$f(t)=textdet(A+tB)$ is continuous for symmetric positive semi-definite matrices $A$ and $B$?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
We have any symmetric positive semi-definite matrices $A$ and $B$
where $textdetB=0$.
A function $f(t)=textdet(A+tB)$ is continuous on $tin[0,1]$?
$f(t_1)-f(t_2)=textdet(A+t_1B)-textdet(A+t_2B)$. From
here, I can't go farther. Thanks in advance.
continuity matrix-calculus positive-semidefinite
asked Sep 9 at 7:16
kayak
578318
add a comment |Â
up vote
3
down vote
favorite
We have any symmetric positive semi-definite matrices $A$ and $B$
where $textdetB=0$.
A function $f(t)=textdet(A+tB)$ is continuous on $tin[0,1]$?
$f(t_1)-f(t_2)=textdet(A+t_1B)-textdet(A+t_2B)$. From
here, I can't go farther. Thanks in advance.
continuity matrix-calculus positive-semidefinite
asked Sep 9 at 7:16
kayak
578318
Do you know any formula for the determinant, or any calculation rule?
– Thomas
Sep 9 at 7:21
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
We have any symmetric positive semi-definite matrices $A$ and $B$
where $textdetB=0$.
A function $f(t)=textdet(A+tB)$ is continuous on $tin[0,1]$?
$f(t_1)-f(t_2)=textdet(A+t_1B)-textdet(A+t_2B)$. From
here, I can't go farther. Thanks in advance.
continuity matrix-calculus positive-semidefinite
asked Sep 9 at 7:16
kayak
578318
We have any symmetric positive semi-definite matrices $A$ and $B$
where $textdetB=0$.
A function $f(t)=textdet(A+tB)$ is continuous on $tin[0,1]$?
$f(t_1)-f(t_2)=textdet(A+t_1B)-textdet(A+t_2B)$. From
here, I can't go farther. Thanks in advance.
continuity matrix-calculus positive-semidefinite
continuity matrix-calculus positive-semidefinite
asked Sep 9 at 7:16
kayak
578318
asked Sep 9 at 7:16
kayak
578318
asked Sep 9 at 7:16
kayak
578318
asked Sep 9 at 7:16
kayak
578318
asked Sep 9 at 7:16
kayak
578318
578318
Do you know any formula for the determinant, or any calculation rule?
– Thomas
Sep 9 at 7:21
add a comment |Â
Do you know any formula for the determinant, or any calculation rule?
– Thomas
Sep 9 at 7:21
Do you know any formula for the determinant, or any calculation rule?
– Thomas
Sep 9 at 7:21
Do you know any formula for the determinant, or any calculation rule?
– Thomas
Sep 9 at 7:21
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Expanding out $f(t)=det(A+tB)$ using the formula for the determinant
shows that $f(t)$ is a polynomial in $t$. All polynomials are continuous
on all of $Bbb R$. This is true for all matrices $A$ and $B$.
answered Sep 9 at 7:21
Lord Shark the Unknown
90.2k955117
Oh I see. Thanks!
– kayak
Sep 9 at 7:22
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Expanding out $f(t)=det(A+tB)$ using the formula for the determinant
shows that $f(t)$ is a polynomial in $t$. All polynomials are continuous
on all of $Bbb R$. This is true for all matrices $A$ and $B$.
answered Sep 9 at 7:21
Lord Shark the Unknown
90.2k955117
Oh I see. Thanks!
– kayak
Sep 9 at 7:22
add a comment |Â
up vote
3
down vote
accepted
Expanding out $f(t)=det(A+tB)$ using the formula for the determinant
shows that $f(t)$ is a polynomial in $t$. All polynomials are continuous
on all of $Bbb R$. This is true for all matrices $A$ and $B$.
answered Sep 9 at 7:21
Lord Shark the Unknown
90.2k955117
Oh I see. Thanks!
– kayak
Sep 9 at 7:22
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Expanding out $f(t)=det(A+tB)$ using the formula for the determinant
shows that $f(t)$ is a polynomial in $t$. All polynomials are continuous
on all of $Bbb R$. This is true for all matrices $A$ and $B$.
answered Sep 9 at 7:21
Lord Shark the Unknown
90.2k955117
Expanding out $f(t)=det(A+tB)$ using the formula for the determinant
shows that $f(t)$ is a polynomial in $t$. All polynomials are continuous
on all of $Bbb R$. This is true for all matrices $A$ and $B$.
answered Sep 9 at 7:21
Lord Shark the Unknown
90.2k955117
answered Sep 9 at 7:21
Lord Shark the Unknown
90.2k955117
answered Sep 9 at 7:21
Lord Shark the Unknown
90.2k955117
answered Sep 9 at 7:21
Lord Shark the Unknown
90.2k955117
90.2k955117
Oh I see. Thanks!
– kayak
Sep 9 at 7:22
add a comment |Â
Oh I see. Thanks!
– kayak
Sep 9 at 7:22
Oh I see. Thanks!
– kayak
Sep 9 at 7:22
Oh I see. Thanks!
– kayak
Sep 9 at 7:22
add a comment |Â
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Do you know any formula for the determinant, or any calculation rule?
– Thomas
Sep 9 at 7:21