vectors.

• \(\mathbb{R}^{n}\) does not have a unique set of orthonormal basis vectors.

• Linearly independent vectors in \(\mathbb{R}^{n}\) are orthonormal.

• Orthonormal vectors in \(\mathbb{R}^{n}\) are linearly independent.

NoteSolution

• \(\mathbb{R}^{n}\) has infinitely many orthonormal bases. One non-standard example for \(\mathbb{R}^{2}\) is \(\left\{\frac{1}{\sqrt{2}}( 1,1) ,\frac{1}{\sqrt{2}}( 1,-1)\right\}\).

• Independence doesn’t imply orthogonality. For example, \(\{( 1,1) ,( 2,1)\}\) are independent but not orthogonal.

• Orthonormality implies independence. To show this, let \(\{v_{1} ,\cdots ,v_{m}\}\) be an orthonormal set, then let \(\begin{aligned} c_{1} v_{1} +\cdots +c_{m} v_{m} & =0 \end{aligned}\). Multiplying both sides by \(v_{i}^{T}\), we get \(c_{i} =0\). Since \(c_{1} =\cdots =c_{m} =0\) is the only combination that sends the linear combination to zero, \(\{v_{1} ,\cdots ,v_{m}\}\) is an independent set of vectors.