Revisit “Inequality between Frobenius and nuclear norm”
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I am reading the following question:
Inequality between Frobenius and nuclear norm
I know $$|X|_* = sum_{i=1}^r sigma_i(X)$$and $$|X|_F = bigg(sum_{i=1}^r sigma^2_i(X)bigg)^{1/2}$$
I try to prove it by Cauchy-Schwarz inequality; however, I cannot see how to use C-S inequality to prove this. Could anyone please give me a small hint? Thanks in advanced.
linear-algebra matrices cauchy-schwarz-inequality nuclear-norm
asked Dec 5 '18 at 22:25
DennyDenny
31729
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$begingroup$
I am reading the following question:
Inequality between Frobenius and nuclear norm
I know $$|X|_* = sum_{i=1}^r sigma_i(X)$$and $$|X|_F = bigg(sum_{i=1}^r sigma^2_i(X)bigg)^{1/2}$$
I try to prove it by Cauchy-Schwarz inequality; however, I cannot see how to use C-S inequality to prove this. Could anyone please give me a small hint? Thanks in advanced.
linear-algebra matrices cauchy-schwarz-inequality nuclear-norm
asked Dec 5 '18 at 22:25
DennyDenny
31729
$endgroup$
add a comment |
$begingroup$
I am reading the following question:
Inequality between Frobenius and nuclear norm
I know $$|X|_* = sum_{i=1}^r sigma_i(X)$$and $$|X|_F = bigg(sum_{i=1}^r sigma^2_i(X)bigg)^{1/2}$$
I try to prove it by Cauchy-Schwarz inequality; however, I cannot see how to use C-S inequality to prove this. Could anyone please give me a small hint? Thanks in advanced.
linear-algebra matrices cauchy-schwarz-inequality nuclear-norm
asked Dec 5 '18 at 22:25
DennyDenny
31729
$endgroup$
I am reading the following question:
Inequality between Frobenius and nuclear norm
I know $$|X|_* = sum_{i=1}^r sigma_i(X)$$and $$|X|_F = bigg(sum_{i=1}^r sigma^2_i(X)bigg)^{1/2}$$
I try to prove it by Cauchy-Schwarz inequality; however, I cannot see how to use C-S inequality to prove this. Could anyone please give me a small hint? Thanks in advanced.
linear-algebra matrices cauchy-schwarz-inequality nuclear-norm
linear-algebra matrices cauchy-schwarz-inequality nuclear-norm
asked Dec 5 '18 at 22:25
DennyDenny
31729
asked Dec 5 '18 at 22:25
DennyDenny
31729
asked Dec 5 '18 at 22:25
DennyDenny
31729
asked Dec 5 '18 at 22:25
DennyDenny
31729
asked Dec 5 '18 at 22:25
DennyDenny
31729
31729
add a comment |
add a comment |
1 Answer
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Cauchy—Schwarz tells you that
$$
sum_{i=1}^r a_i b_i leq left(sum_{i=1}^r a_i^2right)^{1/2}left(sum_{i=1}^r b_i^2right)^{1/2}
$$
Apply it with $a_i = 1$, $b_i = sigma_i(X)$ (for $1leq ileq r$).
answered Dec 5 '18 at 22:38
Clement C.Clement C.
49.8k33886
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Cauchy—Schwarz tells you that
$$
sum_{i=1}^r a_i b_i leq left(sum_{i=1}^r a_i^2right)^{1/2}left(sum_{i=1}^r b_i^2right)^{1/2}
$$
Apply it with $a_i = 1$, $b_i = sigma_i(X)$ (for $1leq ileq r$).
answered Dec 5 '18 at 22:38
Clement C.Clement C.
49.8k33886
$endgroup$
add a comment |
$begingroup$
Cauchy—Schwarz tells you that
$$
sum_{i=1}^r a_i b_i leq left(sum_{i=1}^r a_i^2right)^{1/2}left(sum_{i=1}^r b_i^2right)^{1/2}
$$
Apply it with $a_i = 1$, $b_i = sigma_i(X)$ (for $1leq ileq r$).
answered Dec 5 '18 at 22:38
Clement C.Clement C.
49.8k33886
$endgroup$
add a comment |
$begingroup$
Cauchy—Schwarz tells you that
$$
sum_{i=1}^r a_i b_i leq left(sum_{i=1}^r a_i^2right)^{1/2}left(sum_{i=1}^r b_i^2right)^{1/2}
$$
Apply it with $a_i = 1$, $b_i = sigma_i(X)$ (for $1leq ileq r$).
answered Dec 5 '18 at 22:38
Clement C.Clement C.
49.8k33886
$endgroup$
Cauchy—Schwarz tells you that
$$
sum_{i=1}^r a_i b_i leq left(sum_{i=1}^r a_i^2right)^{1/2}left(sum_{i=1}^r b_i^2right)^{1/2}
$$
Apply it with $a_i = 1$, $b_i = sigma_i(X)$ (for $1leq ileq r$).
answered Dec 5 '18 at 22:38
Clement C.Clement C.
49.8k33886
answered Dec 5 '18 at 22:38
Clement C.Clement C.
49.8k33886
answered Dec 5 '18 at 22:38
Clement C.Clement C.
49.8k33886
answered Dec 5 '18 at 22:38
Clement C.Clement C.
49.8k33886
49.8k33886
add a comment |
add a comment |
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