- #1
twinkerules
- 4
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In my linear algebra class we previously studied how to find a basis and I had no issues with that. Now we are studying the basis of a row space and basis of a column space and I'm struggling to understand the methods being used in the textbook. The textbook uses different methods to find these bases and doesn't explain why or even how to properly do it. The textbook in question is Elementary Linear Algebra by Kolman (9th ed, section 4.9). I've read the chapter at least three times and can't make sense of why they do things certain ways. I decided to attempt some problems and was doing ok until the following two problems. The first problem listed below I was able to solve successfully but the second I got wrong at first and was successful on a second attempt.
Problem number one: Find a basis for the row space of A consisting of vectors that (a) are not necessarily row vectors of A; and (b) are row vectors of A.
A = [tex]
\begin{bmatrix}
1 & 2 & -1\\
1 & 9 & -1\\
-3 & 8 & 3\\
-2 & 3 & 2
\end{bmatrix}
[/tex]
I was able to solve this one successfully by just doing reduced row echelon form (rref) and using any nonzero row as the basis.
Problem number two: Find a basis for the column space of A consisting of vectors that (a) are not necessarily column vectors of A; and (b) are column vectors of A.
A = [tex]
\begin{bmatrix}
1 & -2 & 7 & 0\\
1 & -1 & 4 & 0\\
3 & 2 & -3 & 5\\
2 & 1 & -1 & 3
\end{bmatrix}
[/tex]2. Attempt at a solution.
For this one I tried doing rref but had the wrong answer. Instead, I transposed the matrix given above, did rref, and then got the correct answer if I transposed the columns again as shown below.
[tex]
\begin{bmatrix}
1 & 1 & 3 & 2\\
-2 & -1 & 2 & 1\\
7 & 4 & -3 & -1\\
0 & 0 & 5 & 3
\end{bmatrix} -->rref \begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & \frac{1}{5}\\
0 & 0 & 1 & \frac{3}{5}\\
0 & 0 & 0 & 0
\end{bmatrix} Answer: \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}, \begin{bmatrix} 0\\1\\0\\ \frac{1}{5} \end{bmatrix}, \begin{bmatrix} 0\\0\\1\\ \frac{3}{5} \end{bmatrix}
[/tex]
Can you please explain why the column space problem needs to be transposed before doing rref and why the answer is actually the rows of the matrix which need to be transposed back to columns. I can't find a pattern or answer as to why there is a difference in the procedures being performed. I listed problem one above, even though I got it right, because I don't feel confident that I know why that one was solved the way it was. Can you please provide some contrast between the different procedures being performed and how to know which one to do? I apologize in advance for any formatting errors, I have never used tex prior to this day. Thank you kindly.
Homework Statement
Problem number one: Find a basis for the row space of A consisting of vectors that (a) are not necessarily row vectors of A; and (b) are row vectors of A.
A = [tex]
\begin{bmatrix}
1 & 2 & -1\\
1 & 9 & -1\\
-3 & 8 & 3\\
-2 & 3 & 2
\end{bmatrix}
[/tex]
I was able to solve this one successfully by just doing reduced row echelon form (rref) and using any nonzero row as the basis.
Problem number two: Find a basis for the column space of A consisting of vectors that (a) are not necessarily column vectors of A; and (b) are column vectors of A.
A = [tex]
\begin{bmatrix}
1 & -2 & 7 & 0\\
1 & -1 & 4 & 0\\
3 & 2 & -3 & 5\\
2 & 1 & -1 & 3
\end{bmatrix}
[/tex]2. Attempt at a solution.
For this one I tried doing rref but had the wrong answer. Instead, I transposed the matrix given above, did rref, and then got the correct answer if I transposed the columns again as shown below.
[tex]
\begin{bmatrix}
1 & 1 & 3 & 2\\
-2 & -1 & 2 & 1\\
7 & 4 & -3 & -1\\
0 & 0 & 5 & 3
\end{bmatrix} -->rref \begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & \frac{1}{5}\\
0 & 0 & 1 & \frac{3}{5}\\
0 & 0 & 0 & 0
\end{bmatrix} Answer: \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}, \begin{bmatrix} 0\\1\\0\\ \frac{1}{5} \end{bmatrix}, \begin{bmatrix} 0\\0\\1\\ \frac{3}{5} \end{bmatrix}
[/tex]
Can you please explain why the column space problem needs to be transposed before doing rref and why the answer is actually the rows of the matrix which need to be transposed back to columns. I can't find a pattern or answer as to why there is a difference in the procedures being performed. I listed problem one above, even though I got it right, because I don't feel confident that I know why that one was solved the way it was. Can you please provide some contrast between the different procedures being performed and how to know which one to do? I apologize in advance for any formatting errors, I have never used tex prior to this day. Thank you kindly.