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Non-Properties: (things we’d expect to be true, but aren’t)
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• A2 = 0, with A not equal to a zero matrix
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• AC = BC even though A ̸= B and C not a zero matrix
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This is to be
expected considering the complexity of the elements
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



2
−1
0
 −3 
 1 
 1





 0  x1 +  2  x2 +  −2
1
0
−1






1



 x3 =  2 
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Matrix Derived Spaces
These are spaces, all subspaces/vector spaces, derived from a matrix
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We will look into three, how to find bases for them, etc
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The last is the easiest,
but will take some explanation
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Col(A) is a span, so it’s a subspace
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, an } = {Ax, for all x ∈ Rm } ⊆ Rn
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So, you can view it as a span, or as the ‘output’ of the operator
x → Ax
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We discussed this earlier
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You have to use the
ORIGINAL columns, but the row reduction decides which are necessary to the span (so, the
minimal spanning set)
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Proof:
Fairly simple, and mostly involves reiterating stuff we’ve already seen
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Removing the columns without pivots will not affect the span,
since for any y where you have a solution you’d also have a solution with all the free variables set to zero
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Solving [B|0] will
result only in pivot columns on the left, so the trivial solution, and the columns of B form
a linearly independent spanning set and so a basis
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The row reduction changes their shape
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Example:



−1 −1
0
2
2
1 −1 −1 −1 
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

2
0
1
0 0 0
1
2
And now for another space
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Definition: The Null Space of A is the set
Null(A) = {x such that Ax = 0} ⊆ Rm
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The order of the vectors is essential, however
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We just want the
parametric vector form of the solution of Ax = 0
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Example: Find a basis for Null(A) using the A from before
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It
further reduces to

 
 
 
x1
t+s
1
1


 x2   −t − 3s   −1   −3
1 0 −1 0 −1

 
 
 
 0 1
1 0
3  so the Null space is =  x3  = 
t  =  1  t+ 0

 
 
 
 x4  
0 0
0 1
2
−2s   0   −2
x5
s
0
1


 
1 
1





 −1   −3 


 
so the basis is  1  ,  0 
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You can prove it using the
subspace test if you like
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Its Column Space has
dimension 3 and its Null space has dimension 2
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This is to
be expected: the dimension of the column space is the number of pivots, the dimension of
the null space is the number of free variables, which is the pivot-less rows
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As such,
the dimensions of the column and null spaces sum up to the number of columns in A
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This one is somewhat different from the others
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This is fairly easy to see
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Also, if a given row gets removed during row
reduction (i
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, made into a zero row) that means it’s a linear combination of the remaining
non-zero rows, and so does not contribute to the span
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They are a spanning set (being equal in span to the
originals) and are quite trivially linearly independent
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Notice that we use the rows directly from the reduced form
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The interchange operation can move them around
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Definition: The Rank of A, with A an n × m matrix, is:
Rank(A) = number of pivots of A
= Dim(Col(A))
= Dim(Row(A))
= m − Dim(Null(A))
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First, if a matrix A, n × m, has Rank(A) = n then no problem
[A|b] will have a pivot on the right hand side
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At least
a solution
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Recall that if Null(A) has zero dimension (no free variables) then Null(A) has precisely
ONE element (0)
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Exercises
Section 1
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bd)
Section 1
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bdfh) Note: Coefficient matrix means Left Hand Side, so the coefficient
matrix is the matrix given
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bd), 6,
7
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4: 1
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b), 5
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bdf), 12, 16

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