Prove $M(M+I)^{-1}M succeq frac{1}{2}I$ for any PSD matrix $M succeq I$
$begingroup$
For any $n times n$ symmetrix matrices A, B, we define $A succeq B$ if and only if $v^TAv succeq v^TBv$ for all $v in mathbb{R}^n$
linear-algebra matrices inequality
asked Jan 14 at 23:48
SaveMeFromConvOptSaveMeFromConvOpt
32
$endgroup$
add a comment |
$begingroup$
For any $n times n$ symmetrix matrices A, B, we define $A succeq B$ if and only if $v^TAv succeq v^TBv$ for all $v in mathbb{R}^n$
linear-algebra matrices inequality
asked Jan 14 at 23:48
SaveMeFromConvOptSaveMeFromConvOpt
32
$endgroup$
$begingroup$
Do you mean for all $vin mathbb{R}^n$?
$endgroup$
– Ryan Greyling
Jan 14 at 23:54
$begingroup$
@RyanGreyling Right! Edited
$endgroup$
– SaveMeFromConvOpt
Jan 14 at 23:57
add a comment |
$begingroup$
For any $n times n$ symmetrix matrices A, B, we define $A succeq B$ if and only if $v^TAv succeq v^TBv$ for all $v in mathbb{R}^n$
linear-algebra matrices inequality
asked Jan 14 at 23:48
SaveMeFromConvOptSaveMeFromConvOpt
32
$endgroup$
For any $n times n$ symmetrix matrices A, B, we define $A succeq B$ if and only if $v^TAv succeq v^TBv$ for all $v in mathbb{R}^n$
linear-algebra matrices inequality
linear-algebra matrices inequality
asked Jan 14 at 23:48
SaveMeFromConvOptSaveMeFromConvOpt
32
asked Jan 14 at 23:48
SaveMeFromConvOptSaveMeFromConvOpt
32
edited Jan 14 at 23:57
SaveMeFromConvOpt
asked Jan 14 at 23:48
SaveMeFromConvOptSaveMeFromConvOpt
32
asked Jan 14 at 23:48
SaveMeFromConvOptSaveMeFromConvOpt
32
asked Jan 14 at 23:48
SaveMeFromConvOptSaveMeFromConvOpt
32
32
$begingroup$
Do you mean for all $vin mathbb{R}^n$?
$endgroup$
– Ryan Greyling
Jan 14 at 23:54
$begingroup$
@RyanGreyling Right! Edited
$endgroup$
– SaveMeFromConvOpt
Jan 14 at 23:57
add a comment |
$begingroup$
Do you mean for all $vin mathbb{R}^n$?
$endgroup$
– Ryan Greyling
Jan 14 at 23:54
$begingroup$
@RyanGreyling Right! Edited
$endgroup$
– SaveMeFromConvOpt
Jan 14 at 23:57
$begingroup$
Do you mean for all $vin mathbb{R}^n$?
$endgroup$
– Ryan Greyling
Jan 14 at 23:54
$begingroup$
Do you mean for all $vin mathbb{R}^n$?
$endgroup$
– Ryan Greyling
Jan 14 at 23:54
$begingroup$
@RyanGreyling Right! Edited
$endgroup$
– SaveMeFromConvOpt
Jan 14 at 23:57
$begingroup$
@RyanGreyling Right! Edited
$endgroup$
– SaveMeFromConvOpt
Jan 14 at 23:57
add a comment |
1 Answer
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$begingroup$
Since $M succeq I$, $M$ is PD so it is invertible. We have $M^2 succeq I$ and then $M succeq M^{-1}$. From this and the fact that $M succeq I$ we have $$2M succeq M+I succeq M^{-1}+I$$ The left and right sides of this inequality yield $$M succeq frac{1}{2}I+frac{1}{2}M^{-1}$$ $$M succeq frac{1}{2}M^{-1}(M+I)$$ $$M(M+I)^{-1}M succeq frac{1}{2}I$$
answered Jan 15 at 0:29
Ryan GreylingRyan Greyling
3599
$endgroup$
$begingroup$
When solving problems like this I find that it's easiest to assume that the end statement is true, and then work backwards from there to prove the initial condition. The written proof is just the reverse of this process.
$endgroup$
– Ryan Greyling
Jan 15 at 0:34
1
$begingroup$
Being positive definite implies invertibility. Being PSD does not. However, you can deduce that $M$ is positive definite from $M succeq I$. In particular, if $v neq 0$, then $v^top M v ge v^top I v = |v|^2 > 0$. Similar reasoning will also justify the existence of $(M + I)^{-1}$.
$endgroup$
– Theo Bendit
Jan 15 at 0:42
$begingroup$
Thanks @TheoBendit , edited my answer to clarify that $M$ is PD.
$endgroup$
– Ryan Greyling
Jan 15 at 0:55
add a comment |
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1 Answer
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$begingroup$
Since $M succeq I$, $M$ is PD so it is invertible. We have $M^2 succeq I$ and then $M succeq M^{-1}$. From this and the fact that $M succeq I$ we have $$2M succeq M+I succeq M^{-1}+I$$ The left and right sides of this inequality yield $$M succeq frac{1}{2}I+frac{1}{2}M^{-1}$$ $$M succeq frac{1}{2}M^{-1}(M+I)$$ $$M(M+I)^{-1}M succeq frac{1}{2}I$$
answered Jan 15 at 0:29
Ryan GreylingRyan Greyling
3599
$endgroup$
$begingroup$
When solving problems like this I find that it's easiest to assume that the end statement is true, and then work backwards from there to prove the initial condition. The written proof is just the reverse of this process.
$endgroup$
– Ryan Greyling
Jan 15 at 0:34
1
$begingroup$
Being positive definite implies invertibility. Being PSD does not. However, you can deduce that $M$ is positive definite from $M succeq I$. In particular, if $v neq 0$, then $v^top M v ge v^top I v = |v|^2 > 0$. Similar reasoning will also justify the existence of $(M + I)^{-1}$.
$endgroup$
– Theo Bendit
Jan 15 at 0:42
$begingroup$
Thanks @TheoBendit , edited my answer to clarify that $M$ is PD.
$endgroup$
– Ryan Greyling
Jan 15 at 0:55
add a comment |
$begingroup$
Since $M succeq I$, $M$ is PD so it is invertible. We have $M^2 succeq I$ and then $M succeq M^{-1}$. From this and the fact that $M succeq I$ we have $$2M succeq M+I succeq M^{-1}+I$$ The left and right sides of this inequality yield $$M succeq frac{1}{2}I+frac{1}{2}M^{-1}$$ $$M succeq frac{1}{2}M^{-1}(M+I)$$ $$M(M+I)^{-1}M succeq frac{1}{2}I$$
answered Jan 15 at 0:29
Ryan GreylingRyan Greyling
3599
$endgroup$
$begingroup$
When solving problems like this I find that it's easiest to assume that the end statement is true, and then work backwards from there to prove the initial condition. The written proof is just the reverse of this process.
$endgroup$
– Ryan Greyling
Jan 15 at 0:34
1
$begingroup$
Being positive definite implies invertibility. Being PSD does not. However, you can deduce that $M$ is positive definite from $M succeq I$. In particular, if $v neq 0$, then $v^top M v ge v^top I v = |v|^2 > 0$. Similar reasoning will also justify the existence of $(M + I)^{-1}$.
$endgroup$
– Theo Bendit
Jan 15 at 0:42
$begingroup$
Thanks @TheoBendit , edited my answer to clarify that $M$ is PD.
$endgroup$
– Ryan Greyling
Jan 15 at 0:55
add a comment |
$begingroup$
Since $M succeq I$, $M$ is PD so it is invertible. We have $M^2 succeq I$ and then $M succeq M^{-1}$. From this and the fact that $M succeq I$ we have $$2M succeq M+I succeq M^{-1}+I$$ The left and right sides of this inequality yield $$M succeq frac{1}{2}I+frac{1}{2}M^{-1}$$ $$M succeq frac{1}{2}M^{-1}(M+I)$$ $$M(M+I)^{-1}M succeq frac{1}{2}I$$
answered Jan 15 at 0:29
Ryan GreylingRyan Greyling
3599
$endgroup$
Since $M succeq I$, $M$ is PD so it is invertible. We have $M^2 succeq I$ and then $M succeq M^{-1}$. From this and the fact that $M succeq I$ we have $$2M succeq M+I succeq M^{-1}+I$$ The left and right sides of this inequality yield $$M succeq frac{1}{2}I+frac{1}{2}M^{-1}$$ $$M succeq frac{1}{2}M^{-1}(M+I)$$ $$M(M+I)^{-1}M succeq frac{1}{2}I$$
answered Jan 15 at 0:29
Ryan GreylingRyan Greyling
3599
edited Jan 15 at 0:54
answered Jan 15 at 0:29
Ryan GreylingRyan Greyling
3599
answered Jan 15 at 0:29
Ryan GreylingRyan Greyling
3599
answered Jan 15 at 0:29
Ryan GreylingRyan Greyling
3599
3599
$begingroup$
When solving problems like this I find that it's easiest to assume that the end statement is true, and then work backwards from there to prove the initial condition. The written proof is just the reverse of this process.
$endgroup$
– Ryan Greyling
Jan 15 at 0:34
1
$begingroup$
Being positive definite implies invertibility. Being PSD does not. However, you can deduce that $M$ is positive definite from $M succeq I$. In particular, if $v neq 0$, then $v^top M v ge v^top I v = |v|^2 > 0$. Similar reasoning will also justify the existence of $(M + I)^{-1}$.
$endgroup$
– Theo Bendit
Jan 15 at 0:42
$begingroup$
Thanks @TheoBendit , edited my answer to clarify that $M$ is PD.
$endgroup$
– Ryan Greyling
Jan 15 at 0:55
add a comment |
$begingroup$
When solving problems like this I find that it's easiest to assume that the end statement is true, and then work backwards from there to prove the initial condition. The written proof is just the reverse of this process.
$endgroup$
– Ryan Greyling
Jan 15 at 0:34
1
$begingroup$
Being positive definite implies invertibility. Being PSD does not. However, you can deduce that $M$ is positive definite from $M succeq I$. In particular, if $v neq 0$, then $v^top M v ge v^top I v = |v|^2 > 0$. Similar reasoning will also justify the existence of $(M + I)^{-1}$.
$endgroup$
– Theo Bendit
Jan 15 at 0:42
$begingroup$
Thanks @TheoBendit , edited my answer to clarify that $M$ is PD.
$endgroup$
– Ryan Greyling
Jan 15 at 0:55
$begingroup$
When solving problems like this I find that it's easiest to assume that the end statement is true, and then work backwards from there to prove the initial condition. The written proof is just the reverse of this process.
$endgroup$
– Ryan Greyling
Jan 15 at 0:34
$begingroup$
When solving problems like this I find that it's easiest to assume that the end statement is true, and then work backwards from there to prove the initial condition. The written proof is just the reverse of this process.
$endgroup$
– Ryan Greyling
Jan 15 at 0:34
1
1
$begingroup$
Being positive definite implies invertibility. Being PSD does not. However, you can deduce that $M$ is positive definite from $M succeq I$. In particular, if $v neq 0$, then $v^top M v ge v^top I v = |v|^2 > 0$. Similar reasoning will also justify the existence of $(M + I)^{-1}$.
$endgroup$
– Theo Bendit
Jan 15 at 0:42
$begingroup$
Being positive definite implies invertibility. Being PSD does not. However, you can deduce that $M$ is positive definite from $M succeq I$. In particular, if $v neq 0$, then $v^top M v ge v^top I v = |v|^2 > 0$. Similar reasoning will also justify the existence of $(M + I)^{-1}$.
$endgroup$
– Theo Bendit
Jan 15 at 0:42
$begingroup$
Thanks @TheoBendit , edited my answer to clarify that $M$ is PD.
$endgroup$
– Ryan Greyling
Jan 15 at 0:55
$begingroup$
Thanks @TheoBendit , edited my answer to clarify that $M$ is PD.
$endgroup$
– Ryan Greyling
Jan 15 at 0:55
add a comment |
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$begingroup$
Do you mean for all $vin mathbb{R}^n$?
$endgroup$
– Ryan Greyling
Jan 14 at 23:54
$begingroup$
@RyanGreyling Right! Edited
$endgroup$
– SaveMeFromConvOpt
Jan 14 at 23:57