When we find a basis for a span, why we should transfer the matrix into RREF first?
What's the relationship between basis and RREF?
I have an example in the attachment. Problem 2.(a)
Assignment 4.pdfAssignment 4.pdf
When we find a basis for a span, why we should transfer the matrix into RREF first?
What's the relationship between basis and RREF?
I have an example in the attachment. Problem 2.(a)
Assignment 4.pdfAssignment 4.pdf
You don't need to transform to a matrix! I assume you know the definition of "basis of a vector space", the vector space here, being the span of the given vectors.
A basis must both span the space and be linearly independent. Here we are given that the vectors span the space so we want to find a subset that is linearly independent To do that, start from $\displaystyle a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$ where the "a"s are numbers and the "v"s are the vectors. $\displaystyle a_1= a_2= \cdot\cdot\cdot= a_n= 0$ is an obvious solution. If it is the only solution, then the set of vectors is independent and is a basis. If not, then at least on of the "a" is not 0. Suppose it is $\displaystyle a_i$. Then we can rewrite $\displaystyle a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_iv_i+ \cdot+ \cdot+ \cdot+ a_nv_n= 0$ as $\displaystyle a_iv_i= -a_1v_1- a_2v_2- \cdot\cdot\cdot- \cdot\cdot\cdot- a_nv_n$ and then $\displaystyle v_i= -(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)/a_i$. Since $\displaystyle v_i$ can be written as a linear combination of the other vectors, you can drop $\displaystyle v_i$ and still have the same span. Now check to see if this smaller set is independent. If not, repeat the process.
How I would do this: We are given $\displaystyle \begin{bmatrix}2\\ 3\\ 1\end{bmatrix}$, $\displaystyle \begin{bmatrix}-1 \\ 2 \\ 3 \end{bmatrix}$, and $\displaystyle \begin{bmatrix}0 \\ 7 \\ 7 \end{bmatrix}$ so to check for independence we should look at $\displaystyle a\begin{bmatrix}2\\ 3\\ 1\end{bmatrix}+ b\begin{bmatrix}-1 \\ 2 \\ 3 \end{bmatrix}+ c\begin{bmatrix}0 \\ 7 \\ 7 \end{bmatrix}= \begin{bmatrix}2a- b \\ 3a+ 2b+ 7c \\ a+ 3b+ 7c\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$.
That is equivalent to the three equations 2a- b= 0, 3a+ 2b+ 7c= 0, and a+ 3b+ 7c= 0. From the first equation b= 2a. Putting that into the other two equations, 3a+ 2(2a)+ 7c= 7a+ 7c= 0 and a+ 3(2a)+ 7c= 7a+ 7c= 0. Those are the same equation and are equivalent to c= -a. So we have $\displaystyle a\begin{bmatrix}2\\ 3\\ 1\end{bmatrix}+ 2a\begin{bmatrix}-1 \\ 2 \\ 3 \end{bmatrix}- a\begin{bmatrix}0 \\ 7 \\ 7 \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$ for any a. Therefore the vectors are dependent and cannot be a basis for a three dimensional vector space.
If I had wanted to use matrices to solve those equations, the "matrix of coefficients" would be $\displaystyle \begin{bmatrix}2 & -1 & 0 \\ 3 & 2 & 7 \\ 0 & 7 & 7 \end{bmatrix}$, with each of the given vectors as a column. "Row reducing, I would divide the first row by 2 (to get "1" in the first place) then subtract 3 times the new first row from the second and subtract the first row from the third (to get "0"s below that 2). That gives $\displaystyle \begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 0 & \frac{7}{2} & 7 \\ 0 & \frac{7}{2} & 7\end{bmatrix}$. Notice that the last two rows are the same. If I subtract the second row from the third, multiply the second row by 2/7, then add 1/2 the new second row from the first, I get $\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}$. That is the reduced matrix your teacher had. The fact that the last row is all "0"s tells us that the vectors are dependent.