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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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  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.

A= \left[\begin{array}{ccc}1&2&3\\2&4&6\\3&0&5\end{array}\right] =>\left[\begin{array}{ccc}1&2&3\\0&0&0\\3&0&5\end{array}\right]=\left[\begin{array}{ccc}0&2&4/3\\0&0&0\\3&0&5\end{array}\right]\\\\Row\ Space\ of\ A=\{ [1,2,3],[3,0,5]\},\ \ or\ \ \{[0,2,4/3 ], [3,0,5 ] \}\\\\Column\ 3\ in\ row\ reduced\ Echelon\ form=v3=v2*2/3+v1*5/3.\\\\Column\ space\ of\ A=\{ [1,2,3 ], [2,4,0 ] \}

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.

B= \left[\begin{array}{ccc}1&2&3\\2&4&6\\3&1&0\end{array}\right]=\left[\begin{array}{ccc}0&0&0\\2&4&6\\3&1&0\end{array}\right]=\left[\begin{array}{ccc}0&0&0\\0&10/3&6\\3&1&0\end{array}\right]\\\\Column2=v2=10/18*v3+3*v1\\\\Hence,\ column\ space=\{ [1,2,3 ]^T, [3,6,0 ]^T \}\\

Null space of B = { [1,2,3] }  

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. 

2 5 2
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