Mixing
Linear algebra inequality
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(1) Note that if $| mathbfA mathbfx |_2 = 1$, then
$$| mathbfA^T mathbfA mathbfx |_2 cdot | mathbfx |_2 ge langle mathbfA^T mathbfA mathbfx, mathbfx rangle = | mathbfA mathbfx |_2^2 = 1$$
and thus we have $| mathbfA^T mathbfA mathbfx |_2 ge frac1 mathbfx $.
(2) By (1), if $| mathbfA mathbfx_1 |_2 = cdots = | mathbfA mathbfx_r |_2 = 1$, we have
$$| mathbfA^T mathbfA mathbfx_1 |_2^2 + cdots + | mathbfA^T mathbfA mathbfx_r |_2^2 ge frac1+cdots+frac1.$$
Here is my question:
If $ mathbfAmathbfx_1, cdots, mathbfAmathbfx_r $ is orthonormal, can we obtain a better inequality than (2)?
Thanks.
linear-algebra orthonormal
asked Jul 30 at 4:14
user580055
213
add a comment |Â
up vote
1
down vote
favorite
(1) Note that if $| mathbfA mathbfx |_2 = 1$, then
$$| mathbfA^T mathbfA mathbfx |_2 cdot | mathbfx |_2 ge langle mathbfA^T mathbfA mathbfx, mathbfx rangle = | mathbfA mathbfx |_2^2 = 1$$
and thus we have $| mathbfA^T mathbfA mathbfx |_2 ge frac1 mathbfx $.
(2) By (1), if $| mathbfA mathbfx_1 |_2 = cdots = | mathbfA mathbfx_r |_2 = 1$, we have
$$| mathbfA^T mathbfA mathbfx_1 |_2^2 + cdots + | mathbfA^T mathbfA mathbfx_r |_2^2 ge frac1+cdots+frac1.$$
Here is my question:
If $ mathbfAmathbfx_1, cdots, mathbfAmathbfx_r $ is orthonormal, can we obtain a better inequality than (2)?
Thanks.
linear-algebra orthonormal
asked Jul 30 at 4:14
user580055
213
1
How do you want to make the inequality better? The inequality is sharp as it is.
– Batominovski
Jul 30 at 5:00
When obtaining the inequality (2), the orthogonality of the vectors is not used. I am just wondering how we can exploit the orthogonality to get the sharper bound.
– user580055
Jul 30 at 5:34
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
(1) Note that if $| mathbfA mathbfx |_2 = 1$, then
$$| mathbfA^T mathbfA mathbfx |_2 cdot | mathbfx |_2 ge langle mathbfA^T mathbfA mathbfx, mathbfx rangle = | mathbfA mathbfx |_2^2 = 1$$
and thus we have $| mathbfA^T mathbfA mathbfx |_2 ge frac1 mathbfx $.
(2) By (1), if $| mathbfA mathbfx_1 |_2 = cdots = | mathbfA mathbfx_r |_2 = 1$, we have
$$| mathbfA^T mathbfA mathbfx_1 |_2^2 + cdots + | mathbfA^T mathbfA mathbfx_r |_2^2 ge frac1+cdots+frac1.$$
Here is my question:
If $ mathbfAmathbfx_1, cdots, mathbfAmathbfx_r $ is orthonormal, can we obtain a better inequality than (2)?
Thanks.
linear-algebra orthonormal
asked Jul 30 at 4:14
user580055
213
(1) Note that if $| mathbfA mathbfx |_2 = 1$, then
$$| mathbfA^T mathbfA mathbfx |_2 cdot | mathbfx |_2 ge langle mathbfA^T mathbfA mathbfx, mathbfx rangle = | mathbfA mathbfx |_2^2 = 1$$
and thus we have $| mathbfA^T mathbfA mathbfx |_2 ge frac1 mathbfx $.
(2) By (1), if $| mathbfA mathbfx_1 |_2 = cdots = | mathbfA mathbfx_r |_2 = 1$, we have
$$| mathbfA^T mathbfA mathbfx_1 |_2^2 + cdots + | mathbfA^T mathbfA mathbfx_r |_2^2 ge frac1+cdots+frac1.$$
Here is my question:
If $ mathbfAmathbfx_1, cdots, mathbfAmathbfx_r $ is orthonormal, can we obtain a better inequality than (2)?
Thanks.
linear-algebra orthonormal
asked Jul 30 at 4:14
user580055
213
asked Jul 30 at 4:14
user580055
213
asked Jul 30 at 4:14
user580055
213
asked Jul 30 at 4:14
user580055
213
213
1
How do you want to make the inequality better? The inequality is sharp as it is.
– Batominovski
Jul 30 at 5:00
When obtaining the inequality (2), the orthogonality of the vectors is not used. I am just wondering how we can exploit the orthogonality to get the sharper bound.
– user580055
Jul 30 at 5:34
add a comment |Â
1
How do you want to make the inequality better? The inequality is sharp as it is.
– Batominovski
Jul 30 at 5:00
When obtaining the inequality (2), the orthogonality of the vectors is not used. I am just wondering how we can exploit the orthogonality to get the sharper bound.
– user580055
Jul 30 at 5:34
1
1
How do you want to make the inequality better? The inequality is sharp as it is.
– Batominovski
Jul 30 at 5:00
How do you want to make the inequality better? The inequality is sharp as it is.
– Batominovski
Jul 30 at 5:00
When obtaining the inequality (2), the orthogonality of the vectors is not used. I am just wondering how we can exploit the orthogonality to get the sharper bound.
– user580055
Jul 30 at 5:34
When obtaining the inequality (2), the orthogonality of the vectors is not used. I am just wondering how we can exploit the orthogonality to get the sharper bound.
– user580055
Jul 30 at 5:34
add a comment |Â
1 Answer
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oldest
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up vote
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Consider $A=I$ and $x_i=e_i$ the $i$-th vector of the canonical basis. Then the set $ mathbfAmathbfx_1, cdots, mathbfAmathbfx_r $ is orthonormal, but :
$$| mathbfA^T mathbfA mathbfx_1 |_2^2 + cdots + | mathbfA^T mathbfA mathbfx_r |_2^2 =r= frac1+cdots+frac1.$$
Hence the inequality cannot be "better" even if we take orthogonality into account.
answered Jul 30 at 6:47
nicomezi
3,3871819
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Consider $A=I$ and $x_i=e_i$ the $i$-th vector of the canonical basis. Then the set $ mathbfAmathbfx_1, cdots, mathbfAmathbfx_r $ is orthonormal, but :
$$| mathbfA^T mathbfA mathbfx_1 |_2^2 + cdots + | mathbfA^T mathbfA mathbfx_r |_2^2 =r= frac1+cdots+frac1.$$
Hence the inequality cannot be "better" even if we take orthogonality into account.
answered Jul 30 at 6:47
nicomezi
3,3871819
add a comment |Â
up vote
0
down vote
Consider $A=I$ and $x_i=e_i$ the $i$-th vector of the canonical basis. Then the set $ mathbfAmathbfx_1, cdots, mathbfAmathbfx_r $ is orthonormal, but :
$$| mathbfA^T mathbfA mathbfx_1 |_2^2 + cdots + | mathbfA^T mathbfA mathbfx_r |_2^2 =r= frac1+cdots+frac1.$$
Hence the inequality cannot be "better" even if we take orthogonality into account.
answered Jul 30 at 6:47
nicomezi
3,3871819
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider $A=I$ and $x_i=e_i$ the $i$-th vector of the canonical basis. Then the set $ mathbfAmathbfx_1, cdots, mathbfAmathbfx_r $ is orthonormal, but :
$$| mathbfA^T mathbfA mathbfx_1 |_2^2 + cdots + | mathbfA^T mathbfA mathbfx_r |_2^2 =r= frac1+cdots+frac1.$$
Hence the inequality cannot be "better" even if we take orthogonality into account.
answered Jul 30 at 6:47
nicomezi
3,3871819
Consider $A=I$ and $x_i=e_i$ the $i$-th vector of the canonical basis. Then the set $ mathbfAmathbfx_1, cdots, mathbfAmathbfx_r $ is orthonormal, but :
$$| mathbfA^T mathbfA mathbfx_1 |_2^2 + cdots + | mathbfA^T mathbfA mathbfx_r |_2^2 =r= frac1+cdots+frac1.$$
Hence the inequality cannot be "better" even if we take orthogonality into account.
answered Jul 30 at 6:47
nicomezi
3,3871819
answered Jul 30 at 6:47
nicomezi
3,3871819
answered Jul 30 at 6:47
nicomezi
3,3871819
answered Jul 30 at 6:47
nicomezi
3,3871819
3,3871819
add a comment |Â
add a comment |Â
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1
How do you want to make the inequality better? The inequality is sharp as it is.
– Batominovski
Jul 30 at 5:00
When obtaining the inequality (2), the orthogonality of the vectors is not used. I am just wondering how we can exploit the orthogonality to get the sharper bound.
– user580055
Jul 30 at 5:34