On invertibility of $A+E$ where $||E||_2<$ smallest singular value of $A$ and $||A^{-1}E||_2<1$
$begingroup$
Let $A,E in M_n(mathbb C)$ . Suppose $sigma_min >0$ be the smallest singular value of $A$ and $||E||_2 < sigma_min$. Suppose $||A^{-1}E||_2 <1$. Then how to show that $A+E$ is invertible ?
My work : Going by contradiction; assume ,if possible, $det (A+E)=0$, then $det (I+A^{-1}E)=0$. So $-1$ is an eigenvalue of $A^{-1}E$, so $1=|-1|le ||I+A^{-1}E||_2$, so $||A^{-1}E||_2 <1 le ||I+A^{-1}E||_2$ ; but I am unable to proceed further.
Please help.
NOTE: Here $||M||_2:=sup_{||x||_2=1}||Mx||_2=sigma_max$
linear-algebra matrices eigenvalues-eigenvectors svd singularvalues
asked Jan 12 at 4:53
user521337user521337
1,1791417
$endgroup$
add a comment |
$begingroup$
Let $A,E in M_n(mathbb C)$ . Suppose $sigma_min >0$ be the smallest singular value of $A$ and $||E||_2 < sigma_min$. Suppose $||A^{-1}E||_2 <1$. Then how to show that $A+E$ is invertible ?
My work : Going by contradiction; assume ,if possible, $det (A+E)=0$, then $det (I+A^{-1}E)=0$. So $-1$ is an eigenvalue of $A^{-1}E$, so $1=|-1|le ||I+A^{-1}E||_2$, so $||A^{-1}E||_2 <1 le ||I+A^{-1}E||_2$ ; but I am unable to proceed further.
Please help.
NOTE: Here $||M||_2:=sup_{||x||_2=1}||Mx||_2=sigma_max$
linear-algebra matrices eigenvalues-eigenvectors svd singularvalues
asked Jan 12 at 4:53
user521337user521337
1,1791417
$endgroup$
add a comment |
$begingroup$
Let $A,E in M_n(mathbb C)$ . Suppose $sigma_min >0$ be the smallest singular value of $A$ and $||E||_2 < sigma_min$. Suppose $||A^{-1}E||_2 <1$. Then how to show that $A+E$ is invertible ?
My work : Going by contradiction; assume ,if possible, $det (A+E)=0$, then $det (I+A^{-1}E)=0$. So $-1$ is an eigenvalue of $A^{-1}E$, so $1=|-1|le ||I+A^{-1}E||_2$, so $||A^{-1}E||_2 <1 le ||I+A^{-1}E||_2$ ; but I am unable to proceed further.
Please help.
NOTE: Here $||M||_2:=sup_{||x||_2=1}||Mx||_2=sigma_max$
linear-algebra matrices eigenvalues-eigenvectors svd singularvalues
asked Jan 12 at 4:53
user521337user521337
1,1791417
$endgroup$
Let $A,E in M_n(mathbb C)$ . Suppose $sigma_min >0$ be the smallest singular value of $A$ and $||E||_2 < sigma_min$. Suppose $||A^{-1}E||_2 <1$. Then how to show that $A+E$ is invertible ?
My work : Going by contradiction; assume ,if possible, $det (A+E)=0$, then $det (I+A^{-1}E)=0$. So $-1$ is an eigenvalue of $A^{-1}E$, so $1=|-1|le ||I+A^{-1}E||_2$, so $||A^{-1}E||_2 <1 le ||I+A^{-1}E||_2$ ; but I am unable to proceed further.
Please help.
NOTE: Here $||M||_2:=sup_{||x||_2=1}||Mx||_2=sigma_max$
linear-algebra matrices eigenvalues-eigenvectors svd singularvalues
linear-algebra matrices eigenvalues-eigenvectors svd singularvalues
asked Jan 12 at 4:53
user521337user521337
1,1791417
asked Jan 12 at 4:53
user521337user521337
1,1791417
edited Jan 12 at 6:53
user521337
asked Jan 12 at 4:53
user521337user521337
1,1791417
asked Jan 12 at 4:53
user521337user521337
1,1791417
asked Jan 12 at 4:53
user521337user521337
1,1791417
1,1791417
add a comment |
add a comment |
1 Answer
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$begingroup$
Suppose that $(A+E)(x)=0$, $xneq 0$, this implies that $A(x)=-E(x)$ and $x=-A^{-1}E(x)$, we deduce that $|x|=|A^{-1}E(x)|leq |A^{-1}E||x|<|x|$. Contradiction. This implies that $Ker(A+E)={0}$ and $A+E$ is invertible.
answered Jan 12 at 4:58
Tsemo AristideTsemo Aristide
59.3k11445
$endgroup$
$begingroup$
ah, so simple ... I was thinking the wrong way ... thanks !
$endgroup$
– user521337
Jan 12 at 5:03
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Suppose that $(A+E)(x)=0$, $xneq 0$, this implies that $A(x)=-E(x)$ and $x=-A^{-1}E(x)$, we deduce that $|x|=|A^{-1}E(x)|leq |A^{-1}E||x|<|x|$. Contradiction. This implies that $Ker(A+E)={0}$ and $A+E$ is invertible.
answered Jan 12 at 4:58
Tsemo AristideTsemo Aristide
59.3k11445
$endgroup$
$begingroup$
ah, so simple ... I was thinking the wrong way ... thanks !
$endgroup$
– user521337
Jan 12 at 5:03
add a comment |
$begingroup$
Suppose that $(A+E)(x)=0$, $xneq 0$, this implies that $A(x)=-E(x)$ and $x=-A^{-1}E(x)$, we deduce that $|x|=|A^{-1}E(x)|leq |A^{-1}E||x|<|x|$. Contradiction. This implies that $Ker(A+E)={0}$ and $A+E$ is invertible.
answered Jan 12 at 4:58
Tsemo AristideTsemo Aristide
59.3k11445
$endgroup$
$begingroup$
ah, so simple ... I was thinking the wrong way ... thanks !
$endgroup$
– user521337
Jan 12 at 5:03
add a comment |
$begingroup$
Suppose that $(A+E)(x)=0$, $xneq 0$, this implies that $A(x)=-E(x)$ and $x=-A^{-1}E(x)$, we deduce that $|x|=|A^{-1}E(x)|leq |A^{-1}E||x|<|x|$. Contradiction. This implies that $Ker(A+E)={0}$ and $A+E$ is invertible.
answered Jan 12 at 4:58
Tsemo AristideTsemo Aristide
59.3k11445
$endgroup$
Suppose that $(A+E)(x)=0$, $xneq 0$, this implies that $A(x)=-E(x)$ and $x=-A^{-1}E(x)$, we deduce that $|x|=|A^{-1}E(x)|leq |A^{-1}E||x|<|x|$. Contradiction. This implies that $Ker(A+E)={0}$ and $A+E$ is invertible.
answered Jan 12 at 4:58
Tsemo AristideTsemo Aristide
59.3k11445
answered Jan 12 at 4:58
Tsemo AristideTsemo Aristide
59.3k11445
answered Jan 12 at 4:58
Tsemo AristideTsemo Aristide
59.3k11445
answered Jan 12 at 4:58
Tsemo AristideTsemo Aristide
59.3k11445
59.3k11445
$begingroup$
ah, so simple ... I was thinking the wrong way ... thanks !
$endgroup$
– user521337
Jan 12 at 5:03
add a comment |
$begingroup$
ah, so simple ... I was thinking the wrong way ... thanks !
$endgroup$
– user521337
Jan 12 at 5:03
$begingroup$
ah, so simple ... I was thinking the wrong way ... thanks !
$endgroup$
– user521337
Jan 12 at 5:03
$begingroup$
ah, so simple ... I was thinking the wrong way ... thanks !
$endgroup$
– user521337
Jan 12 at 5:03
add a comment |
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