How to put a matrix in Reduced Echelon Form

Question

5. (6 pts) Find a basis for the following subspace: $$ \left\{ \begin{bmatrix} r - s - 2t - u \\ -r + 2s + 5t + 2u \\ s + 3t + u \\ 3r + s + 5u \end{bmatrix} \in \mathbb{R}^4 : \text{$r, s, t$ and $u$ are scalars} \right\}. $$

Original image here.

So I have this matrix: $$\begin{pmatrix} 1 &-1 &-2 &-1 \\ -1 &2& 5 &2 \\ 0 &1 &3 &1 \\ 3 &1 & 0 & 5 \end{pmatrix} $$ If I compute the the matrix in reduced echelon form, I get: $$\begin{pmatrix} 1&0&1&0\\ 0&1&0&3\\ 0&0&1&-2/3\\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Please note that i want to put the matrix in reduced echelon form , and not row reduced echelon form.

I know that the three rules of reduced echelon form are:

1) The first non zero number of a row is 1 (leading entry).

2) An all zero row is placed at the end of the matrix.

3)A leading entry of a row is placed to the right of a leading entry of the previous row.

If it were to be in row reduced echelon form , then another property would hold, or else that each column that contains a leading entry, all other entries in the same column are zeros.

So is my matrix correctly put in a row echelon form ?

Answer 1

I think you are confused on terminology. Your matrix is correctly put into one of its row echelon forms. Note that these are not unique, as we could add the second row to the first row and the result would still be in row echelon form, that is, \begin{pmatrix} 1&1&1&3\\ 0&1&0&3\\ 0&0&1&-2/3\\ 0 & 0 & 0 & 0 \end{pmatrix} Is still in row echelon form. In order to get to the reduced row echelon form (sometimes called reduced echelon form), we must, as you said, make $0$'s in every column which contains a leading $1$. In your example, we can subtract the third row from the first \begin{pmatrix} 1&0&1&0\\ 0&1&0&3\\ 0&0&1&-2/3\\ 0 & 0 & 0 & 0 \end{pmatrix} to obtain \begin{pmatrix} 1&0&0&2/3\\ 0&1&0&3\\ 0&0&1&-2/3\\ 0 & 0 & 0 & 0 \end{pmatrix} Which is now in reduced row echelon form. Hopefully that clears up any confusion you had.

Answer 2

With reference to the original problem we need to find a basis for

$$\begin{bmatrix} r-s-2t-u \\ -r+2s+5t+2u \\ s+3t+u \\ 3r+s+5u \end{bmatrix}=\begin{pmatrix} 1 &-1 &-2 &-1 \\ -1 &2& 5 &2 \\ 0 &1 &3 &1 \\ 3 &1 & 0 & 5 \end{pmatrix}\begin{pmatrix} r \\ s \\ t \\ u \end{pmatrix}$$

note that the LHS is a vector $\in \mathbb{R^4}$ and in the RHS the given coefficients are arranged by columns, therefore we need to find a basis for the column space of the matrix in on the RHS.

Then we have

$$\begin{pmatrix} 1 &-1 &-2 &-1 \\ -1 &2& 5 &2 \\ 0 &1 &3 &1 \\ 3 &1 & 0 & 5 \end{pmatrix}\to \begin{pmatrix} 1 &-1 &-2 &-1 \\ 0 &1& 3 &1 \\ 0 &1 &3 &1 \\ 3 &1 & 0 & 5 \end{pmatrix}\to \begin{pmatrix} 1 &-1 &-2 &-1 \\ 0 &1& 3 &1 \\ 0 &1 &3 &1 \\ 0 &4 & 6 & 8 \end{pmatrix}\to \begin{pmatrix} 1 &-1 &-2 &-1 \\ 0 &1& 3 &1 \\ 0 &0 &0 &0 \\ 0 &0 & -6 & 4 \end{pmatrix}\to \begin{pmatrix} 1 &-1 &-2 &-1 \\ 0 &1& 3 &1 \\ 0 &0 & -6 & 4\\ 0 &0 &0 &0 \end{pmatrix}$$

which is the RREF. From here we can conclude that a basis is given by the first three vectors of the original matrix that is

$$\{(1,-1,0,3),(-1,2,1,1),(-2,5,3,0)\}$$

Note that we can also continue and obtain

$$\to \begin{pmatrix} 1 &0 &1 &0 \\ 0 &1& 3 &1 \\ 0 &0 & 1 & -\frac23\\ 0 &0 &0 &0 \end{pmatrix}\to \begin{pmatrix} 1 &0 &0 &\frac23 \\ 0 &1& 0 &3 \\ 0 &0 & 1 & -\frac23\\ 0 &0 &0 &0 \end{pmatrix}$$

the latter is the Gauss-Jordan form of the RREF but it is not needed to establish a basis for the given subspace.