Matrix which reverses Gram-Schmidt - Linear Algebra

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SUMMARY

The discussion focuses on the Gram-Schmidt process in linear algebra, specifically the challenge of reversing the transformation to express orthogonal vectors (q vectors) in terms of the original linearly independent vectors (a vectors). The user proposes that if vectors a1, a2, and a3 are linearly independent, then the orthogonal vectors should directly correspond to the original vectors, i.e., q1 = a1, q2 = a2, and q3 = a3. The complexity arises from the requirement that the orthogonal vectors must also be of unit length, complicating the transformation process.

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  • Understanding of the Gram-Schmidt process in linear algebra
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  • Familiarity with vector normalization techniques
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Homework Statement
Let a1, a2, a3 be linearly independent vectors in R 3 , and let q1, q2, q3 be the vectors obtained from a1, a2, a3 by the Gram-Schmidt algorithm.

Define the linear transformation T : R 3 → R 3 by T(q1) = a1, T(q2) = a2 and T(q3) = a3.

For the choice of basis {q1, q2, q3} both for the input and output spaces, find the matrix MT which represents the linear transformation T for this choice of basis.
Relevant Equations
projection equation used in Gram-Schmidt: p=((a^t)(b))/((a^t)(a)) (a)
My idea was to write out the formulas for the orthogonal q vectors in terms of the input vectors using the basics of gram-schmidt. Then, I would rewrite those equations suhc that the a vectors were written in terms of the q vectors. And then, try to find some matrix which would capture the needed transformation. However, the q vectors must be written in terms of multiple a vectors, as well as other q vectors. Additionally, the need to make all of the orthogonal vecotrs of unit length makes it even more complicated. There must be another way.
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If a1-3 are linearly independent vectors, shouldn't q1 = a1, q2 = a2, q3 = a3?

Because Gram-Schmidt method is about substracting the part of your vector that's represented in previous vectors. But there are no such parts since they are linearly independent.
 

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