Find a Basis for the Solution Space of a
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Definition : A Subspace of 
�is any set "H" that contains the zero vector; is closed under vector addition; and is closed under scalar multiplication.
Definition : The Column Space of a matrix "A" is the set "Col A "of all linear combinations of the columns of "A".
Definition : The Null Space of a matrix "A" is the set
�"Nul A" of all solutions to the equation 
.
Definition : A basis for a subspace "H" of �is a linearly independent set in 'H" that spans "H".
Example 1 : Determine if "w" is in the subspace of 
�spanned by 
� and 
.
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The vector "w" is NOT in the subspace because "w" can not be constructed from a linear combination of the spanning set of vectors.
Example 2 : Find bases for both "Col A" & "Nul A", and determine if the vector "p" is in either space.
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The equation 
�has a solution so "p" is in "Col A".
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Only the first two columns of "A" are pivot columns. Therefore, a basis for "Col A" is the set { 
, 
} of the first two columns of "A".
To find a basis for "Nul A", solve .
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Thus, the vector: 
�is a basis for "Nul A".
Evidently, "p" is NOT in "Nul A". Otherwise, it would be a scalar multiple of the vector "n".
Alternatively, "p" must satisfy the equation 
�to be in "Nul A".
There are 4 basic subspaces associated with the
mxn martix "A". These subspaces lie in � or 
. Two of these subspaces: Col (A), and Nul ( 
) lie in . The other pair: Nul ( 
) and Row (A) lie in . Subspace pairs are Orthogonal Complements . That is, al vectors lying in one are orthogonal or perpendicular to all vectors in the other.
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Example 3 : Let "A" be an mxn matrix. Show that Nul (A) is the orthogonal complement of Row (A).
Therefore, every row of "A" is perpendicular or orthogonal to every vector in the null space of "A". Since rows of "A" span "row space", Nul (A) must be the orthogonal complement of Row (A).
Example4 : Let "A" be an mxn matrix. Show that Col (A) is the orthogonal complement of 
.
The rows of �are the columns of A. Thus, since "y" is in , the columns of A are orthogonal to all vectors lying in . Since columns of "A" span "column space", Col (A) must be the orthogonal complement of .
All of �and all of� are �thus each neatly separated into 2 orthogonal subspaces.
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Suppose we need a minimum of "r" vectors to span
Col (A), then dim[Col (A)] = r. Since Col (A) and
constitute all of , then dim[ ] = m-r. If we need "r" column vectors to span Col (A), we also need "r" vectors to span Row (A). Thus, dim[Row (A)]=r and therefore, dim[Null (A)] = n-r.
Example5 : Let 
. Use this matrix to exemplify the concepts of orthogonal subspace pairs.
Span Col (A): { 
}.���� dim[Col (A)] = 2
Span Nul (A): { 
}. �� dim[ Nul (A)] = 2
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Span Row (A): { 
}.���� � dim[Row (A)] = 2
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Span Nul ( ): { 
}.����� � dim[ Nul ( )] = 1
Col (A) is orthogonal to Nul ( ) and Row (A) is orthogonal to Nul ( ), which can be confirmed by showing that the vectors that span each are perpendicular to one another.
Finally, dim[Col (A)] + dim[Nul ( )] = 3 =m and dim[Row(A)] + dim[Row(A)] = 4, which checks because matrix "A" has dimensions: 3x4.
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Find a Basis for the Solution Space of a
Source: https://www.math.drexel.edu/~jwd25/LM_SPRING_07/lectures/Lecture4A.html