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For the column space, you need to look at the columns in the RREF that have leading $1$'s. the column space will be the span of the columns from your original matrix which have a leading $1$ in the RREF (i.e. the column space is the span of columns $1$, $2$, and $3$).
1. You can just show, that rank of matrix A is equal to 2 (you actually did it by reducing A). So if rank is 2 the dimension of a column space is also two. It means that there are two elements in the basis of A column space. And you only need to notice that vectors (1, 4, 3)T and (3, 4, 3)T are linearly independent, so they form a basis of a ...
1. Do row reduction on the transpose of the given matrix to get the row space of the transpose which will inturn give you the column space of A A. Share. Cite. Follow. answered Mar 28, 2017 at 7:45. Parish.
1. This is the formal definition: Let A be an m × n matrix: -The column space (or range) of A,is the set of all linear combinations of the column vectors of A. -The null space of A, denoted by N(A), is the set of all vectors such that Ax = 0. Share. Cite.
Because the matrix is already in row-echelon form: The number of leading $1$'s (three) is the rank; in fact, the columns containing leading $1$'s (i.e., the first, third, and sixth columns) form a basis of the column space. The number of columns not containing leading $1$'s (four) is the dimension of the null space (a.k.a. the nullity).
13. Start with a matrix whose columns are the vectors you have. Then reduce this matrix to row-echelon form. A basis for the columnspace of the original matrix is given by the columns that correspond to the pivots in the row-echelon form. What you are doing does not really make sense because elementary row operations preserve the column space ...
Similarly, the first column must be $-2$ times the second, that is, $4$ times the third. Conveniently for us, the standard basis vectors $(0,0,1,0)^T$ and $(0,0,0,1)^T$ are a valid extension of the kernel basis, so for the third and fourth columns we can take any pair of basis vectors for the image (column space) of the matrix.
0. Finding the nullspace is in my opinion easier, because it amounts to solving an equation. More precisely, since the nullspace is the set of all vectors x x that the operation represented by matrix A A sends to the vector 0 0, then you have the equation : x ∈ N(A) ⇔ Ax = 0 x ∈ N (A) ⇔ A x = 0. Here your example is pretty simple ...
After finding a basis for the row space, by row reduction, so that its dimension was 3, we could have immediately said that the column space had the same dimension, 3, and that the dimension of the null space was 4- 3= 1 without any more computation.
A basis for the column space lets you to completely describe the image of the linear transformation represented by the matrix. Given n> 1 and 0 <k <n, we know that Rn has infinitely many different subspaces of dimension k, so being able to completely describe an image is a great deal more information than just stating its dimension.