Can matrices with dependent columns being QR factorization?

Question

The problem comes from the $18.06$ Linear Algebra by MIT Open Courseware.

The answer:

I am very confused. According to the definition,

Matrix A -> QR means that A has independent columns. BUT it is obviously that the matrix B is singular in the problem. But it can be diagnoalized with 3 independent eigenvectors. How could that happen? Could you explain ?

Thanks!

I got this exact same question at the exact same problem set & same step. — Gintas_, Jan 19 '23 at 13:47

Siong Thye Goh · Accepted Answer · 2016-05-25T05:50:50.263

6

If $A$ is singular, it can still exhibit a $QR$ decomposition, the trade off is $R$ is singular as well. Some of the diagonal entries are $0$.

Also, any symmetrical matrices can be diagonalized. Being singular means that some of the eigenvalues are $0$.

edited May 25 '16 at 05:50

answered May 25 '16 at 05:42

Siong Thye Goh

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Er. So any matrix can exhibit QR factorization? There is no requirement in QR factorization? – stander Qiu May 25 '16 at 05:53
Yes, any square matrices exhibit QR factorization. – Siong Thye Goh May 25 '16 at 06:06

Can matrices with dependent columns being QR factorization?

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