Cauchy-Schwarz inequality to prove $A$ is spd
$begingroup$
Let $vec v$ be a non-zero vector in $Bbb R^n$
such that $|vec v| = 1$ and let $A = I −βvec vvec v^T$,
with $β > 0$.
(a) Show that if $β ≤ 1$, then $A$ is spd.
Hint: use Cauchy-Schwarz inequality
Proving its symmetric is of course very straightforward. However, I am struggling with figuring out why if $β ≤ 1$ then $A$ is spd.
linear-algebra cauchy-schwarz-inequality
edited Dec 14 '18 at 9:48
Tianlalu
3,08421138
asked Dec 14 '18 at 7:22
TheFlyingPandaaTheFlyingPandaa
1
$endgroup$
add a comment |
$begingroup$
Let $vec v$ be a non-zero vector in $Bbb R^n$
such that $|vec v| = 1$ and let $A = I −βvec vvec v^T$,
with $β > 0$.
(a) Show that if $β ≤ 1$, then $A$ is spd.
Hint: use Cauchy-Schwarz inequality
Proving its symmetric is of course very straightforward. However, I am struggling with figuring out why if $β ≤ 1$ then $A$ is spd.
linear-algebra cauchy-schwarz-inequality
edited Dec 14 '18 at 9:48
Tianlalu
3,08421138
asked Dec 14 '18 at 7:22
TheFlyingPandaaTheFlyingPandaa
1
$endgroup$
add a comment |
$begingroup$
Let $vec v$ be a non-zero vector in $Bbb R^n$
such that $|vec v| = 1$ and let $A = I −βvec vvec v^T$,
with $β > 0$.
(a) Show that if $β ≤ 1$, then $A$ is spd.
Hint: use Cauchy-Schwarz inequality
Proving its symmetric is of course very straightforward. However, I am struggling with figuring out why if $β ≤ 1$ then $A$ is spd.
linear-algebra cauchy-schwarz-inequality
edited Dec 14 '18 at 9:48
Tianlalu
3,08421138
asked Dec 14 '18 at 7:22
TheFlyingPandaaTheFlyingPandaa
1
$endgroup$
Let $vec v$ be a non-zero vector in $Bbb R^n$
such that $|vec v| = 1$ and let $A = I −βvec vvec v^T$,
with $β > 0$.
(a) Show that if $β ≤ 1$, then $A$ is spd.
Hint: use Cauchy-Schwarz inequality
Proving its symmetric is of course very straightforward. However, I am struggling with figuring out why if $β ≤ 1$ then $A$ is spd.
linear-algebra cauchy-schwarz-inequality
linear-algebra cauchy-schwarz-inequality
edited Dec 14 '18 at 9:48
Tianlalu
3,08421138
asked Dec 14 '18 at 7:22
TheFlyingPandaaTheFlyingPandaa
1
edited Dec 14 '18 at 9:48
Tianlalu
3,08421138
asked Dec 14 '18 at 7:22
TheFlyingPandaaTheFlyingPandaa
1
edited Dec 14 '18 at 9:48
Tianlalu
3,08421138
edited Dec 14 '18 at 9:48
Tianlalu
3,08421138
edited Dec 14 '18 at 9:48
Tianlalu
3,08421138
3,08421138
asked Dec 14 '18 at 7:22
TheFlyingPandaaTheFlyingPandaa
1
asked Dec 14 '18 at 7:22
TheFlyingPandaaTheFlyingPandaa
1
asked Dec 14 '18 at 7:22
TheFlyingPandaaTheFlyingPandaa
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
$x^{T}Ax=x^{T}x-beta x^{T}vv^{T}x=|x|^{2}-beta |v^{T}x|^{2} geq 0$ because $|v^{T}x|leq |v||x|=|x|$ by Cauchy - Schwarz inequality.
answered Dec 14 '18 at 8:08
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
$x^{T}Ax=x^{T}x-beta x^{T}vv^{T}x=|x|^{2}-beta |v^{T}x|^{2} geq 0$ because $|v^{T}x|leq |v||x|=|x|$ by Cauchy - Schwarz inequality.
answered Dec 14 '18 at 8:08
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
add a comment |
$begingroup$
$x^{T}Ax=x^{T}x-beta x^{T}vv^{T}x=|x|^{2}-beta |v^{T}x|^{2} geq 0$ because $|v^{T}x|leq |v||x|=|x|$ by Cauchy - Schwarz inequality.
answered Dec 14 '18 at 8:08
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
add a comment |
$begingroup$
$x^{T}Ax=x^{T}x-beta x^{T}vv^{T}x=|x|^{2}-beta |v^{T}x|^{2} geq 0$ because $|v^{T}x|leq |v||x|=|x|$ by Cauchy - Schwarz inequality.
answered Dec 14 '18 at 8:08
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
$x^{T}Ax=x^{T}x-beta x^{T}vv^{T}x=|x|^{2}-beta |v^{T}x|^{2} geq 0$ because $|v^{T}x|leq |v||x|=|x|$ by Cauchy - Schwarz inequality.
answered Dec 14 '18 at 8:08
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
answered Dec 14 '18 at 8:08
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
answered Dec 14 '18 at 8:08
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
answered Dec 14 '18 at 8:08
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
64.5k42665
add a comment |
add a comment |
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