Mixing
How to show $|x-y|_2^2 leq |x|_2^2+2|x^Ty|$?
1
$begingroup$
Let $x,y in mathbb{R}^n$.
How can I show the following
$$|x-y|_2^2 leq |x|_2^2+2|x^Ty|$$
The above has been used by the authors of the following paper on page 8, in first line
Online Principal Component Analysis.
Also, I think using the above for $M , N in mathbb{R}^{d times k}$ the following is true.
$$|M-N|_F^2 leq |M|_F^2+2|text{tr}(M^TN)|$$
where $|cdot|_F$ is Frobenius norm.
linear-algebra inequality trace cauchy-schwarz-inequality
asked Jan 3 at 14:25
SaeedSaeed
913310
$endgroup$
add a comment |
1
$begingroup$
Let $x,y in mathbb{R}^n$.
How can I show the following
$$|x-y|_2^2 leq |x|_2^2+2|x^Ty|$$
The above has been used by the authors of the following paper on page 8, in first line
Online Principal Component Analysis.
Also, I think using the above for $M , N in mathbb{R}^{d times k}$ the following is true.
$$|M-N|_F^2 leq |M|_F^2+2|text{tr}(M^TN)|$$
where $|cdot|_F$ is Frobenius norm.
linear-algebra inequality trace cauchy-schwarz-inequality
asked Jan 3 at 14:25
SaeedSaeed
913310
$endgroup$
$begingroup$
that doesn't seem to be true. take x, y such that $x^{T}y=0$
$endgroup$
– zimbra314
Jan 3 at 15:13
$begingroup$
@ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
$endgroup$
– Saeed
Jan 3 at 15:23
$begingroup$
Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
$endgroup$
– zimbra314
Jan 3 at 15:55
add a comment |
1
1
1
$begingroup$
Let $x,y in mathbb{R}^n$.
How can I show the following
$$|x-y|_2^2 leq |x|_2^2+2|x^Ty|$$
The above has been used by the authors of the following paper on page 8, in first line
Online Principal Component Analysis.
Also, I think using the above for $M , N in mathbb{R}^{d times k}$ the following is true.
$$|M-N|_F^2 leq |M|_F^2+2|text{tr}(M^TN)|$$
where $|cdot|_F$ is Frobenius norm.
linear-algebra inequality trace cauchy-schwarz-inequality
asked Jan 3 at 14:25
SaeedSaeed
913310
$endgroup$
Let $x,y in mathbb{R}^n$.
How can I show the following
$$|x-y|_2^2 leq |x|_2^2+2|x^Ty|$$
The above has been used by the authors of the following paper on page 8, in first line
Online Principal Component Analysis.
Also, I think using the above for $M , N in mathbb{R}^{d times k}$ the following is true.
$$|M-N|_F^2 leq |M|_F^2+2|text{tr}(M^TN)|$$
where $|cdot|_F$ is Frobenius norm.
linear-algebra inequality trace cauchy-schwarz-inequality
linear-algebra inequality trace cauchy-schwarz-inequality
asked Jan 3 at 14:25
SaeedSaeed
913310
asked Jan 3 at 14:25
SaeedSaeed
913310
edited Jan 3 at 15:15
Saeed
asked Jan 3 at 14:25
SaeedSaeed
913310
asked Jan 3 at 14:25
SaeedSaeed
913310
asked Jan 3 at 14:25
SaeedSaeed
913310
913310
$begingroup$
that doesn't seem to be true. take x, y such that $x^{T}y=0$
$endgroup$
– zimbra314
Jan 3 at 15:13
$begingroup$
@ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
$endgroup$
– Saeed
Jan 3 at 15:23
$begingroup$
Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
$endgroup$
– zimbra314
Jan 3 at 15:55
add a comment |
$begingroup$
that doesn't seem to be true. take x, y such that $x^{T}y=0$
$endgroup$
– zimbra314
Jan 3 at 15:13
$begingroup$
@ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
$endgroup$
– Saeed
Jan 3 at 15:23
$begingroup$
Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
$endgroup$
– zimbra314
Jan 3 at 15:55
$begingroup$
that doesn't seem to be true. take x, y such that $x^{T}y=0$
$endgroup$
– zimbra314
Jan 3 at 15:13
$begingroup$
that doesn't seem to be true. take x, y such that $x^{T}y=0$
$endgroup$
– zimbra314
Jan 3 at 15:13
$begingroup$
@ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
$endgroup$
– Saeed
Jan 3 at 15:23
$begingroup$
@ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
$endgroup$
– Saeed
Jan 3 at 15:23
$begingroup$
Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
$endgroup$
– zimbra314
Jan 3 at 15:55
$begingroup$
Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
$endgroup$
– zimbra314
Jan 3 at 15:55
add a comment |
1 Answer
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$begingroup$
You'll have hard time to prove:
$$|x-y|_2^2 leq |x|_2^2+2|x^Ty|$$
Take $x = 0$ and $y neq 0$, you get:
$$|y|_2^2 le 0$$
which is unlikely.
answered Jan 3 at 15:04
mathcounterexamples.netmathcounterexamples.net
25.8k21954
$endgroup$
$begingroup$
But in the following paper which is a well-known paper on page 8, in first line, the author claims this cs-www.cs.yale.edu/homes/el327/papers/opca.pdf. Maybe, for $x=0$ $y$ has to be zero.
$endgroup$
– Saeed
Jan 3 at 15:12
$begingroup$
There is probably a typo in the paper. Take $n = 2$, $x=(0.1,0.0)$ and $y=(0,1)$. The inequality clearly doesn't hold.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:24
$begingroup$
But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
$endgroup$
– Saeed
Jan 3 at 15:28
$begingroup$
Sorry. I won't spend time on it. The inequality is clearly wrong. Just have a look at the counterexample I provided in the comment above. And make the computation ==> $0.1^2 + 1^2 nleq 0.1^2$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:32
add a comment |
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1
$begingroup$
You'll have hard time to prove:
$$|x-y|_2^2 leq |x|_2^2+2|x^Ty|$$
Take $x = 0$ and $y neq 0$, you get:
$$|y|_2^2 le 0$$
which is unlikely.
answered Jan 3 at 15:04
mathcounterexamples.netmathcounterexamples.net
25.8k21954
$endgroup$
$begingroup$
But in the following paper which is a well-known paper on page 8, in first line, the author claims this cs-www.cs.yale.edu/homes/el327/papers/opca.pdf. Maybe, for $x=0$ $y$ has to be zero.
$endgroup$
– Saeed
Jan 3 at 15:12
$begingroup$
There is probably a typo in the paper. Take $n = 2$, $x=(0.1,0.0)$ and $y=(0,1)$. The inequality clearly doesn't hold.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:24
$begingroup$
But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
$endgroup$
– Saeed
Jan 3 at 15:28
$begingroup$
Sorry. I won't spend time on it. The inequality is clearly wrong. Just have a look at the counterexample I provided in the comment above. And make the computation ==> $0.1^2 + 1^2 nleq 0.1^2$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:32
add a comment |
1
$begingroup$
You'll have hard time to prove:
$$|x-y|_2^2 leq |x|_2^2+2|x^Ty|$$
Take $x = 0$ and $y neq 0$, you get:
$$|y|_2^2 le 0$$
which is unlikely.
answered Jan 3 at 15:04
mathcounterexamples.netmathcounterexamples.net
25.8k21954
$endgroup$
$begingroup$
But in the following paper which is a well-known paper on page 8, in first line, the author claims this cs-www.cs.yale.edu/homes/el327/papers/opca.pdf. Maybe, for $x=0$ $y$ has to be zero.
$endgroup$
– Saeed
Jan 3 at 15:12
$begingroup$
There is probably a typo in the paper. Take $n = 2$, $x=(0.1,0.0)$ and $y=(0,1)$. The inequality clearly doesn't hold.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:24
$begingroup$
But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
$endgroup$
– Saeed
Jan 3 at 15:28
$begingroup$
Sorry. I won't spend time on it. The inequality is clearly wrong. Just have a look at the counterexample I provided in the comment above. And make the computation ==> $0.1^2 + 1^2 nleq 0.1^2$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:32
add a comment |
1
1
1
$begingroup$
You'll have hard time to prove:
$$|x-y|_2^2 leq |x|_2^2+2|x^Ty|$$
Take $x = 0$ and $y neq 0$, you get:
$$|y|_2^2 le 0$$
which is unlikely.
answered Jan 3 at 15:04
mathcounterexamples.netmathcounterexamples.net
25.8k21954
$endgroup$
You'll have hard time to prove:
$$|x-y|_2^2 leq |x|_2^2+2|x^Ty|$$
Take $x = 0$ and $y neq 0$, you get:
$$|y|_2^2 le 0$$
which is unlikely.
answered Jan 3 at 15:04
mathcounterexamples.netmathcounterexamples.net
25.8k21954
answered Jan 3 at 15:04
mathcounterexamples.netmathcounterexamples.net
25.8k21954
answered Jan 3 at 15:04
mathcounterexamples.netmathcounterexamples.net
25.8k21954
answered Jan 3 at 15:04
mathcounterexamples.netmathcounterexamples.net
25.8k21954
25.8k21954
$begingroup$
But in the following paper which is a well-known paper on page 8, in first line, the author claims this cs-www.cs.yale.edu/homes/el327/papers/opca.pdf. Maybe, for $x=0$ $y$ has to be zero.
$endgroup$
– Saeed
Jan 3 at 15:12
$begingroup$
There is probably a typo in the paper. Take $n = 2$, $x=(0.1,0.0)$ and $y=(0,1)$. The inequality clearly doesn't hold.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:24
$begingroup$
But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
$endgroup$
– Saeed
Jan 3 at 15:28
$begingroup$
Sorry. I won't spend time on it. The inequality is clearly wrong. Just have a look at the counterexample I provided in the comment above. And make the computation ==> $0.1^2 + 1^2 nleq 0.1^2$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:32
add a comment |
$begingroup$
But in the following paper which is a well-known paper on page 8, in first line, the author claims this cs-www.cs.yale.edu/homes/el327/papers/opca.pdf. Maybe, for $x=0$ $y$ has to be zero.
$endgroup$
– Saeed
Jan 3 at 15:12
$begingroup$
There is probably a typo in the paper. Take $n = 2$, $x=(0.1,0.0)$ and $y=(0,1)$. The inequality clearly doesn't hold.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:24
$begingroup$
But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
$endgroup$
– Saeed
Jan 3 at 15:28
$begingroup$
Sorry. I won't spend time on it. The inequality is clearly wrong. Just have a look at the counterexample I provided in the comment above. And make the computation ==> $0.1^2 + 1^2 nleq 0.1^2$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:32
$begingroup$
But in the following paper which is a well-known paper on page 8, in first line, the author claims this cs-www.cs.yale.edu/homes/el327/papers/opca.pdf. Maybe, for $x=0$ $y$ has to be zero.
$endgroup$
– Saeed
Jan 3 at 15:12
$begingroup$
But in the following paper which is a well-known paper on page 8, in first line, the author claims this cs-www.cs.yale.edu/homes/el327/papers/opca.pdf. Maybe, for $x=0$ $y$ has to be zero.
$endgroup$
– Saeed
Jan 3 at 15:12
$begingroup$
There is probably a typo in the paper. Take $n = 2$, $x=(0.1,0.0)$ and $y=(0,1)$. The inequality clearly doesn't hold.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:24
$begingroup$
There is probably a typo in the paper. Take $n = 2$, $x=(0.1,0.0)$ and $y=(0,1)$. The inequality clearly doesn't hold.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:24
$begingroup$
But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
$endgroup$
– Saeed
Jan 3 at 15:28
$begingroup$
But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
$endgroup$
– Saeed
Jan 3 at 15:28
$begingroup$
Sorry. I won't spend time on it. The inequality is clearly wrong. Just have a look at the counterexample I provided in the comment above. And make the computation ==> $0.1^2 + 1^2 nleq 0.1^2$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:32
$begingroup$
Sorry. I won't spend time on it. The inequality is clearly wrong. Just have a look at the counterexample I provided in the comment above. And make the computation ==> $0.1^2 + 1^2 nleq 0.1^2$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:32
add a comment |
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$begingroup$
that doesn't seem to be true. take x, y such that $x^{T}y=0$
$endgroup$
– zimbra314
Jan 3 at 15:13
$begingroup$
@ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
$endgroup$
– Saeed
Jan 3 at 15:23
$begingroup$
Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
$endgroup$
– zimbra314
Jan 3 at 15:55