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Let $U$ be an $n \times n$ orthogonal matrix. Show that the rows of $U$ form an orthonormal basis of $\mathbb{R}^{n}$ .

This implies that, the rows of $U$ form an orthonormal basis for $\mathbb{R}^{n}$ .

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 2

Orthogonal Sets

Vectors

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Okay, This question gives us an orthogonal matrix, and it wants us to prove that the columns of this matrix form an Ortho normal basis for our end. So we'll start by writing you as this matrix, which is spanned by these vectors u one u two all the way up to U N. And if we're just asked to prove that these vectors form a north a normal basis, we need to prove that they form a basis and that they have magnitude one. And there are Huggel. So first we know that based on definition oven orthogonal matrix you I dotted with you. Jay is equal to zero. If I is not equal to J or you I dotted with you. I is equal to one because we're given that we have orthogonal matrices here. So this condition right here tells us that the dot product of any to different vectors in this matrix would be zero. So, for example, these are just some examples. And if two vectors orthogonal, that means that they do not lie on the same line. So these vectors air all linearly independent. So therefore this'll set is linearly independent. So this forms a basis in orthogonal basis Also, and then we just need to prove that these vectors are also normal. So they have magnitude one. And to do that, the magnitude of use of I would just be you survived dotted with itself or the magnitude squared would just be one. So therefore, the magnitude of each of these vectors is also one. So it's so it's an orthogonal basis and a normal basis. So therefore, it's a North a normal basis, and we're done.

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