Mixing
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How to find $dim( operatorname{span}(M)+ operatorname{span}(N))$ and $dim(operatorname{span}(Mcap N))$?
$begingroup$
$M={(1,2,1),(3,1,5),(3,-4,7)}$, $N={(1,0,-1),(1,2,1)}$
Let $A= operatorname{span}(M)$ and $B= operatorname{span}(N)$
I want to know how to find $;(a);dim(A+B)$ and $(b);dim(Acap B)$
begin{bmatrix}
1 & 3 & 3 \
2 & 1 & -4 \
1 & 5 & 7
end{bmatrix}$ sim$
begin{bmatrix}
1 & 3 & 3 \
0 & 1 & 2 \
0 & 0 & 0
end{bmatrix}
Since, only the first two vectors are linearly independent, we have $dim(A)=2$
Clearly, $dim(B)=2$ since it is a linearly independent set(They are not multiple of each other).
I want to make use of:
$dim(A+B)= dim(A)+dim(B)-dim(Acap B)$
So how do I find either of $dim(A+B)$ or $dim(Acap B)$ to get the other ?
Does making a single matrix with the vectors in $M$ and $N$ and calculating its rank help?
begin{bmatrix}
1 & 3 & 3 & 1 &1\
2 & 1 & -4&0 &2 \
1 & 5 & 7 &-1&1
end{bmatrix}
$sim$
begin{bmatrix}
1 & 3 & 3 & 1 &1\
0 & -5 & -10&-2 &0 \
0 & 0 & 0 &14&0
end{bmatrix}
So this gives a rank of 3
linear-algebra vector-spaces
asked Jan 27 at 11:36
AbhayAbhay
3789
$endgroup$
|
show 8 more comments
$begingroup$
$M={(1,2,1),(3,1,5),(3,-4,7)}$, $N={(1,0,-1),(1,2,1)}$
Let $A= operatorname{span}(M)$ and $B= operatorname{span}(N)$
I want to know how to find $;(a);dim(A+B)$ and $(b);dim(Acap B)$
begin{bmatrix}
1 & 3 & 3 \
2 & 1 & -4 \
1 & 5 & 7
end{bmatrix}$ sim$
begin{bmatrix}
1 & 3 & 3 \
0 & 1 & 2 \
0 & 0 & 0
end{bmatrix}
Since, only the first two vectors are linearly independent, we have $dim(A)=2$
Clearly, $dim(B)=2$ since it is a linearly independent set(They are not multiple of each other).
I want to make use of:
$dim(A+B)= dim(A)+dim(B)-dim(Acap B)$
So how do I find either of $dim(A+B)$ or $dim(Acap B)$ to get the other ?
Does making a single matrix with the vectors in $M$ and $N$ and calculating its rank help?
begin{bmatrix}
1 & 3 & 3 & 1 &1\
2 & 1 & -4&0 &2 \
1 & 5 & 7 &-1&1
end{bmatrix}
$sim$
begin{bmatrix}
1 & 3 & 3 & 1 &1\
0 & -5 & -10&-2 &0 \
0 & 0 & 0 &14&0
end{bmatrix}
So this gives a rank of 3
linear-algebra vector-spaces
asked Jan 27 at 11:36
AbhayAbhay
3789
$endgroup$
$begingroup$
@DietrichBurde I row reduced to find that only first and second vectors are linearly independent. So doesn't that mean its dimension is 2?
$endgroup$
– Abhay
Jan 27 at 11:40
$begingroup$
Yes, it does. But where is this written? "So, $dim(A)=2$" is a very poor proof.
$endgroup$
– Dietrich Burde
Jan 27 at 11:40
$begingroup$
I have written the matrices in my post
$endgroup$
– Abhay
Jan 27 at 11:41
$begingroup$
You have written the matrices, yes, but no word about why the rank is not full.
$endgroup$
– Dietrich Burde
Jan 27 at 11:42
$begingroup$
To find $dim(A+B)$ you can put the generating vectors of $A$ and $B$ in the columns of a matrix and do the same thing as before.
$endgroup$
– Kolja
Jan 27 at 11:42
|
show 8 more comments
$begingroup$
$M={(1,2,1),(3,1,5),(3,-4,7)}$, $N={(1,0,-1),(1,2,1)}$
Let $A= operatorname{span}(M)$ and $B= operatorname{span}(N)$
I want to know how to find $;(a);dim(A+B)$ and $(b);dim(Acap B)$
begin{bmatrix}
1 & 3 & 3 \
2 & 1 & -4 \
1 & 5 & 7
end{bmatrix}$ sim$
begin{bmatrix}
1 & 3 & 3 \
0 & 1 & 2 \
0 & 0 & 0
end{bmatrix}
Since, only the first two vectors are linearly independent, we have $dim(A)=2$
Clearly, $dim(B)=2$ since it is a linearly independent set(They are not multiple of each other).
I want to make use of:
$dim(A+B)= dim(A)+dim(B)-dim(Acap B)$
So how do I find either of $dim(A+B)$ or $dim(Acap B)$ to get the other ?
Does making a single matrix with the vectors in $M$ and $N$ and calculating its rank help?
begin{bmatrix}
1 & 3 & 3 & 1 &1\
2 & 1 & -4&0 &2 \
1 & 5 & 7 &-1&1
end{bmatrix}
$sim$
begin{bmatrix}
1 & 3 & 3 & 1 &1\
0 & -5 & -10&-2 &0 \
0 & 0 & 0 &14&0
end{bmatrix}
So this gives a rank of 3
linear-algebra vector-spaces
asked Jan 27 at 11:36
AbhayAbhay
3789
$endgroup$
$M={(1,2,1),(3,1,5),(3,-4,7)}$, $N={(1,0,-1),(1,2,1)}$
Let $A= operatorname{span}(M)$ and $B= operatorname{span}(N)$
I want to know how to find $;(a);dim(A+B)$ and $(b);dim(Acap B)$
begin{bmatrix}
1 & 3 & 3 \
2 & 1 & -4 \
1 & 5 & 7
end{bmatrix}$ sim$
begin{bmatrix}
1 & 3 & 3 \
0 & 1 & 2 \
0 & 0 & 0
end{bmatrix}
Since, only the first two vectors are linearly independent, we have $dim(A)=2$
Clearly, $dim(B)=2$ since it is a linearly independent set(They are not multiple of each other).
I want to make use of:
$dim(A+B)= dim(A)+dim(B)-dim(Acap B)$
So how do I find either of $dim(A+B)$ or $dim(Acap B)$ to get the other ?
Does making a single matrix with the vectors in $M$ and $N$ and calculating its rank help?
begin{bmatrix}
1 & 3 & 3 & 1 &1\
2 & 1 & -4&0 &2 \
1 & 5 & 7 &-1&1
end{bmatrix}
$sim$
begin{bmatrix}
1 & 3 & 3 & 1 &1\
0 & -5 & -10&-2 &0 \
0 & 0 & 0 &14&0
end{bmatrix}
So this gives a rank of 3
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 27 at 11:36
AbhayAbhay
3789
asked Jan 27 at 11:36
AbhayAbhay
3789
edited Jan 27 at 11:58
Abhay
asked Jan 27 at 11:36
AbhayAbhay
3789
asked Jan 27 at 11:36
AbhayAbhay
3789
asked Jan 27 at 11:36
AbhayAbhay
3789
3789
$begingroup$
@DietrichBurde I row reduced to find that only first and second vectors are linearly independent. So doesn't that mean its dimension is 2?
$endgroup$
– Abhay
Jan 27 at 11:40
$begingroup$
Yes, it does. But where is this written? "So, $dim(A)=2$" is a very poor proof.
$endgroup$
– Dietrich Burde
Jan 27 at 11:40
$begingroup$
I have written the matrices in my post
$endgroup$
– Abhay
Jan 27 at 11:41
$begingroup$
You have written the matrices, yes, but no word about why the rank is not full.
$endgroup$
– Dietrich Burde
Jan 27 at 11:42
$begingroup$
To find $dim(A+B)$ you can put the generating vectors of $A$ and $B$ in the columns of a matrix and do the same thing as before.
$endgroup$
– Kolja
Jan 27 at 11:42
|
show 8 more comments
$begingroup$
@DietrichBurde I row reduced to find that only first and second vectors are linearly independent. So doesn't that mean its dimension is 2?
$endgroup$
– Abhay
Jan 27 at 11:40
$begingroup$
Yes, it does. But where is this written? "So, $dim(A)=2$" is a very poor proof.
$endgroup$
– Dietrich Burde
Jan 27 at 11:40
$begingroup$
I have written the matrices in my post
$endgroup$
– Abhay
Jan 27 at 11:41
$begingroup$
You have written the matrices, yes, but no word about why the rank is not full.
$endgroup$
– Dietrich Burde
Jan 27 at 11:42
$begingroup$
To find $dim(A+B)$ you can put the generating vectors of $A$ and $B$ in the columns of a matrix and do the same thing as before.
$endgroup$
– Kolja
Jan 27 at 11:42
$begingroup$
@DietrichBurde I row reduced to find that only first and second vectors are linearly independent. So doesn't that mean its dimension is 2?
$endgroup$
– Abhay
Jan 27 at 11:40
$begingroup$
@DietrichBurde I row reduced to find that only first and second vectors are linearly independent. So doesn't that mean its dimension is 2?
$endgroup$
– Abhay
Jan 27 at 11:40
$begingroup$
Yes, it does. But where is this written? "So, $dim(A)=2$" is a very poor proof.
$endgroup$
– Dietrich Burde
Jan 27 at 11:40
$begingroup$
Yes, it does. But where is this written? "So, $dim(A)=2$" is a very poor proof.
$endgroup$
– Dietrich Burde
Jan 27 at 11:40
$begingroup$
I have written the matrices in my post
$endgroup$
– Abhay
Jan 27 at 11:41
$begingroup$
I have written the matrices in my post
$endgroup$
– Abhay
Jan 27 at 11:41
$begingroup$
You have written the matrices, yes, but no word about why the rank is not full.
$endgroup$
– Dietrich Burde
Jan 27 at 11:42
$begingroup$
You have written the matrices, yes, but no word about why the rank is not full.
$endgroup$
– Dietrich Burde
Jan 27 at 11:42
$begingroup$
To find $dim(A+B)$ you can put the generating vectors of $A$ and $B$ in the columns of a matrix and do the same thing as before.
$endgroup$
– Kolja
Jan 27 at 11:42
$begingroup$
To find $dim(A+B)$ you can put the generating vectors of $A$ and $B$ in the columns of a matrix and do the same thing as before.
$endgroup$
– Kolja
Jan 27 at 11:42
|
show 8 more comments
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$begingroup$
@DietrichBurde I row reduced to find that only first and second vectors are linearly independent. So doesn't that mean its dimension is 2?
$endgroup$
– Abhay
Jan 27 at 11:40
$begingroup$
Yes, it does. But where is this written? "So, $dim(A)=2$" is a very poor proof.
$endgroup$
– Dietrich Burde
Jan 27 at 11:40
$begingroup$
I have written the matrices in my post
$endgroup$
– Abhay
Jan 27 at 11:41
$begingroup$
You have written the matrices, yes, but no word about why the rank is not full.
$endgroup$
– Dietrich Burde
Jan 27 at 11:42
$begingroup$
To find $dim(A+B)$ you can put the generating vectors of $A$ and $B$ in the columns of a matrix and do the same thing as before.
$endgroup$
– Kolja
Jan 27 at 11:42