How can I find rank of $A=sum_{i=1}^4 x_ix_i^T$ without actually finding the matrix $A$?
$begingroup$
Suppose $A=sumlimits_{i=1}^4 x_ix_i^T$ where
$x_1=(1,-1,1,0)^T,x_2=(1,1,0,1)^T,x_3=(1,3,1,0)^T$ and $x_4=(1,1,1,0)^T$.
How can I find the rank of $A$ without explicitly finding the matrix $A$ and then transforming $A$ to an echelon matrix?
I know that $operatorname{rank}(A)=3$ proceeding the usual way.
We have $operatorname{rank}(x_ix_i^T)=1$ for each $i$, but that doesn't help me much here. Also, $x_1=2x_4-x_3$. Does that imply there are $3$ linearly independent rows/columns in $A$?
As this question was supposed to be solved in the exam with limited time at hand, I am left to wonder if it can be solved rather quickly.
linear-algebra matrix-rank
asked Jan 3 at 14:41
StubbornAtomStubbornAtom
6,29831440
$endgroup$
add a comment |
$begingroup$
Suppose $A=sumlimits_{i=1}^4 x_ix_i^T$ where
$x_1=(1,-1,1,0)^T,x_2=(1,1,0,1)^T,x_3=(1,3,1,0)^T$ and $x_4=(1,1,1,0)^T$.
How can I find the rank of $A$ without explicitly finding the matrix $A$ and then transforming $A$ to an echelon matrix?
I know that $operatorname{rank}(A)=3$ proceeding the usual way.
We have $operatorname{rank}(x_ix_i^T)=1$ for each $i$, but that doesn't help me much here. Also, $x_1=2x_4-x_3$. Does that imply there are $3$ linearly independent rows/columns in $A$?
As this question was supposed to be solved in the exam with limited time at hand, I am left to wonder if it can be solved rather quickly.
linear-algebra matrix-rank
asked Jan 3 at 14:41
StubbornAtomStubbornAtom
6,29831440
$endgroup$
add a comment |
$begingroup$
Suppose $A=sumlimits_{i=1}^4 x_ix_i^T$ where
$x_1=(1,-1,1,0)^T,x_2=(1,1,0,1)^T,x_3=(1,3,1,0)^T$ and $x_4=(1,1,1,0)^T$.
How can I find the rank of $A$ without explicitly finding the matrix $A$ and then transforming $A$ to an echelon matrix?
I know that $operatorname{rank}(A)=3$ proceeding the usual way.
We have $operatorname{rank}(x_ix_i^T)=1$ for each $i$, but that doesn't help me much here. Also, $x_1=2x_4-x_3$. Does that imply there are $3$ linearly independent rows/columns in $A$?
As this question was supposed to be solved in the exam with limited time at hand, I am left to wonder if it can be solved rather quickly.
linear-algebra matrix-rank
asked Jan 3 at 14:41
StubbornAtomStubbornAtom
6,29831440
$endgroup$
Suppose $A=sumlimits_{i=1}^4 x_ix_i^T$ where
$x_1=(1,-1,1,0)^T,x_2=(1,1,0,1)^T,x_3=(1,3,1,0)^T$ and $x_4=(1,1,1,0)^T$.
How can I find the rank of $A$ without explicitly finding the matrix $A$ and then transforming $A$ to an echelon matrix?
I know that $operatorname{rank}(A)=3$ proceeding the usual way.
We have $operatorname{rank}(x_ix_i^T)=1$ for each $i$, but that doesn't help me much here. Also, $x_1=2x_4-x_3$. Does that imply there are $3$ linearly independent rows/columns in $A$?
As this question was supposed to be solved in the exam with limited time at hand, I am left to wonder if it can be solved rather quickly.
linear-algebra matrix-rank
linear-algebra matrix-rank
asked Jan 3 at 14:41
StubbornAtomStubbornAtom
6,29831440
asked Jan 3 at 14:41
StubbornAtomStubbornAtom
6,29831440
asked Jan 3 at 14:41
StubbornAtomStubbornAtom
6,29831440
asked Jan 3 at 14:41
StubbornAtomStubbornAtom
6,29831440
asked Jan 3 at 14:41
StubbornAtomStubbornAtom
6,29831440
6,29831440
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The matrix $A$ can be written as $A=XX^T$ where $X=[x_1 x_2 x_3 x_4]$. Now use the fact that $operatorname{rank}XX^T=operatorname{rank}X$, so all you need to do is the find the rank of $X$.
answered Jan 3 at 15:13
A.Γ.A.Γ.
22.9k32656
$endgroup$
$begingroup$
Thank you. I overlooked this decomposition.
$endgroup$
– StubbornAtom
Jan 3 at 15:26
add a comment |
$begingroup$
The rank of $A$ is by definition the dimension of the range, and the range is spanned by ${x_1, x_2, x_3, x_4}$. Thus it suffices to find the dimension of this span.
Since both $x_1, x_3, x_4$ has $0$ in the fourth entries while $x_2$ does not. Thus if you can show that $operatorname{span}{x_1, x_3, x_4}$ is two dimensional, then the span of ${x_1, x_2, x_3, x_4}$ is three dimensional.
You know already that $operatorname{span}{x_1, x_3, x_4}$ is at most two dimensional. That it is exactly two is easy, since obviously (e.g.) ${x_1, x_4}$ is linearly independent.
answered Jan 3 at 14:45
Arctic CharArctic Char
301114
$endgroup$
add a comment |
$begingroup$
Observe that for any vector $c$,
$$Ac = sum_{i=1}^{4} x_ix_i^Tc = sum_{i=1}^{4} (x^Tc)x_i$$
The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).
answered Jan 3 at 14:49
tchtch
833310
$endgroup$
$begingroup$
How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
$endgroup$
– StubbornAtom
Jan 3 at 15:04
1
$begingroup$
$x_i^Tc$ is a scalar, so you can move it around
$endgroup$
– tch
Jan 3 at 15:14
$begingroup$
Oh I see $c$ is a vector.
$endgroup$
– StubbornAtom
Jan 3 at 15:20
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The matrix $A$ can be written as $A=XX^T$ where $X=[x_1 x_2 x_3 x_4]$. Now use the fact that $operatorname{rank}XX^T=operatorname{rank}X$, so all you need to do is the find the rank of $X$.
answered Jan 3 at 15:13
A.Γ.A.Γ.
22.9k32656
$endgroup$
$begingroup$
Thank you. I overlooked this decomposition.
$endgroup$
– StubbornAtom
Jan 3 at 15:26
add a comment |
$begingroup$
The matrix $A$ can be written as $A=XX^T$ where $X=[x_1 x_2 x_3 x_4]$. Now use the fact that $operatorname{rank}XX^T=operatorname{rank}X$, so all you need to do is the find the rank of $X$.
answered Jan 3 at 15:13
A.Γ.A.Γ.
22.9k32656
$endgroup$
$begingroup$
Thank you. I overlooked this decomposition.
$endgroup$
– StubbornAtom
Jan 3 at 15:26
add a comment |
$begingroup$
The matrix $A$ can be written as $A=XX^T$ where $X=[x_1 x_2 x_3 x_4]$. Now use the fact that $operatorname{rank}XX^T=operatorname{rank}X$, so all you need to do is the find the rank of $X$.
answered Jan 3 at 15:13
A.Γ.A.Γ.
22.9k32656
$endgroup$
The matrix $A$ can be written as $A=XX^T$ where $X=[x_1 x_2 x_3 x_4]$. Now use the fact that $operatorname{rank}XX^T=operatorname{rank}X$, so all you need to do is the find the rank of $X$.
answered Jan 3 at 15:13
A.Γ.A.Γ.
22.9k32656
answered Jan 3 at 15:13
A.Γ.A.Γ.
22.9k32656
answered Jan 3 at 15:13
A.Γ.A.Γ.
22.9k32656
answered Jan 3 at 15:13
A.Γ.A.Γ.
22.9k32656
22.9k32656
$begingroup$
Thank you. I overlooked this decomposition.
$endgroup$
– StubbornAtom
Jan 3 at 15:26
add a comment |
$begingroup$
Thank you. I overlooked this decomposition.
$endgroup$
– StubbornAtom
Jan 3 at 15:26
$begingroup$
Thank you. I overlooked this decomposition.
$endgroup$
– StubbornAtom
Jan 3 at 15:26
$begingroup$
Thank you. I overlooked this decomposition.
$endgroup$
– StubbornAtom
Jan 3 at 15:26
add a comment |
$begingroup$
The rank of $A$ is by definition the dimension of the range, and the range is spanned by ${x_1, x_2, x_3, x_4}$. Thus it suffices to find the dimension of this span.
Since both $x_1, x_3, x_4$ has $0$ in the fourth entries while $x_2$ does not. Thus if you can show that $operatorname{span}{x_1, x_3, x_4}$ is two dimensional, then the span of ${x_1, x_2, x_3, x_4}$ is three dimensional.
You know already that $operatorname{span}{x_1, x_3, x_4}$ is at most two dimensional. That it is exactly two is easy, since obviously (e.g.) ${x_1, x_4}$ is linearly independent.
answered Jan 3 at 14:45
Arctic CharArctic Char
301114
$endgroup$
add a comment |
$begingroup$
The rank of $A$ is by definition the dimension of the range, and the range is spanned by ${x_1, x_2, x_3, x_4}$. Thus it suffices to find the dimension of this span.
Since both $x_1, x_3, x_4$ has $0$ in the fourth entries while $x_2$ does not. Thus if you can show that $operatorname{span}{x_1, x_3, x_4}$ is two dimensional, then the span of ${x_1, x_2, x_3, x_4}$ is three dimensional.
You know already that $operatorname{span}{x_1, x_3, x_4}$ is at most two dimensional. That it is exactly two is easy, since obviously (e.g.) ${x_1, x_4}$ is linearly independent.
answered Jan 3 at 14:45
Arctic CharArctic Char
301114
$endgroup$
add a comment |
$begingroup$
The rank of $A$ is by definition the dimension of the range, and the range is spanned by ${x_1, x_2, x_3, x_4}$. Thus it suffices to find the dimension of this span.
Since both $x_1, x_3, x_4$ has $0$ in the fourth entries while $x_2$ does not. Thus if you can show that $operatorname{span}{x_1, x_3, x_4}$ is two dimensional, then the span of ${x_1, x_2, x_3, x_4}$ is three dimensional.
You know already that $operatorname{span}{x_1, x_3, x_4}$ is at most two dimensional. That it is exactly two is easy, since obviously (e.g.) ${x_1, x_4}$ is linearly independent.
answered Jan 3 at 14:45
Arctic CharArctic Char
301114
$endgroup$
The rank of $A$ is by definition the dimension of the range, and the range is spanned by ${x_1, x_2, x_3, x_4}$. Thus it suffices to find the dimension of this span.
Since both $x_1, x_3, x_4$ has $0$ in the fourth entries while $x_2$ does not. Thus if you can show that $operatorname{span}{x_1, x_3, x_4}$ is two dimensional, then the span of ${x_1, x_2, x_3, x_4}$ is three dimensional.
You know already that $operatorname{span}{x_1, x_3, x_4}$ is at most two dimensional. That it is exactly two is easy, since obviously (e.g.) ${x_1, x_4}$ is linearly independent.
answered Jan 3 at 14:45
Arctic CharArctic Char
301114
answered Jan 3 at 14:45
Arctic CharArctic Char
301114
answered Jan 3 at 14:45
Arctic CharArctic Char
301114
answered Jan 3 at 14:45
Arctic CharArctic Char
301114
301114
add a comment |
add a comment |
$begingroup$
Observe that for any vector $c$,
$$Ac = sum_{i=1}^{4} x_ix_i^Tc = sum_{i=1}^{4} (x^Tc)x_i$$
The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).
answered Jan 3 at 14:49
tchtch
833310
$endgroup$
$begingroup$
How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
$endgroup$
– StubbornAtom
Jan 3 at 15:04
1
$begingroup$
$x_i^Tc$ is a scalar, so you can move it around
$endgroup$
– tch
Jan 3 at 15:14
$begingroup$
Oh I see $c$ is a vector.
$endgroup$
– StubbornAtom
Jan 3 at 15:20
add a comment |
$begingroup$
Observe that for any vector $c$,
$$Ac = sum_{i=1}^{4} x_ix_i^Tc = sum_{i=1}^{4} (x^Tc)x_i$$
The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).
answered Jan 3 at 14:49
tchtch
833310
$endgroup$
$begingroup$
How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
$endgroup$
– StubbornAtom
Jan 3 at 15:04
1
$begingroup$
$x_i^Tc$ is a scalar, so you can move it around
$endgroup$
– tch
Jan 3 at 15:14
$begingroup$
Oh I see $c$ is a vector.
$endgroup$
– StubbornAtom
Jan 3 at 15:20
add a comment |
$begingroup$
Observe that for any vector $c$,
$$Ac = sum_{i=1}^{4} x_ix_i^Tc = sum_{i=1}^{4} (x^Tc)x_i$$
The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).
answered Jan 3 at 14:49
tchtch
833310
$endgroup$
Observe that for any vector $c$,
$$Ac = sum_{i=1}^{4} x_ix_i^Tc = sum_{i=1}^{4} (x^Tc)x_i$$
The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).
answered Jan 3 at 14:49
tchtch
833310
answered Jan 3 at 14:49
tchtch
833310
answered Jan 3 at 14:49
tchtch
833310
answered Jan 3 at 14:49
tchtch
833310
833310
$begingroup$
How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
$endgroup$
– StubbornAtom
Jan 3 at 15:04
1
$begingroup$
$x_i^Tc$ is a scalar, so you can move it around
$endgroup$
– tch
Jan 3 at 15:14
$begingroup$
Oh I see $c$ is a vector.
$endgroup$
– StubbornAtom
Jan 3 at 15:20
add a comment |
$begingroup$
How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
$endgroup$
– StubbornAtom
Jan 3 at 15:04
1
$begingroup$
$x_i^Tc$ is a scalar, so you can move it around
$endgroup$
– tch
Jan 3 at 15:14
$begingroup$
Oh I see $c$ is a vector.
$endgroup$
– StubbornAtom
Jan 3 at 15:20
$begingroup$
How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
$endgroup$
– StubbornAtom
Jan 3 at 15:04
$begingroup$
How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
$endgroup$
– StubbornAtom
Jan 3 at 15:04
1
1
$begingroup$
$x_i^Tc$ is a scalar, so you can move it around
$endgroup$
– tch
Jan 3 at 15:14
$begingroup$
$x_i^Tc$ is a scalar, so you can move it around
$endgroup$
– tch
Jan 3 at 15:14
$begingroup$
Oh I see $c$ is a vector.
$endgroup$
– StubbornAtom
Jan 3 at 15:20
$begingroup$
Oh I see $c$ is a vector.
$endgroup$
– StubbornAtom
Jan 3 at 15:20
add a comment |
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