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Let $U$ be a square matrix with orthonormal columns. Explain why $U$ is invertible. (Mention the theorems you use.)

So $U^{T} U=I$ implies that $U$ is invertible.

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 2

Orthogonal Sets

Vectors

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So this question says that let you next, you'll be a square matrix with Martin. Um Oh, alter normal columns on. Explain why you is invaluable. So if you is a square metrics with Autonoma columns, um and therefore the columns off you would be auto. Go now, because you with the screamer trips with autonomous columns on if the columns of you are to colonel, there must be linear independence nearly independence according to tear Um, for so the columns The columns of you you are auto Go now, Andi linearly independence. But the question says explain why you is invaluable. So since there Linnean since the columns of your Linnean independence on the metrics square there for the metrics has full rank, I would guess that I just square my tricks with four runs or in vegetable according to the vegetable metric story. Um, so since, um, you can say that there, Caesar, Caesar linen, Independence metrics as for around for one. So we can say that you is impossible because the metrics as a full rank Yeah, or we can just say so. Therefore, you're s about C. U is equal to one on. Therefore, this implies that using possible I'm sorry about this so enough for using vegetable

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