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Let $u,v,x \in \mathbb R^d$ be three unit vectors. I found a very complicated proof that $(v^Tx)^2-(u^Tx)^2 \leq 1-(u^Tv)^2$. That is $\lVert uu^T-vv^T\rVert^2_2 = 1-(u^Tv)^2$, or that $f(v,x)\leq f(v,u)+f(u,x)$ where $f(v,u)=1-(v^Tu)^2$ is the squared sine of the angle between $u$ and $v$.

Is there a one-liner proof?

Say, using the (spherical?) law of cosines or the Haversine formula? Induced norm for positive semi-definite matrices?

Edit: Thank you all for the quick answers. I am confused by the counter examples. I tried to cite Lemma 27 in a paper (I think from STOC'15): https://arxiv.org/abs/1606.05225 The eigenvalue of $||uu^T-vv^T||_2$ seems to be correct by How to find the eigenvalues of $xx^T-yy^T$

Edit 2: I assumed that all the conjectures in the question are equivalent but maybe I was wrong. I took the Schur Decomposition $USU^T$ of $uu^T-vv^T$ to get $\max_{||x||=1} ||(uu^T-vv^T)x||^2=||USU^Tx||^2=S_{1,1}^2$ and assumed it is the same $\max_{||x||=1} |x^T(uu^T-vv^T)x|^2=|x^TUSU^Tx|=S_{1,1}^2$. Then I noticed that $|x^T(uu^T-vv^T)x|=|(x^Tu)^2-(v^Tx)|$. Not sure what went wrong.

Let $u,v,x \in \mathbb R^d$ be three unit vectors. I found a very complicated proof that $(v^Tx)^2-(u^Tx)^2 \leq 1-(u^Tv)^2$. That is $\lVert uu^T-vv^T\rVert^2_2 = 1-(u^Tv)^2$, or that $f(v,x)\leq f(v,u)+f(u,x)$ where $f(v,u)=1-(v^Tu)^2$ is the squared sine of the angle between $u$ and $v$.

Is there a one-liner proof?

Say, using the (spherical?) law of cosines or the Haversine formula? Induced norm for positive semi-definite matrices?

Edit: Thank you all for the quick answers. I am confused by the counter examples. I tried to cite Lemma 27 in a paper (I think from STOC'15): https://arxiv.org/abs/1606.05225 The eigenvalue of $||uu^T-vv^T||_2$ seems to be correct by How to find the eigenvalues of $xx^T-yy^T$

Edit 2: I assumed that all the conjectures in the question are equivalent but maybe I was wrong. I took the Schur Decomposition $USU^T$ of $uu^T-vv^T$ to get $\max_{||x||=1} ||(uu^T-vv^T)x||^2=||USU^Tx||^2=S_{1,1}^2$ and assumed it is the same $\max_{||x||=1} |x^T(uu^T-vv^T)x|^2=|x^TUSU^Tx|=S_{1,1}^2$. Then I noticed that $|x^T(uu^T-vv^T)x|^2=|(x^Tu)^2-(v^Tx)|$. Not sure what went wrong.

Summary: As GH from Mo noted below, I forgot a squared root in the right hand side and the statement is wrong. Hope to get your help also in the fixed version here

Let $u,v,x \in \mathbb R^d$ be three unit vectors. I found a very complicated proof that $(v^Tx)^2-(u^Tx)^2 \leq 1-(u^Tv)^2$. That is $\lVert uu^T-vv^T\rVert^2_2 = 1-(u^Tv)^2$, or that $f(v,x)\leq f(v,u)+f(u,x)$ where $f(v,u)=1-(v^Tu)^2$ is the squared sine of the angle between $u$ and $v$.

Is there a one-liner proof?

Say, using the (spherical?) law of cosines or the Haversine formula? Induced norm for positive semi-definite matrices?

Edit: Thank you all for the quick answers. I am confused by the counter examples. I tried to cite Lemma 27 in a paper (I think from STOC'15): https://arxiv.org/abs/1606.05225 The eigenvalue of $||uu^T-vv^T||_2$ seems to be correct by How to find the eigenvalues of $xx^T-yy^T$

Let $u,v,x \in \mathbb R^d$ be three unit vectors. I found a very complicated proof that $(v^Tx)^2-(u^Tx)^2 \leq 1-(u^Tv)^2$. That is $\lVert uu^T-vv^T\rVert^2_2 = 1-(u^Tv)^2$, or that $f(v,x)\leq f(v,u)+f(u,x)$ where $f(v,u)=1-(v^Tu)^2$ is the squared sine of the angle between $u$ and $v$.

Is there a one-liner proof?

Say, using the (spherical?) law of cosines or the Haversine formula? Induced norm for positive semi-definite matrices?

Edit: Thank you all for the quick answers. I am confused by the counter examples. I tried to cite Lemma 27 in a paper (I think from STOC'15): https://arxiv.org/abs/1606.05225 The eigenvalue of $||uu^T-vv^T||_2$ seems to be correct by How to find the eigenvalues of $xx^T-yy^T$

Edit 2: I assumed that all the conjectures in the question are equivalent but maybe I was wrong. I took the Schur Decomposition $USU^T$ of $uu^T-vv^T$ to get $\max_{||x||=1} ||(uu^T-vv^T)x||^2=||USU^Tx||^2=S_{1,1}^2$ and assumed it is the same $\max_{||x||=1} |x^T(uu^T-vv^T)x|^2=|x^TUSU^Tx|=S_{1,1}^2$. Then I noticed that $|x^T(uu^T-vv^T)x|=|(x^Tu)^2-(v^Tx)|$. Not sure what went wrong.

Let $u,v,x \in \mathbb R^d$ be three unit vectors. I found a very complicated proof that $(v^Tx)^2-(u^Tx)^2 \leq 1-(u^Tv)^2$. That is $\lVert uu^T-vv^T\rVert^2_2 = 1-(u^Tv)^2$, or that $f(v,x)\leq f(v,u)+f(u,x)$ where $f(v,u)=1-(v^Tu)^2$ is the squared sine of the angle between $u$ and $v$.

Is there a one-liner proof?

Say, using the (spherical?) law of cosines or the Haversine formula? Induced norm for positive semi-definite matrices?

Let $u,v,x \in \mathbb R^d$ be three unit vectors. I found a very complicated proof that $(v^Tx)^2-(u^Tx)^2 \leq 1-(u^Tv)^2$. That is $\lVert uu^T-vv^T\rVert^2_2 = 1-(u^Tv)^2$, or that $f(v,x)\leq f(v,u)+f(u,x)$ where $f(v,u)=1-(v^Tu)^2$ is the squared sine of the angle between $u$ and $v$.

Is there a one-liner proof?

Say, using the (spherical?) law of cosines or the Haversine formula? Induced norm for positive semi-definite matrices?

Edit: Thank you all for the quick answers. I am confused by the counter examples. I tried to cite Lemma 27 in a paper (I think from STOC'15): https://arxiv.org/abs/1606.05225 The eigenvalue of $||uu^T-vv^T||_2$ seems to be correct by How to find the eigenvalues of $xx^T-yy^T$