Show that $langle x,y rangle = 0$. [duplicate]
This question is an exact duplicate of:
Proving orthogonality of vectors in a complex vector space
1 answer
Suppose that $left (X, langle cdot,cdot rangle right)$ be a complex inner product space. Let $x,y in X$ be such that $|alpha x + beta y |^2 = |alpha x|^2 + |beta y|^2$ for all pairs $alpha, beta in Bbb C$ $($ where $|cdot|$ is the norm induced by the inner product on $X$ $).$ Then $langle x,y rangle = 0$ i.e. $x perp y.$
How do I show that? Please help me in this regard.
Thank you very much.
linear-algebra inner-product-space orthogonality
asked Dec 3 '18 at 8:21
Dbchatto67
52215
marked as duplicate by Martin R, Brahadeesh, Lord_Farin, Xander Henderson, Lord Shark the Unknown Dec 3 '18 at 22:19
This question was marked as an exact duplicate of an existing question.
add a comment |
This question is an exact duplicate of:
Proving orthogonality of vectors in a complex vector space
1 answer
Suppose that $left (X, langle cdot,cdot rangle right)$ be a complex inner product space. Let $x,y in X$ be such that $|alpha x + beta y |^2 = |alpha x|^2 + |beta y|^2$ for all pairs $alpha, beta in Bbb C$ $($ where $|cdot|$ is the norm induced by the inner product on $X$ $).$ Then $langle x,y rangle = 0$ i.e. $x perp y.$
How do I show that? Please help me in this regard.
Thank you very much.
linear-algebra inner-product-space orthogonality
asked Dec 3 '18 at 8:21
Dbchatto67
52215
marked as duplicate by Martin R, Brahadeesh, Lord_Farin, Xander Henderson, Lord Shark the Unknown Dec 3 '18 at 22:19
This question was marked as an exact duplicate of an existing question.
add a comment |
This question is an exact duplicate of:
Proving orthogonality of vectors in a complex vector space
1 answer
Suppose that $left (X, langle cdot,cdot rangle right)$ be a complex inner product space. Let $x,y in X$ be such that $|alpha x + beta y |^2 = |alpha x|^2 + |beta y|^2$ for all pairs $alpha, beta in Bbb C$ $($ where $|cdot|$ is the norm induced by the inner product on $X$ $).$ Then $langle x,y rangle = 0$ i.e. $x perp y.$
How do I show that? Please help me in this regard.
Thank you very much.
linear-algebra inner-product-space orthogonality
asked Dec 3 '18 at 8:21
Dbchatto67
52215
This question is an exact duplicate of:
Proving orthogonality of vectors in a complex vector space
1 answer
Suppose that $left (X, langle cdot,cdot rangle right)$ be a complex inner product space. Let $x,y in X$ be such that $|alpha x + beta y |^2 = |alpha x|^2 + |beta y|^2$ for all pairs $alpha, beta in Bbb C$ $($ where $|cdot|$ is the norm induced by the inner product on $X$ $).$ Then $langle x,y rangle = 0$ i.e. $x perp y.$
How do I show that? Please help me in this regard.
Thank you very much.
This question is an exact duplicate of:
Proving orthogonality of vectors in a complex vector space
1 answer
linear-algebra inner-product-space orthogonality
linear-algebra inner-product-space orthogonality
asked Dec 3 '18 at 8:21
Dbchatto67
52215
asked Dec 3 '18 at 8:21
Dbchatto67
52215
edited Dec 3 '18 at 8:26
asked Dec 3 '18 at 8:21
Dbchatto67
52215
asked Dec 3 '18 at 8:21
Dbchatto67
52215
asked Dec 3 '18 at 8:21
Dbchatto67
52215
52215
marked as duplicate by Martin R, Brahadeesh, Lord_Farin, Xander Henderson, Lord Shark the Unknown Dec 3 '18 at 22:19
This question was marked as an exact duplicate of an existing question.
marked as duplicate by Martin R, Brahadeesh, Lord_Farin, Xander Henderson, Lord Shark the Unknown Dec 3 '18 at 22:19
This question was marked as an exact duplicate of an existing question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Taking $alpha =beta=1$ and expanding $langle (x+y),(x+y) rangle$ we get $Re langle x,y rangle =0$. Take $alpha =1,beta =i$ to $Im langle x,y rangle =0$.
answered Dec 3 '18 at 8:24
Kavi Rama Murthy
51.2k31855
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Taking $alpha =beta=1$ and expanding $langle (x+y),(x+y) rangle$ we get $Re langle x,y rangle =0$. Take $alpha =1,beta =i$ to $Im langle x,y rangle =0$.
answered Dec 3 '18 at 8:24
Kavi Rama Murthy
51.2k31855
add a comment |
Taking $alpha =beta=1$ and expanding $langle (x+y),(x+y) rangle$ we get $Re langle x,y rangle =0$. Take $alpha =1,beta =i$ to $Im langle x,y rangle =0$.
answered Dec 3 '18 at 8:24
Kavi Rama Murthy
51.2k31855
add a comment |
Taking $alpha =beta=1$ and expanding $langle (x+y),(x+y) rangle$ we get $Re langle x,y rangle =0$. Take $alpha =1,beta =i$ to $Im langle x,y rangle =0$.
answered Dec 3 '18 at 8:24
Kavi Rama Murthy
51.2k31855
Taking $alpha =beta=1$ and expanding $langle (x+y),(x+y) rangle$ we get $Re langle x,y rangle =0$. Take $alpha =1,beta =i$ to $Im langle x,y rangle =0$.
answered Dec 3 '18 at 8:24
Kavi Rama Murthy
51.2k31855
answered Dec 3 '18 at 8:24
Kavi Rama Murthy
51.2k31855
answered Dec 3 '18 at 8:24
Kavi Rama Murthy
51.2k31855
answered Dec 3 '18 at 8:24
Kavi Rama Murthy
51.2k31855
51.2k31855
add a comment |
add a comment |
