How to show $|x-y|_2^2 leq |x|_2^2+2|x^Ty|$?












1














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







share|cite|improve this question
























  • that doesn't seem to be true. take x, y such that $x^{T}y=0$
    – zimbra314
    yesterday










  • @ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
    – Saeed
    yesterday










  • Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
    – zimbra314
    yesterday
















1














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







share|cite|improve this question
























  • that doesn't seem to be true. take x, y such that $x^{T}y=0$
    – zimbra314
    yesterday










  • @ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
    – Saeed
    yesterday










  • Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
    – zimbra314
    yesterday














1












1








1







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked yesterday









Saeed

698310




698310












  • that doesn't seem to be true. take x, y such that $x^{T}y=0$
    – zimbra314
    yesterday










  • @ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
    – Saeed
    yesterday










  • Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
    – zimbra314
    yesterday


















  • that doesn't seem to be true. take x, y such that $x^{T}y=0$
    – zimbra314
    yesterday










  • @ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
    – Saeed
    yesterday










  • Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
    – zimbra314
    yesterday
















that doesn't seem to be true. take x, y such that $x^{T}y=0$
– zimbra314
yesterday




that doesn't seem to be true. take x, y such that $x^{T}y=0$
– zimbra314
yesterday












@ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
– Saeed
yesterday




@ zimbra314: Could you take a look at the paper that I have cited. It is a well-known paper.
– Saeed
yesterday












Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
– zimbra314
yesterday




Evenif the inequality was true (which it is not), equation 8 on page 7 will change its inequality sign so equation 8 is wrong.
– zimbra314
yesterday










1 Answer
1






active

oldest

votes


















1














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.






share|cite|improve this answer





















  • 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.
    – Saeed
    yesterday










  • 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.
    – mathcounterexamples.net
    yesterday










  • But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
    – Saeed
    yesterday










  • 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 ==> $0.1^2 + 1^2 nleq 0.1^2$.
    – mathcounterexamples.net
    yesterday













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1 Answer
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1 Answer
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active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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.






share|cite|improve this answer





















  • 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.
    – Saeed
    yesterday










  • 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.
    – mathcounterexamples.net
    yesterday










  • But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
    – Saeed
    yesterday










  • 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 ==> $0.1^2 + 1^2 nleq 0.1^2$.
    – mathcounterexamples.net
    yesterday


















1














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.






share|cite|improve this answer





















  • 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.
    – Saeed
    yesterday










  • 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.
    – mathcounterexamples.net
    yesterday










  • But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
    – Saeed
    yesterday










  • 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 ==> $0.1^2 + 1^2 nleq 0.1^2$.
    – mathcounterexamples.net
    yesterday
















1












1








1






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









mathcounterexamples.net

25k21853




25k21853












  • 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.
    – Saeed
    yesterday










  • 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.
    – mathcounterexamples.net
    yesterday










  • But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
    – Saeed
    yesterday










  • 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 ==> $0.1^2 + 1^2 nleq 0.1^2$.
    – mathcounterexamples.net
    yesterday




















  • 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.
    – Saeed
    yesterday










  • 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.
    – mathcounterexamples.net
    yesterday










  • But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
    – Saeed
    yesterday










  • 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 ==> $0.1^2 + 1^2 nleq 0.1^2$.
    – mathcounterexamples.net
    yesterday


















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.
– Saeed
yesterday




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.
– Saeed
yesterday












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.
– mathcounterexamples.net
yesterday




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.
– mathcounterexamples.net
yesterday












But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
– Saeed
yesterday




But it uses at equation (8), could you take a look, I have spent a day on that to figure it out.
– Saeed
yesterday












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 ==> $0.1^2 + 1^2 nleq 0.1^2$.
– mathcounterexamples.net
yesterday






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 ==> $0.1^2 + 1^2 nleq 0.1^2$.
– mathcounterexamples.net
yesterday




















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