1. ## basis and image

Find the basis for the image of the matrix B

B = 4 2 2 0
0 3 -1 1
1 -3 4 2

So first i transposed it B'

4 0 1
2 3 -3
2 -1 4
0 1 2

then i reduced it to echelon form
divide top row by 4

1 0 1/4
2 3 -3
2 -1 4
0 1 2

row 2 - 2 x row1
row 3 - 2 x row 1

1 0 1/4
0 3 -7/2
0 -1 7/2
0 1 2

1 0 1/4
0 1 -7/6
0 1 -7/2
0 1 2

row 3 - row 2
row 4 - row 2

1 0 1/4
0 1 -7/6
0 0 -14/6
0 0 19/6

1 0 1/4
0 1 -7/6
0 0 1
0 0 1

row 4 - 3

1 0 1/4
0 1 -7/6
0 0 1

So these are the basis, is this right? Have i done the ecehlon form right?

2. What happened to the last row? But what you have done is correct, so far. If you were to complete the row reduction you would have, as reduced echelon form,
$\displaystyle \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 0 & 0\end{array}\right]$
so that your basis vectors are <1, 0 , 0>, <0, 1, 0>, and <0, 0, 1>, the standard basis for $\displaystyle R^3$. That is because this matrix maps $\displaystyle R^4$ into $\displaystyle R^3$ and its null space has dimension 1 so its image is all of $\displaystyle R^3$.

3. Originally Posted by HallsofIvy
What happened to the last row? But what you have done is correct, so far. If you were to complete the row reduction you would have, as reduced echelon form,
$\displaystyle \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 0 & 0\end{array}\right]$
so that your basis vectors are <1, 0 , 0>, <0, 1, 0>, and <0, 0, 1>, the standard basis for $\displaystyle R^3$. That is because this matrix maps $\displaystyle R^4$ into $\displaystyle R^3$ and its null space has dimension 1 so its image is all of $\displaystyle R^3$.
I thought it was just echelon form you needed to put in, not row reduction form.

4. I'm not sure what you mean by "row reduction form". "Row reduction" is the method you use to reduce a matrix to "echelon" or "reduced echelon" form.

In any case, either "echelon form" or "reduced echelon form" (which is probably what you mean by "row reduction form") of an n by n matrix is still an n by n matrix.

5. Originally Posted by HallsofIvy
I'm not sure what you mean by "row reduction form". "Row reduction" is the method you use to reduce a matrix to "echelon" or "reduced echelon" form.

In any case, either "echelon form" or "reduced echelon form" (which is probably what you mean by "row reduction form") of an n by n matrix is still an n by n matrix.
Yeah sorry i thought it just needed to be in echelon form not reduced echelon form.

6. It doesn't have to be. That was not my point. The "echelon form" would be
$\displaystyle \begin{bmatrix}1 & 0 & \frac{1}{4} \\ 0 & 1 & -\frac{7}{6} \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$
which still has all four rows. Of course the 0 vector is never a member of a basis so a basis for the image is
$\displaystyle \{\begin{bmatrix}1 \\ 0 \\ \frac{1}{4}\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ -\frac{7}{6}\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\}$.

7. Originally Posted by HallsofIvy
It doesn't have to be. That was not my point. The "echelon form" would be
$\displaystyle \begin{bmatrix}1 & 0 & \frac{1}{4} \\ 0 & 1 & -\frac{7}{6} \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$
which still has all four rows. Of course the 0 vector is never a member of a basis so a basis for the image is
$\displaystyle \{\begin{bmatrix}1 \\ 0 \\ \frac{1}{4}\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ -\frac{7}{6}\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\}$.
Yeah i see now thanks