Alg1231W2T4 - Column space of a matrix

Wednesday, 25 July 2018

8:46 PM

Machine generated alternative text:
For the matrix 
consider the set 
An element of S has the form A 
3 
-4 
1 
1 
-1 
2 
-8 
10 
-6 
s 
Ax for some 
y 
, which can be written as the linear combination of vectors 
So indeed S deserves name of the column space 
O of A, as it consists of all linear combinations ot columns 
Recall: the Maple notation for a vector in R is< a,b,c >. 
O of A.

 

For the matrix 
the set 
is the column space of A. The vector v 
has [select all that apply] 
a unique solution 
infinitely many solutions 
Cl no solution. 
A 
2 
2 
2 
-3 
-1 
3 
2 
3 
5 
S 
b e R : b = Ax for some 
y 
belongs to this set whenever the augmented matrix 
2 
2 
2 
-3 
-1 
3 
By row reducing, we find that saying v e S is equivalent to and z satisfying a linear condition of the form ax + by + cz 
In this case the column space of A is a plane tnrougn tne origin 
Note: 
• the Maple syntax for the equation ax + by + cz = 0 is 
• make sure to explicitly specify multiplication using an asterisk * 
0 _ Specifically, this condition is equivalent to

 

Machine generated alternative text:
For the matrix 
the set 
is the column space of A. The vector v 
has [select all that apply] 
a unique solution 
infinitely many solutions 
Cl no solutions 
2 
8 
4 
4 
16 
8 
3 
12 
6 
S 
b e R : b = Ax for some 
xeR3} 
y 
belongs to this set whenever the augmented matrix 
2 
8 
4 
{Z-æx=o, 
4 
16 
8 
12 y 
There are ftvo linear conditions that must be satisfied; namely: 
Note: enter the conditions above in the form of a set 
In this case the column space of A is a 
line 
(e.g. 
O with parametric vector form x = Ab where 
Recall: the Maple notation for a vector in IR3 is< a,b,c

 

Machine generated alternative text:
For the matrix 
consider the set 
s 
: b = Ax 
3 
Let's show that S is indeed a subspace of R 
1 
-4 
1 
So a + b e S 
tor some 
4 
-2 
2 
-6 
10 
-4 
-1 
4 
-1 
for some 
i) S contains the zero-vector. Since 
Recall: the Maple syntax for a vector in R 
ii) S is closed under addition If a, b e S 
Ax where 
is< a,b,c,d 
then: 
• a = Ax tor some 
. b = Ay 
tor some 
O x e IR and 
4 
O y €zlR 
Thus a + b = Ax + Ay = Az where z 
iii) S is closed under scalar multiplication Suppose b e S. 
and so, if * e R , then = M Ax) = Az where z 
o e IR4 
Then b = Ax 
Rx 
xeR4} 
4 
O x e IR, 
O e IR and so Abe S. 
3 
Thus; using the Subspace Theorem; we can conclude that S is a vector subspace of R

 

Machine generated alternative text:
For the matrix 
consider the set 
A 
4 
-3 
4 
2 
-1 
4 
The vector 
The vector 
The vector 
4 
-3 
4 
2 
-1 
4 
6 
-4 
8 
e S since 
e S since 
e S since 
4 
-3 
4 
2 
-1 
4 
6 
-4 
8 
s 
Ax where 
Ay where 
Az where 
Ax for some 
xeR2} 
Recall: the Maple syntax for a vector in R 
is< a,b

 

 

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