4x4 matrix A is...
1 4 5 -9
-1 -2 -1 3
-2 -4 0 4
0 -4 -8 12
I have reduced it to...
1 4 5 -9
0 2 4 -6
0 0 2 -2
0 0 0 0 ........... So how do i find the row column and null space?
for the null space you are going to have one vector since there is 1 zero row,
first you should reduce the matrix as much as you can because this will make the calculation easier,
1 0 0 0
0 1 0 -1
0 0 1 -1
0 0 0 0
you will have to let x_4 = k, some real number k
then all the x's will be either 0 or k
x1 = 0, x2 = k = x3
so you will get a nullspace of {0,1,1,1}
try following
http://www.math.jhu.edu/~jmb/note/rowcol.pdf
the row and column space are easy,
the row space are just the remaining non-zero rows after reduction to row-ech form (so you should continue the reduction until you have number 1 down the diagonal)
it will be {1 0 0 0}, {0 1 0 -1}, { 0 0 1 -1 }
then for the column space you take each each column of the original matrix corresponding to a pivot (a leading 1 on the diagonal)
so the column space for this matrix will be the first 3 columns of the original matrix,
{1,-1,-2,0}, {4,-2,-4,-4}, {5,-1,0,-8}
How did you go from ..
1 4 5 -9
0 2 4 -6
0 0 2 -2
0 0 0 0
to
1 0 0 0
0 1 0 -1
0 0 1 -1
0 0 0 0
Thank you so much.
and are.. still row spaces? can these 3 be a final answer?
1 4 5 -9
0 2 4 -6
0 0 2 -2
I don't think they are still the row space vectors, they arn't reduced completely so they are still linear combinations of each other.
In the reduced form they are clearly linearly independent vectors which means they are not a constant multiple of one another,
For example you have a zero row because the bottom row was a constant multiple, or linear combination of two other rows.
So you have to reduce the matrix completely,
You should try doing it yourself, it's really important that you can think about what the next step is and practice, especially in exams.
I'll give you a hint,
Take the top row and minus 2 times the second row
After that its pretty smooth
You don't really need to completely row reduce the matrix- although that will generally make the calculations easier.
The "row space" of a matrix is the vector space spanned by its rows. Generally, when you are asked to "find a space", you need to find a basis for the space. So you want to determine a set of independent vectors that still span that space. Having reduced the matrix to echelon form,
$\displaystyle \begin{bmatrix}1 & 4 & -5 & 9 \\ 0 & 2 & 4 & -6\\ 0 & 0 & 2 & -2 \\ 0 & 0 & 0 & 0\end{bmatrix}$
so the row space is the three dimensional space spanned by <1, 4, -5, 9>, <0, 2, 4, -6>, and <0, 0, 2, -2>.
The "column space" of a matrix is the same thing for the columns and the simplest way to do that take the transpose, so that columns become rows, and reduce that. The null space of linear transformation, A, is the space of all vectors, v, such that Av= 0. Since row reduction does not change that, it is sufficient to look at the row reduced matrix:
$\displaystyle \begin{bmatrix}1 & 4 & -5 & 9 \\ 0 & 2 & 4 & -6\\ 0 & 0 & 2 & -2 \\ 0 & 0 & 0 & 0\end{bmatrix}\begin{bmatrix}a \\ b \\ c \\ c\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0\end{bmatrix}$
which is equivalent to the equations a+ 4b- 5c+ 9d= 0, 2b+ 4c- 6d= 0, 2c- 2d= 0. These are three equations in four unknown values so, assuming no dependencies, we would expect to be able to write three of the variables in terms of the fourth. Because of the row-reduction, they are very easy to solve. Here, the last equation gives c= d. Putting that into the third equation givesd 2b+4c- 6c= 2b- 3d= 0 so b= d. Putting that into the first equation gives a+ 4d- 5d+ 9d= a+ 8d= 0 so a= -8d. You can write any vector in the null space as <a, b, c, d>= <-8d, d, d, d>= d<-8, 1, 1, 1>. Now, what is a basis for the null space?