Mixing
To find the dimension of the intersection of $N(A)$ and $R(B)$.
$begingroup$
Let $A=$
$$begin{pmatrix}1 & 2 & 0 \ -1 & 5 & 2end{pmatrix}$$, $B=$
$$begin{pmatrix}1 & 2 \ -1 & 0 \ 3 & 1end{pmatrix}$$
$N(A)$ the null space of $A$ and $R(B)$ the range space of $B$. Then find the dimension of the intersection of $N(A)$ and $R(B)$.
dim$N(A) = 1, $ dim$R(B) = 2$
Now I want use the formula
Dimension of intersection $=$ dimension of $N(A) +$ dimension of $R(B)$ - dimension of $(N(A) + R(B))$.
But I'm stuck, as I don't know how to find dimension of $(N(A) + R(B))$.
linear-algebra
asked Jan 31 at 3:20
MathsaddictMathsaddict
3669
$endgroup$
add a comment |
$begingroup$
Let $A=$
$$begin{pmatrix}1 & 2 & 0 \ -1 & 5 & 2end{pmatrix}$$, $B=$
$$begin{pmatrix}1 & 2 \ -1 & 0 \ 3 & 1end{pmatrix}$$
$N(A)$ the null space of $A$ and $R(B)$ the range space of $B$. Then find the dimension of the intersection of $N(A)$ and $R(B)$.
dim$N(A) = 1, $ dim$R(B) = 2$
Now I want use the formula
Dimension of intersection $=$ dimension of $N(A) +$ dimension of $R(B)$ - dimension of $(N(A) + R(B))$.
But I'm stuck, as I don't know how to find dimension of $(N(A) + R(B))$.
linear-algebra
asked Jan 31 at 3:20
MathsaddictMathsaddict
3669
$endgroup$
add a comment |
$begingroup$
Let $A=$
$$begin{pmatrix}1 & 2 & 0 \ -1 & 5 & 2end{pmatrix}$$, $B=$
$$begin{pmatrix}1 & 2 \ -1 & 0 \ 3 & 1end{pmatrix}$$
$N(A)$ the null space of $A$ and $R(B)$ the range space of $B$. Then find the dimension of the intersection of $N(A)$ and $R(B)$.
dim$N(A) = 1, $ dim$R(B) = 2$
Now I want use the formula
Dimension of intersection $=$ dimension of $N(A) +$ dimension of $R(B)$ - dimension of $(N(A) + R(B))$.
But I'm stuck, as I don't know how to find dimension of $(N(A) + R(B))$.
linear-algebra
asked Jan 31 at 3:20
MathsaddictMathsaddict
3669
$endgroup$
Let $A=$
$$begin{pmatrix}1 & 2 & 0 \ -1 & 5 & 2end{pmatrix}$$, $B=$
$$begin{pmatrix}1 & 2 \ -1 & 0 \ 3 & 1end{pmatrix}$$
$N(A)$ the null space of $A$ and $R(B)$ the range space of $B$. Then find the dimension of the intersection of $N(A)$ and $R(B)$.
dim$N(A) = 1, $ dim$R(B) = 2$
Now I want use the formula
Dimension of intersection $=$ dimension of $N(A) +$ dimension of $R(B)$ - dimension of $(N(A) + R(B))$.
But I'm stuck, as I don't know how to find dimension of $(N(A) + R(B))$.
linear-algebra
linear-algebra
asked Jan 31 at 3:20
MathsaddictMathsaddict
3669
asked Jan 31 at 3:20
MathsaddictMathsaddict
3669
edited Jan 31 at 3:27
Mathsaddict
asked Jan 31 at 3:20
MathsaddictMathsaddict
3669
asked Jan 31 at 3:20
MathsaddictMathsaddict
3669
asked Jan 31 at 3:20
MathsaddictMathsaddict
3669
3669
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$N(A)=span{(4,-2,7)}$
$R(B)=span{(1,-1,3), (2,0,1)}$
$(4,-2,7)=2(1,-1,3)+(2,0,1)$
So, $N(A)cap R(B) = N(A)$
answered Jan 31 at 3:41
OfflawOfflaw
3189
$endgroup$
$begingroup$
Can't we take rows of$B$ as span of $B$? i.e.$R(B) = span {(1,2), (-1,0)}$? Since both are linearly independent.
$endgroup$
– Mathsaddict
Jan 31 at 4:10
$begingroup$
Rows are not members of $mathbb{R}^3$. So we can't.
$endgroup$
– Offlaw
Jan 31 at 8:04
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$N(A)=span{(4,-2,7)}$
$R(B)=span{(1,-1,3), (2,0,1)}$
$(4,-2,7)=2(1,-1,3)+(2,0,1)$
So, $N(A)cap R(B) = N(A)$
answered Jan 31 at 3:41
OfflawOfflaw
3189
$endgroup$
$begingroup$
Can't we take rows of$B$ as span of $B$? i.e.$R(B) = span {(1,2), (-1,0)}$? Since both are linearly independent.
$endgroup$
– Mathsaddict
Jan 31 at 4:10
$begingroup$
Rows are not members of $mathbb{R}^3$. So we can't.
$endgroup$
– Offlaw
Jan 31 at 8:04
add a comment |
$begingroup$
$N(A)=span{(4,-2,7)}$
$R(B)=span{(1,-1,3), (2,0,1)}$
$(4,-2,7)=2(1,-1,3)+(2,0,1)$
So, $N(A)cap R(B) = N(A)$
answered Jan 31 at 3:41
OfflawOfflaw
3189
$endgroup$
$begingroup$
Can't we take rows of$B$ as span of $B$? i.e.$R(B) = span {(1,2), (-1,0)}$? Since both are linearly independent.
$endgroup$
– Mathsaddict
Jan 31 at 4:10
$begingroup$
Rows are not members of $mathbb{R}^3$. So we can't.
$endgroup$
– Offlaw
Jan 31 at 8:04
add a comment |
$begingroup$
$N(A)=span{(4,-2,7)}$
$R(B)=span{(1,-1,3), (2,0,1)}$
$(4,-2,7)=2(1,-1,3)+(2,0,1)$
So, $N(A)cap R(B) = N(A)$
answered Jan 31 at 3:41
OfflawOfflaw
3189
$endgroup$
$N(A)=span{(4,-2,7)}$
$R(B)=span{(1,-1,3), (2,0,1)}$
$(4,-2,7)=2(1,-1,3)+(2,0,1)$
So, $N(A)cap R(B) = N(A)$
answered Jan 31 at 3:41
OfflawOfflaw
3189
answered Jan 31 at 3:41
OfflawOfflaw
3189
answered Jan 31 at 3:41
OfflawOfflaw
3189
answered Jan 31 at 3:41
OfflawOfflaw
3189
3189
$begingroup$
Can't we take rows of$B$ as span of $B$? i.e.$R(B) = span {(1,2), (-1,0)}$? Since both are linearly independent.
$endgroup$
– Mathsaddict
Jan 31 at 4:10
$begingroup$
Rows are not members of $mathbb{R}^3$. So we can't.
$endgroup$
– Offlaw
Jan 31 at 8:04
add a comment |
$begingroup$
Can't we take rows of$B$ as span of $B$? i.e.$R(B) = span {(1,2), (-1,0)}$? Since both are linearly independent.
$endgroup$
– Mathsaddict
Jan 31 at 4:10
$begingroup$
Rows are not members of $mathbb{R}^3$. So we can't.
$endgroup$
– Offlaw
Jan 31 at 8:04
$begingroup$
Can't we take rows of$B$ as span of $B$? i.e.$R(B) = span {(1,2), (-1,0)}$? Since both are linearly independent.
$endgroup$
– Mathsaddict
Jan 31 at 4:10
$begingroup$
Can't we take rows of$B$ as span of $B$? i.e.$R(B) = span {(1,2), (-1,0)}$? Since both are linearly independent.
$endgroup$
– Mathsaddict
Jan 31 at 4:10
$begingroup$
Rows are not members of $mathbb{R}^3$. So we can't.
$endgroup$
– Offlaw
Jan 31 at 8:04
$begingroup$
Rows are not members of $mathbb{R}^3$. So we can't.
$endgroup$
– Offlaw
Jan 31 at 8:04
add a comment |
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