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Find a matrix v =(1,2,3) in the row space and column space. Find another matrix with v in the nullspace and column space. which pairs of subspaces can v not be in ?

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by sweetysiri92

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by sweetysiri92

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1)

We have to find a matrix A of size 3 X 3 ( for example ) which has the vector v = [1,2,3] in the row space as well as column space.

So choose v as the first row. Next, write transpose of v in the 1st column. Then write 2nd row as a multiple of 1st row. Then write 3rd row so that it is not dependent on 1st row. After that perform row reduction operations to get a echelon form.

Null space of A = [ 2,4,6] as it is linearly dependent on 1st and 3rd rows.

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2) Find a matrix with v = [1,2,3] in null space and column space.

Write v as the first column. Write v as the first row as well. But it has to be dependent on 2nd and/or third row, so that it becomes O during row reduction operation. write the last two elements in the matrix in third row and perform row reduction in such a way that a column other than 1 becomes dependent.

Null space of B = { [1,2,3] }

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3)

There are four subspaces. Column space, row space, null space and left null space. Null space is the set of rows in A which are linearly dependent on others. Left null space is the null space of Matrix A^T (transpose of A).

A vector can not possibly be in Column space and left null space.

A vector can not be possibly in row space and null space.

Because these pairs are complementary to each other.

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We can also write remaining two elements in matrix A as x and y, after writing v as first row and column and a multiple of v as 2nd row. Then find their values so that one desired column or row becomes linearly dependent or independent.

We have to find a matrix A of size 3 X 3 ( for example ) which has the vector v = [1,2,3] in the row space as well as column space.

So choose v as the first row. Next, write transpose of v in the 1st column. Then write 2nd row as a multiple of 1st row. Then write 3rd row so that it is not dependent on 1st row. After that perform row reduction operations to get a echelon form.

Null space of A = [ 2,4,6] as it is linearly dependent on 1st and 3rd rows.

=========================================

2) Find a matrix with v = [1,2,3] in null space and column space.

Write v as the first column. Write v as the first row as well. But it has to be dependent on 2nd and/or third row, so that it becomes O during row reduction operation. write the last two elements in the matrix in third row and perform row reduction in such a way that a column other than 1 becomes dependent.

Null space of B = { [1,2,3] }

=====================

3)

There are four subspaces. Column space, row space, null space and left null space. Null space is the set of rows in A which are linearly dependent on others. Left null space is the null space of Matrix A^T (transpose of A).

A vector can not possibly be in Column space and left null space.

A vector can not be possibly in row space and null space.

Because these pairs are complementary to each other.

===========================

We can also write remaining two elements in matrix A as x and y, after writing v as first row and column and a multiple of v as 2nd row. Then find their values so that one desired column or row becomes linearly dependent or independent.