2011年3月23日 · The column space and the row space of a matrix is the span of all its columns and the span of all its rows respectively. Is it true that the column space...

2011年11月15日 · Here's the question I'm trying to answer: Suppose a 6 x 8 matrix A has four pivot columns. What is dim Nul A? Is Col A = R4? Why or why not? I know that dim Nul A = 4 because there are 4 columns that are not pivot columns. But I'm unsure about the column space. I know that the four pivot...

2012年4月16日 · 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:

2012年8月27日 · It is the space of all vectors that can be written as a linear combination of vectors in A. Col A exists only if A is a matrix- it is the space spanned by the columns of A thought of as separate vectors.

2011年3月5日 · Hello, i have a problem that asks me to put in reduced row echelon form and I have no clue how to even begin. Let A be a 4X5 matrix. If a1,a2,a4 are linearly independent and a3=a1-3a2 a5=a1-2a2-3a4 determine the reduced row echelon form of A. Any help is greatly appreeciated. Thank You...

2010年3月5日 · That means that another way to think of the "span" of a matrix A is as the "range" of the function f (x)= Ax. And, of course, the kernel is the set of x such that f (x)= Ax= 0. A more common name for the "span" of a matrix is the "column space" or "image" of the matrix.

2011年3月20日 · Would this be a correct reason for why row reduction method gives linearly independent column vectors of a matrix that are basis of column space

2015年3月25日 · I used the column space algorithm to help me determine which vectors in the set are linearly independent. The answer in the back of the book contains vectors that are not even in U, the answer proposed is: Basis = { (-1,5,6) , ( 0,1,-1 ), ( 0,0, 1) } ( did they do a simplification of somesort?). Can someone please check my work and tell me if I'm correct.

2011年3月9日 · Let A be an m x n matrix. Show that the columns of A are linearly independent iff the homogeneous system Ax=0 has just the trivial solution. I think I need to use the fact that a homogeneous system of n linear equations in n unknowns has a nontrivial solution iff rank(A) < n

2013年11月9日 · This n-linear function is an alternating form. This means that whenever two columns of a matrix are identical, or more generally some column can be expressed as a linear combination of the other columns (i.e. the columns of the matrix form a linearly dependent set), its determinant is 0.