Question: Matrices

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TheWrathOfMath

Junior Member
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Are the following matrices an example for two row equivalent matrices with a different column space?

1 1 2 1
1 3 8 0
2 4 10 1
3 5 12 0


1 0 -1 0
0 1 3 0
0 0 0 0
0 0 0 0​
 
TheWrathOfMath said:
Are the following matrices an example for two row equivalent matrices with a different column space?

1 1 2 1
1 3 8 0
2 4 10 1
3 5 12 0


1 0 -1 0
0 1 3 0
0 0 0 0
0 0 0 0​
Click to expand...
What do you think ( justify your answer through definitions) ?
 
1 1 2 1
1 3 8 0
2 4 10 1 = A
3 5 12 0


1 0 -1 0
0 1 3 0
0 0 0 0 = B
0 0 0 0

Col(A)= sp{(1 1 2 3) (1 3 4 5) (2 8 10 12) (1 0 1 0)}
Col(B)=sp{(1 0 0 0) (0 1 0 0) (-1 3 0 0) (0 0 0 0)} = sp{(1 0 0 0) (0 1 0 0) (-1 3 0 0)}

The spans are not equal, therefore Row(A)=/=Row(B).

A and B are row equivalent since we can get from A to B by using a finite number of elementary operations;
A and B have the same matrix in reduced row echelon form, hence they are row equivalent.

A general question: how do I know whether or not two spans are equal to one another?
Is there an easy way to know? Perhaps using systems of linear equations?

In this question it took me a while to check that the spans are not equal.
 
How to you know the span(A) =/= span(B). Did you find a vector that is in span(A) but not in span(B) for example?

I'm confused that you said that the spans are not equal but then you asked A general question: how do I know whether or not two spans are equal to one another?
 
Steven G said:
How to you know the span(A) =/= span(B). Did you find a vector that is in span(A) but not in span(B) for example?

I'm confused that you said that the spans are not equal but then you asked A general question: how do I know whether or not two spans are equal to one another?
Click to expand...
(2 8 10 12) is not in Col(B) since it cannot be written as a linear combination of the vectors in Col(B).
(The fourth element of the vector is 12, while all the fourth elements of the vectors in Col(B) are zero.
 
Steven G said:
How to you know the span(A) =/= span(B). Did you find a vector that is in span(A) but not in span(B) for example?

I'm confused that you said that the spans are not equal but then you asked A general question: how do I know whether or not two spans are equal to one another?
Click to expand...
I meant to ask if there is a quicker way to do it? In order to prove that spans are equal, I need to check if every vector in one of the spans can be written as a linear combination of the other vectors in the other span.
 
TheWrathOfMath said:
I meant to ask if there is a quicker way to do it? In order to prove that spans are equal, I need to check if every vector in one of the spans can be written as a linear combination of the other vectors in the other span.
Click to expand...
No!, not at all. Suppose span(A) = R^3 and span(B) = {(a, b, c) | a, b and c are all real numbers and c=0}.

Now EVERY vector in span(B) can be written as a linear combination of vectors in A. In my language I would say that span(B) is contained (or is a subset) of span(A). Does span(A) = span(B)
 
Last edited:
TheWrathOfMath said:
1 1 2 1
1 3 8 0
2 4 10 1 = A
3 5 12 0


1 0 -1 0
0 1 3 0
0 0 0 0 = B
0 0 0 0

Col(A)= sp{(1 1 2 3) (1 3 4 5) (2 8 10 12) (1 0 1 0)}
Col(B)=sp{(1 0 0 0) (0 1 0 0) (-1 3 0 0) (0 0 0 0)} = sp{(1 0 0 0) (0 1 0 0) (-1 3 0 0)}

The spans are not equal, therefore Row(A)=/=Row(B).

A and B are row equivalent since we can get from A to B by using a finite number of elementary operations;
A and B have the same matrix in reduced row echelon form, hence they are row equivalent.

A general question: how do I know whether or not two spans are equal to one another?
Is there an easy way to know? Perhaps using systems of linear equations?

In this question it took me a while to check that the spans are not equal.
Click to expand...
Just to fix a mistake:

The spans are not equal, therefore Col(A)=/=Col(B).
 
Steven G said:
No!, not at all. Suppose span(A) = R^3 and span(B) = {(a, b, c) | a, b and c are all real numbers and c=0}.

Now EVERY vector in span(B) can be written as a linear combination of vectors in A. In my language I would say that span(B) is contained (or is a subset) of span(A)
Click to expand...
Right. But not every vector in span(A) can be written as a linear combination of span(B)...
 
TheWrathOfMath said:
Right. But not every vector in span(A) can be written as a linear combination of span(B)...
Click to expand...
So your method is not valid. You said In order to prove that spans are equal, I need to check if every vector in one of the spans can be written as a linear combination of the other vectors in the other span.
 
Steven G said:
So your method is not valid. You said In order to prove that spans are equal, I need to check if every vector in one of the spans can be written as a linear combination of the other vectors in the other span.
Click to expand...
Let's label "one of the spans" as A and "the other span" as B.
I should clarify that I will need to check if every vector in A can be written as a linear combination of the other vectors in B and vice versa (if every vector in B can be written as a linear combination of the other vectors in A).
 
Steven G said:
So your method is not valid. You said In order to prove that spans are equal, I need to check if every vector in one of the spans can be written as a linear combination of the other vectors in the other span.
Click to expand...
Perhaps a better way to say it is that I can prove it using double containment?
 
Steven G said:
It is better to say that A is a subset of B AND B is a subset of A. No need to mention linear combinations.
Click to expand...
Now I'll just have to figure out how to prove that X is a subset of Y.
But thank you very much for taking the time to reply.
I highly appreciate your immense help :)
 
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