figuring out orthonormal basis for a matrix?
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For $T in mathcal{L}(mathbb{R}^2)$ given by $T(x,y) = (x, -y)$, its basis is $ mathcal{M}(T) = begin{pmatrix}1 & 0\0 & -1end{pmatrix}$. How would I find the orthonormal basis for this? Is $frac{1}{sqrt{2}}(x + y), frac{1}{sqrt{2}}(x - y)$ one? How would I figure this out?
linear-algebra
asked Nov 15 at 6:14
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For $T in mathcal{L}(mathbb{R}^2)$ given by $T(x,y) = (x, -y)$, its basis is $ mathcal{M}(T) = begin{pmatrix}1 & 0\0 & -1end{pmatrix}$. How would I find the orthonormal basis for this? Is $frac{1}{sqrt{2}}(x + y), frac{1}{sqrt{2}}(x - y)$ one? How would I figure this out?
linear-algebra
asked Nov 15 at 6:14
user589759
111
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user589759 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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It is the usual basis ${(1,0), (0,1)}$.
– Kavi Rama Murthy
Nov 15 at 6:20
add a comment |
up vote
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down vote
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up vote
0
down vote
favorite
For $T in mathcal{L}(mathbb{R}^2)$ given by $T(x,y) = (x, -y)$, its basis is $ mathcal{M}(T) = begin{pmatrix}1 & 0\0 & -1end{pmatrix}$. How would I find the orthonormal basis for this? Is $frac{1}{sqrt{2}}(x + y), frac{1}{sqrt{2}}(x - y)$ one? How would I figure this out?
linear-algebra
asked Nov 15 at 6:14
user589759
111
New contributor
user589759 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For $T in mathcal{L}(mathbb{R}^2)$ given by $T(x,y) = (x, -y)$, its basis is $ mathcal{M}(T) = begin{pmatrix}1 & 0\0 & -1end{pmatrix}$. How would I find the orthonormal basis for this? Is $frac{1}{sqrt{2}}(x + y), frac{1}{sqrt{2}}(x - y)$ one? How would I figure this out?
linear-algebra
linear-algebra
asked Nov 15 at 6:14
user589759
111
New contributor
user589759 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 6:14
user589759
111
New contributor
user589759 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 6:14
user589759
111
New contributor
user589759 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 6:14
user589759
111
asked Nov 15 at 6:14
user589759
111
111
New contributor
user589759 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user589759 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user589759 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It is the usual basis ${(1,0), (0,1)}$.
– Kavi Rama Murthy
Nov 15 at 6:20
add a comment |
It is the usual basis ${(1,0), (0,1)}$.
– Kavi Rama Murthy
Nov 15 at 6:20
It is the usual basis ${(1,0), (0,1)}$.
– Kavi Rama Murthy
Nov 15 at 6:20
It is the usual basis ${(1,0), (0,1)}$.
– Kavi Rama Murthy
Nov 15 at 6:20
add a comment |
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user589759 is a new contributor. Be nice, and check out our Code of Conduct.
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It is the usual basis ${(1,0), (0,1)}$.
– Kavi Rama Murthy
Nov 15 at 6:20