Mixing
Show that if $langle X,AXrangle = 0$, then $AX = 0$
$begingroup$
Let $A$ be an $ntimes n$ symmetric positive semidefinite matrix. Let $X in mathbb{R}^n$
Show that if $langle X, AXrangle = 0,$ then $AX = 0$.
This seems like it's really simple, but I must be missing a trick.
We have $operatorname{tr}(X^TAX) = 0$. The trace is just the sum of the eigenvalues, so could I say something about the eigenvalues all having to be zero? Would that help with the proof?
linear-algebra
asked Apr 28 '13 at 0:48
user70864user70864
414
$endgroup$
add a comment |
$begingroup$
Let $A$ be an $ntimes n$ symmetric positive semidefinite matrix. Let $X in mathbb{R}^n$
Show that if $langle X, AXrangle = 0,$ then $AX = 0$.
This seems like it's really simple, but I must be missing a trick.
We have $operatorname{tr}(X^TAX) = 0$. The trace is just the sum of the eigenvalues, so could I say something about the eigenvalues all having to be zero? Would that help with the proof?
linear-algebra
asked Apr 28 '13 at 0:48
user70864user70864
414
$endgroup$
$begingroup$
Please clarify. It appears that $X, AX$ are column vectors, however your inner product appears to be for matrices. If $x,y$ are column vectors, then $x^ty$ is a number; there's no trace to be taken.
$endgroup$
– vadim123
Apr 28 '13 at 0:52
$begingroup$
Your question is unclear. Is $X$ a matrix or a vector?
$endgroup$
– Julien
Apr 28 '13 at 0:53
1
$begingroup$
In any case, you might want to use that $A$ has a square root, i.e. $B$ symmetric positive semidefinite such that $A=B^2$. You don't have to go this way, but it makes things very easy. Then $(X,AX)=(BX,BX)=0$ if and only if $BX=0$, using separation if the inner product.
$endgroup$
– Julien
Apr 28 '13 at 0:55
1
$begingroup$
sorry, yes I copied the wrong inner product from the book. It doesn't specify which one to use. $X$ is vector in R^n
$endgroup$
– user70864
Apr 28 '13 at 0:57
$begingroup$
So $(X,Y)=X^tY$, I guess, the Euclidean inner product. Then the square root thing works.
$endgroup$
– Julien
Apr 28 '13 at 1:03
add a comment |
$begingroup$
Let $A$ be an $ntimes n$ symmetric positive semidefinite matrix. Let $X in mathbb{R}^n$
Show that if $langle X, AXrangle = 0,$ then $AX = 0$.
This seems like it's really simple, but I must be missing a trick.
We have $operatorname{tr}(X^TAX) = 0$. The trace is just the sum of the eigenvalues, so could I say something about the eigenvalues all having to be zero? Would that help with the proof?
linear-algebra
asked Apr 28 '13 at 0:48
user70864user70864
414
$endgroup$
Let $A$ be an $ntimes n$ symmetric positive semidefinite matrix. Let $X in mathbb{R}^n$
Show that if $langle X, AXrangle = 0,$ then $AX = 0$.
This seems like it's really simple, but I must be missing a trick.
We have $operatorname{tr}(X^TAX) = 0$. The trace is just the sum of the eigenvalues, so could I say something about the eigenvalues all having to be zero? Would that help with the proof?
linear-algebra
linear-algebra
asked Apr 28 '13 at 0:48
user70864user70864
414
asked Apr 28 '13 at 0:48
user70864user70864
414
edited Apr 28 '13 at 1:01
user70864
asked Apr 28 '13 at 0:48
user70864user70864
414
asked Apr 28 '13 at 0:48
user70864user70864
414
asked Apr 28 '13 at 0:48
user70864user70864
414
414
$begingroup$
Please clarify. It appears that $X, AX$ are column vectors, however your inner product appears to be for matrices. If $x,y$ are column vectors, then $x^ty$ is a number; there's no trace to be taken.
$endgroup$
– vadim123
Apr 28 '13 at 0:52
$begingroup$
Your question is unclear. Is $X$ a matrix or a vector?
$endgroup$
– Julien
Apr 28 '13 at 0:53
1
$begingroup$
In any case, you might want to use that $A$ has a square root, i.e. $B$ symmetric positive semidefinite such that $A=B^2$. You don't have to go this way, but it makes things very easy. Then $(X,AX)=(BX,BX)=0$ if and only if $BX=0$, using separation if the inner product.
$endgroup$
– Julien
Apr 28 '13 at 0:55
1
$begingroup$
sorry, yes I copied the wrong inner product from the book. It doesn't specify which one to use. $X$ is vector in R^n
$endgroup$
– user70864
Apr 28 '13 at 0:57
$begingroup$
So $(X,Y)=X^tY$, I guess, the Euclidean inner product. Then the square root thing works.
$endgroup$
– Julien
Apr 28 '13 at 1:03
add a comment |
$begingroup$
Please clarify. It appears that $X, AX$ are column vectors, however your inner product appears to be for matrices. If $x,y$ are column vectors, then $x^ty$ is a number; there's no trace to be taken.
$endgroup$
– vadim123
Apr 28 '13 at 0:52
$begingroup$
Your question is unclear. Is $X$ a matrix or a vector?
$endgroup$
– Julien
Apr 28 '13 at 0:53
1
$begingroup$
In any case, you might want to use that $A$ has a square root, i.e. $B$ symmetric positive semidefinite such that $A=B^2$. You don't have to go this way, but it makes things very easy. Then $(X,AX)=(BX,BX)=0$ if and only if $BX=0$, using separation if the inner product.
$endgroup$
– Julien
Apr 28 '13 at 0:55
1
$begingroup$
sorry, yes I copied the wrong inner product from the book. It doesn't specify which one to use. $X$ is vector in R^n
$endgroup$
– user70864
Apr 28 '13 at 0:57
$begingroup$
So $(X,Y)=X^tY$, I guess, the Euclidean inner product. Then the square root thing works.
$endgroup$
– Julien
Apr 28 '13 at 1:03
$begingroup$
Please clarify. It appears that $X, AX$ are column vectors, however your inner product appears to be for matrices. If $x,y$ are column vectors, then $x^ty$ is a number; there's no trace to be taken.
$endgroup$
– vadim123
Apr 28 '13 at 0:52
$begingroup$
Please clarify. It appears that $X, AX$ are column vectors, however your inner product appears to be for matrices. If $x,y$ are column vectors, then $x^ty$ is a number; there's no trace to be taken.
$endgroup$
– vadim123
Apr 28 '13 at 0:52
$begingroup$
Your question is unclear. Is $X$ a matrix or a vector?
$endgroup$
– Julien
Apr 28 '13 at 0:53
$begingroup$
Your question is unclear. Is $X$ a matrix or a vector?
$endgroup$
– Julien
Apr 28 '13 at 0:53
1
1
$begingroup$
In any case, you might want to use that $A$ has a square root, i.e. $B$ symmetric positive semidefinite such that $A=B^2$. You don't have to go this way, but it makes things very easy. Then $(X,AX)=(BX,BX)=0$ if and only if $BX=0$, using separation if the inner product.
$endgroup$
– Julien
Apr 28 '13 at 0:55
$begingroup$
In any case, you might want to use that $A$ has a square root, i.e. $B$ symmetric positive semidefinite such that $A=B^2$. You don't have to go this way, but it makes things very easy. Then $(X,AX)=(BX,BX)=0$ if and only if $BX=0$, using separation if the inner product.
$endgroup$
– Julien
Apr 28 '13 at 0:55
1
1
$begingroup$
sorry, yes I copied the wrong inner product from the book. It doesn't specify which one to use. $X$ is vector in R^n
$endgroup$
– user70864
Apr 28 '13 at 0:57
$begingroup$
sorry, yes I copied the wrong inner product from the book. It doesn't specify which one to use. $X$ is vector in R^n
$endgroup$
– user70864
Apr 28 '13 at 0:57
$begingroup$
So $(X,Y)=X^tY$, I guess, the Euclidean inner product. Then the square root thing works.
$endgroup$
– Julien
Apr 28 '13 at 1:03
$begingroup$
So $(X,Y)=X^tY$, I guess, the Euclidean inner product. Then the square root thing works.
$endgroup$
– Julien
Apr 28 '13 at 1:03
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since $A$ is symmetric, ${bf R}^n$ has a basis consisting of eigenvectors of $A$. Express $X$ as a linear combination of these eigenvectors, then calculate $X^tAX$, then use the hypothesis about $A$ being positive semi-definite.
answered Apr 28 '13 at 1:40
Gerry MyersonGerry Myerson
146k8147299
$endgroup$
add a comment |
$begingroup$
Hint: Because it is positive semi-definite, you have
$$
left<x,Axright> = x^TAxgeq 0
$$
Considering the properties of a symmetric positive semi-definite matrix, what does this tell you about
$$
x^TA^TAx = x^TA^2x
$$
answered Apr 28 '13 at 2:11
Glen OGlen O
8,9931528
$endgroup$
add a comment |
$begingroup$
$$rank(X^TAX) = rank(X)$$
Proof: Null space of X and $X^TAX$ is exactly the same, and hence the rank.
$$Xy = 0 implies X^TAXy = 0$$ and
$$X^TAXy = 0 implies (Xy)^TAXy = 0 implies Xy = 0$$
Rank is also same
Therefore, this is a property for positive definite matrix A for arbitrary matrix X
In our case $$X^TAX=0 implies rank(X^TAX) = 0 = rank(X)$$
Therefore, X is a zero matrix and therefore AX=0
answered Jan 13 at 11:29
AbbasAbbas
1
$endgroup$
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
$begingroup$
Since $A$ is symmetric, ${bf R}^n$ has a basis consisting of eigenvectors of $A$. Express $X$ as a linear combination of these eigenvectors, then calculate $X^tAX$, then use the hypothesis about $A$ being positive semi-definite.
answered Apr 28 '13 at 1:40
Gerry MyersonGerry Myerson
146k8147299
$endgroup$
add a comment |
$begingroup$
Since $A$ is symmetric, ${bf R}^n$ has a basis consisting of eigenvectors of $A$. Express $X$ as a linear combination of these eigenvectors, then calculate $X^tAX$, then use the hypothesis about $A$ being positive semi-definite.
answered Apr 28 '13 at 1:40
Gerry MyersonGerry Myerson
146k8147299
$endgroup$
add a comment |
$begingroup$
Since $A$ is symmetric, ${bf R}^n$ has a basis consisting of eigenvectors of $A$. Express $X$ as a linear combination of these eigenvectors, then calculate $X^tAX$, then use the hypothesis about $A$ being positive semi-definite.
answered Apr 28 '13 at 1:40
Gerry MyersonGerry Myerson
146k8147299
$endgroup$
Since $A$ is symmetric, ${bf R}^n$ has a basis consisting of eigenvectors of $A$. Express $X$ as a linear combination of these eigenvectors, then calculate $X^tAX$, then use the hypothesis about $A$ being positive semi-definite.
answered Apr 28 '13 at 1:40
Gerry MyersonGerry Myerson
146k8147299
answered Apr 28 '13 at 1:40
Gerry MyersonGerry Myerson
146k8147299
answered Apr 28 '13 at 1:40
Gerry MyersonGerry Myerson
146k8147299
answered Apr 28 '13 at 1:40
Gerry MyersonGerry Myerson
146k8147299
146k8147299
add a comment |
add a comment |
$begingroup$
Hint: Because it is positive semi-definite, you have
$$
left<x,Axright> = x^TAxgeq 0
$$
Considering the properties of a symmetric positive semi-definite matrix, what does this tell you about
$$
x^TA^TAx = x^TA^2x
$$
answered Apr 28 '13 at 2:11
Glen OGlen O
8,9931528
$endgroup$
add a comment |
$begingroup$
Hint: Because it is positive semi-definite, you have
$$
left<x,Axright> = x^TAxgeq 0
$$
Considering the properties of a symmetric positive semi-definite matrix, what does this tell you about
$$
x^TA^TAx = x^TA^2x
$$
answered Apr 28 '13 at 2:11
Glen OGlen O
8,9931528
$endgroup$
add a comment |
$begingroup$
Hint: Because it is positive semi-definite, you have
$$
left<x,Axright> = x^TAxgeq 0
$$
Considering the properties of a symmetric positive semi-definite matrix, what does this tell you about
$$
x^TA^TAx = x^TA^2x
$$
answered Apr 28 '13 at 2:11
Glen OGlen O
8,9931528
$endgroup$
Hint: Because it is positive semi-definite, you have
$$
left<x,Axright> = x^TAxgeq 0
$$
Considering the properties of a symmetric positive semi-definite matrix, what does this tell you about
$$
x^TA^TAx = x^TA^2x
$$
answered Apr 28 '13 at 2:11
Glen OGlen O
8,9931528
answered Apr 28 '13 at 2:11
Glen OGlen O
8,9931528
answered Apr 28 '13 at 2:11
Glen OGlen O
8,9931528
answered Apr 28 '13 at 2:11
Glen OGlen O
8,9931528
8,9931528
add a comment |
add a comment |
$begingroup$
$$rank(X^TAX) = rank(X)$$
Proof: Null space of X and $X^TAX$ is exactly the same, and hence the rank.
$$Xy = 0 implies X^TAXy = 0$$ and
$$X^TAXy = 0 implies (Xy)^TAXy = 0 implies Xy = 0$$
Rank is also same
Therefore, this is a property for positive definite matrix A for arbitrary matrix X
In our case $$X^TAX=0 implies rank(X^TAX) = 0 = rank(X)$$
Therefore, X is a zero matrix and therefore AX=0
answered Jan 13 at 11:29
AbbasAbbas
1
$endgroup$
add a comment |
$begingroup$
$$rank(X^TAX) = rank(X)$$
Proof: Null space of X and $X^TAX$ is exactly the same, and hence the rank.
$$Xy = 0 implies X^TAXy = 0$$ and
$$X^TAXy = 0 implies (Xy)^TAXy = 0 implies Xy = 0$$
Rank is also same
Therefore, this is a property for positive definite matrix A for arbitrary matrix X
In our case $$X^TAX=0 implies rank(X^TAX) = 0 = rank(X)$$
Therefore, X is a zero matrix and therefore AX=0
answered Jan 13 at 11:29
AbbasAbbas
1
$endgroup$
add a comment |
$begingroup$
$$rank(X^TAX) = rank(X)$$
Proof: Null space of X and $X^TAX$ is exactly the same, and hence the rank.
$$Xy = 0 implies X^TAXy = 0$$ and
$$X^TAXy = 0 implies (Xy)^TAXy = 0 implies Xy = 0$$
Rank is also same
Therefore, this is a property for positive definite matrix A for arbitrary matrix X
In our case $$X^TAX=0 implies rank(X^TAX) = 0 = rank(X)$$
Therefore, X is a zero matrix and therefore AX=0
answered Jan 13 at 11:29
AbbasAbbas
1
$endgroup$
$$rank(X^TAX) = rank(X)$$
Proof: Null space of X and $X^TAX$ is exactly the same, and hence the rank.
$$Xy = 0 implies X^TAXy = 0$$ and
$$X^TAXy = 0 implies (Xy)^TAXy = 0 implies Xy = 0$$
Rank is also same
Therefore, this is a property for positive definite matrix A for arbitrary matrix X
In our case $$X^TAX=0 implies rank(X^TAX) = 0 = rank(X)$$
Therefore, X is a zero matrix and therefore AX=0
answered Jan 13 at 11:29
AbbasAbbas
1
answered Jan 13 at 11:29
AbbasAbbas
1
answered Jan 13 at 11:29
AbbasAbbas
1
answered Jan 13 at 11:29
AbbasAbbas
1
1
add a comment |
add a comment |
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$begingroup$
Please clarify. It appears that $X, AX$ are column vectors, however your inner product appears to be for matrices. If $x,y$ are column vectors, then $x^ty$ is a number; there's no trace to be taken.
$endgroup$
– vadim123
Apr 28 '13 at 0:52
$begingroup$
Your question is unclear. Is $X$ a matrix or a vector?
$endgroup$
– Julien
Apr 28 '13 at 0:53
1
$begingroup$
In any case, you might want to use that $A$ has a square root, i.e. $B$ symmetric positive semidefinite such that $A=B^2$. You don't have to go this way, but it makes things very easy. Then $(X,AX)=(BX,BX)=0$ if and only if $BX=0$, using separation if the inner product.
$endgroup$
– Julien
Apr 28 '13 at 0:55
1
$begingroup$
sorry, yes I copied the wrong inner product from the book. It doesn't specify which one to use. $X$ is vector in R^n
$endgroup$
– user70864
Apr 28 '13 at 0:57
$begingroup$
So $(X,Y)=X^tY$, I guess, the Euclidean inner product. Then the square root thing works.
$endgroup$
– Julien
Apr 28 '13 at 1:03