In a question I am given this information:

$$ U = \begin {bmatrix} 1 & 2 & 4\\ 0 & 1 & 1\\ 0 & 0 & 3\\ \end {bmatrix} $$

$$L \mathbf y = \begin {bmatrix} 3\\ 13\\ 4\\ \end {bmatrix} $$

$$\mathbf y = \begin {bmatrix} 2\\ 2\\ 3\\ \end {bmatrix} $$

The question asks to solve the the solution $ A \mathbf x = \mathbf b$. I got the values : $\mathbf x = \begin {bmatrix} -2 \\ 0 \\ 1\\ \end {bmatrix}$.

I need to find a basis for the $col(A)$ and the $row(A)$.

To find the basis for the row space I can simply use the rows of the U matrix, however, to find the column space I need A. Is there a way to find A or am I missing some theory that allows me to find the column space with the information I am given.

**Question is from UofT MAT223 December 2017 Final Exam (cite given for anti -plagarism purposes)


Hmm, let's think. If the matrix has an LU factorization, then it is of full rank. So, the column space of $ A $ is ...

  • $\begingroup$ The column space of A would be the 3 standard basis? $\endgroup$ – R83nLK82 Apr 12 '18 at 2:07
  • $\begingroup$ No, that would a basis for the column space. The column space is a space. Not a set of vectors. $\endgroup$ – Robert van de Geijn Apr 15 '18 at 18:24

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