Invertibility of $(textbf{A}^Ttextbf{A}+epsilon textbf{I})$?












1












$begingroup$


I'm given a problem:



$sigma_1 geq sigma_2 geq ... geq sigma_r$ are the nonzero singular values of $textbf{A}inmathbb{R}^{Mtimes N}$. If $epsilon neq 0$ is a real scalar, s.t. $|epsilon| < sigma^{2}_r$, show that $(textbf{A}^Ttextbf{A}+epsilon textbf{I})$ is invertible.



I found the resources Why is $A^TA$ invertible if $A$ has independent columns? and Matrix inverse of $A + epsilon I$, where $A$ is invertible



But I'm not sure how useful they are. The first is in the case where A has independent columns, which is not necessarily true here, and the second presumes A is invertible. I believe that $textbf{A}^Ttextbf{A}$ is invertible by definition, but I'm not sure if I can just plug $textbf{A}^Ttextbf{A}$ in everywhere that post uses A and follow through. That also wouldn't help me understand the problem, just blindly substitute into a solution.



Can anyone help me understand WHY $(textbf{A}^Ttextbf{A}+epsilon textbf{I})$ is invertible? And/or point me in the right direction to construct a proof of it?






linear-algebra inverse







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$endgroup$












  • $begingroup$
    Just a thought, you can diagonalize $A^TA$, so it shouldnt be too hard to analyze everything after a suitable change of basis
    $endgroup$
    – qbert
    Dec 5 '18 at 21:59
















1












$begingroup$


I'm given a problem:



$sigma_1 geq sigma_2 geq ... geq sigma_r$ are the nonzero singular values of $textbf{A}inmathbb{R}^{Mtimes N}$. If $epsilon neq 0$ is a real scalar, s.t. $|epsilon| < sigma^{2}_r$, show that $(textbf{A}^Ttextbf{A}+epsilon textbf{I})$ is invertible.



I found the resources Why is $A^TA$ invertible if $A$ has independent columns? and Matrix inverse of $A + epsilon I$, where $A$ is invertible



But I'm not sure how useful they are. The first is in the case where A has independent columns, which is not necessarily true here, and the second presumes A is invertible. I believe that $textbf{A}^Ttextbf{A}$ is invertible by definition, but I'm not sure if I can just plug $textbf{A}^Ttextbf{A}$ in everywhere that post uses A and follow through. That also wouldn't help me understand the problem, just blindly substitute into a solution.



Can anyone help me understand WHY $(textbf{A}^Ttextbf{A}+epsilon textbf{I})$ is invertible? And/or point me in the right direction to construct a proof of it?






linear-algebra inverse







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just a thought, you can diagonalize $A^TA$, so it shouldnt be too hard to analyze everything after a suitable change of basis
    $endgroup$
    – qbert
    Dec 5 '18 at 21:59














1












1








1





$begingroup$


I'm given a problem:



$sigma_1 geq sigma_2 geq ... geq sigma_r$ are the nonzero singular values of $textbf{A}inmathbb{R}^{Mtimes N}$. If $epsilon neq 0$ is a real scalar, s.t. $|epsilon| < sigma^{2}_r$, show that $(textbf{A}^Ttextbf{A}+epsilon textbf{I})$ is invertible.



I found the resources Why is $A^TA$ invertible if $A$ has independent columns? and Matrix inverse of $A + epsilon I$, where $A$ is invertible



But I'm not sure how useful they are. The first is in the case where A has independent columns, which is not necessarily true here, and the second presumes A is invertible. I believe that $textbf{A}^Ttextbf{A}$ is invertible by definition, but I'm not sure if I can just plug $textbf{A}^Ttextbf{A}$ in everywhere that post uses A and follow through. That also wouldn't help me understand the problem, just blindly substitute into a solution.



Can anyone help me understand WHY $(textbf{A}^Ttextbf{A}+epsilon textbf{I})$ is invertible? And/or point me in the right direction to construct a proof of it?






linear-algebra inverse







share|cite|improve this question











$endgroup$




I'm given a problem:



$sigma_1 geq sigma_2 geq ... geq sigma_r$ are the nonzero singular values of $textbf{A}inmathbb{R}^{Mtimes N}$. If $epsilon neq 0$ is a real scalar, s.t. $|epsilon| < sigma^{2}_r$, show that $(textbf{A}^Ttextbf{A}+epsilon textbf{I})$ is invertible.



I found the resources Why is $A^TA$ invertible if $A$ has independent columns? and Matrix inverse of $A + epsilon I$, where $A$ is invertible



But I'm not sure how useful they are. The first is in the case where A has independent columns, which is not necessarily true here, and the second presumes A is invertible. I believe that $textbf{A}^Ttextbf{A}$ is invertible by definition, but I'm not sure if I can just plug $textbf{A}^Ttextbf{A}$ in everywhere that post uses A and follow through. That also wouldn't help me understand the problem, just blindly substitute into a solution.



Can anyone help me understand WHY $(textbf{A}^Ttextbf{A}+epsilon textbf{I})$ is invertible? And/or point me in the right direction to construct a proof of it?






linear-algebra inverse




linear-algebra inverse






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 8:20









Tommi Brander

956922




956922










asked Dec 5 '18 at 21:24









W. MacTurkW. MacTurk

135




135












  • $begingroup$
    Just a thought, you can diagonalize $A^TA$, so it shouldnt be too hard to analyze everything after a suitable change of basis
    $endgroup$
    – qbert
    Dec 5 '18 at 21:59


















  • $begingroup$
    Just a thought, you can diagonalize $A^TA$, so it shouldnt be too hard to analyze everything after a suitable change of basis
    $endgroup$
    – qbert
    Dec 5 '18 at 21:59
















$begingroup$
Just a thought, you can diagonalize $A^TA$, so it shouldnt be too hard to analyze everything after a suitable change of basis
$endgroup$
– qbert
Dec 5 '18 at 21:59




$begingroup$
Just a thought, you can diagonalize $A^TA$, so it shouldnt be too hard to analyze everything after a suitable change of basis
$endgroup$
– qbert
Dec 5 '18 at 21:59










2 Answers
2






active

oldest

votes


















2












$begingroup$

Write out $A$ in its SVD, $A=USigma V^T$. Then we have



$$A^TA+epsilon I = VSigma^2 V^T+epsilon I = VSigma^2V^T+epsilon VV^T = V(Sigma^2+epsilon I)V^T.$$
From this, we have that the eigenvalues are exactly $sigma_i^2+epsilon$ for $i=1dots r$ and $epsilon$ for $i=r+1dots n$. These are all nonzero, so the matrix is invertible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You moved the transpose in the last step.
    $endgroup$
    – Ian
    Dec 5 '18 at 21:49










  • $begingroup$
    Thanks. Fixed..
    $endgroup$
    – whpowell96
    Dec 5 '18 at 21:51










  • $begingroup$
    Thank you! Is there a straightforward way to find the limit of $(textbf{A}^Ttextbf{A}+epsilon textbf{I})^{−1}textbf{A}^T$ as $epsilon$ goes to 0? Obviously this reduces to $(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$, but I'm not sure where to go from here without specific values for A
    $endgroup$
    – W. MacTurk
    Dec 6 '18 at 1:58



















1












$begingroup$

You can show that $A^top A$ is positive semi-definite (specifically, that it has nonnegative eigenvalues $sigma_1^2, ldots, sigma_r^2, 0, ldots, 0$).



[It is not always invertible. Specifically, if some of its eigenvalues are zero, then it is not invertible.]



Knowing this fact about $A^top A$, can you explicitly write down the eigenvalues of $A^top A + epsilon I$? What values of $epsilon$ make this matrix invertible or non-invertible?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wouldn't the eigenvalues just be $sigma_{1}^{2}+epsilon , sigma_{2}^{2}+epsilon , ...sigma_{r}^{2}+epsilon$, with trailing eigenvalues of $epsilon$ if $textbf{A}^Ttextbf{A}$ was not invertible?
    $endgroup$
    – W. MacTurk
    Dec 7 '18 at 3:24











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Write out $A$ in its SVD, $A=USigma V^T$. Then we have



$$A^TA+epsilon I = VSigma^2 V^T+epsilon I = VSigma^2V^T+epsilon VV^T = V(Sigma^2+epsilon I)V^T.$$
From this, we have that the eigenvalues are exactly $sigma_i^2+epsilon$ for $i=1dots r$ and $epsilon$ for $i=r+1dots n$. These are all nonzero, so the matrix is invertible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You moved the transpose in the last step.
    $endgroup$
    – Ian
    Dec 5 '18 at 21:49










  • $begingroup$
    Thanks. Fixed..
    $endgroup$
    – whpowell96
    Dec 5 '18 at 21:51










  • $begingroup$
    Thank you! Is there a straightforward way to find the limit of $(textbf{A}^Ttextbf{A}+epsilon textbf{I})^{−1}textbf{A}^T$ as $epsilon$ goes to 0? Obviously this reduces to $(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$, but I'm not sure where to go from here without specific values for A
    $endgroup$
    – W. MacTurk
    Dec 6 '18 at 1:58
















2












$begingroup$

Write out $A$ in its SVD, $A=USigma V^T$. Then we have



$$A^TA+epsilon I = VSigma^2 V^T+epsilon I = VSigma^2V^T+epsilon VV^T = V(Sigma^2+epsilon I)V^T.$$
From this, we have that the eigenvalues are exactly $sigma_i^2+epsilon$ for $i=1dots r$ and $epsilon$ for $i=r+1dots n$. These are all nonzero, so the matrix is invertible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You moved the transpose in the last step.
    $endgroup$
    – Ian
    Dec 5 '18 at 21:49










  • $begingroup$
    Thanks. Fixed..
    $endgroup$
    – whpowell96
    Dec 5 '18 at 21:51










  • $begingroup$
    Thank you! Is there a straightforward way to find the limit of $(textbf{A}^Ttextbf{A}+epsilon textbf{I})^{−1}textbf{A}^T$ as $epsilon$ goes to 0? Obviously this reduces to $(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$, but I'm not sure where to go from here without specific values for A
    $endgroup$
    – W. MacTurk
    Dec 6 '18 at 1:58














2












2








2





$begingroup$

Write out $A$ in its SVD, $A=USigma V^T$. Then we have



$$A^TA+epsilon I = VSigma^2 V^T+epsilon I = VSigma^2V^T+epsilon VV^T = V(Sigma^2+epsilon I)V^T.$$
From this, we have that the eigenvalues are exactly $sigma_i^2+epsilon$ for $i=1dots r$ and $epsilon$ for $i=r+1dots n$. These are all nonzero, so the matrix is invertible.






share|cite|improve this answer











$endgroup$



Write out $A$ in its SVD, $A=USigma V^T$. Then we have



$$A^TA+epsilon I = VSigma^2 V^T+epsilon I = VSigma^2V^T+epsilon VV^T = V(Sigma^2+epsilon I)V^T.$$
From this, we have that the eigenvalues are exactly $sigma_i^2+epsilon$ for $i=1dots r$ and $epsilon$ for $i=r+1dots n$. These are all nonzero, so the matrix is invertible.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 21:51

























answered Dec 5 '18 at 21:47









whpowell96whpowell96

56615




56615












  • $begingroup$
    You moved the transpose in the last step.
    $endgroup$
    – Ian
    Dec 5 '18 at 21:49










  • $begingroup$
    Thanks. Fixed..
    $endgroup$
    – whpowell96
    Dec 5 '18 at 21:51










  • $begingroup$
    Thank you! Is there a straightforward way to find the limit of $(textbf{A}^Ttextbf{A}+epsilon textbf{I})^{−1}textbf{A}^T$ as $epsilon$ goes to 0? Obviously this reduces to $(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$, but I'm not sure where to go from here without specific values for A
    $endgroup$
    – W. MacTurk
    Dec 6 '18 at 1:58


















  • $begingroup$
    You moved the transpose in the last step.
    $endgroup$
    – Ian
    Dec 5 '18 at 21:49










  • $begingroup$
    Thanks. Fixed..
    $endgroup$
    – whpowell96
    Dec 5 '18 at 21:51










  • $begingroup$
    Thank you! Is there a straightforward way to find the limit of $(textbf{A}^Ttextbf{A}+epsilon textbf{I})^{−1}textbf{A}^T$ as $epsilon$ goes to 0? Obviously this reduces to $(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$, but I'm not sure where to go from here without specific values for A
    $endgroup$
    – W. MacTurk
    Dec 6 '18 at 1:58
















$begingroup$
You moved the transpose in the last step.
$endgroup$
– Ian
Dec 5 '18 at 21:49




$begingroup$
You moved the transpose in the last step.
$endgroup$
– Ian
Dec 5 '18 at 21:49












$begingroup$
Thanks. Fixed..
$endgroup$
– whpowell96
Dec 5 '18 at 21:51




$begingroup$
Thanks. Fixed..
$endgroup$
– whpowell96
Dec 5 '18 at 21:51












$begingroup$
Thank you! Is there a straightforward way to find the limit of $(textbf{A}^Ttextbf{A}+epsilon textbf{I})^{−1}textbf{A}^T$ as $epsilon$ goes to 0? Obviously this reduces to $(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$, but I'm not sure where to go from here without specific values for A
$endgroup$
– W. MacTurk
Dec 6 '18 at 1:58




$begingroup$
Thank you! Is there a straightforward way to find the limit of $(textbf{A}^Ttextbf{A}+epsilon textbf{I})^{−1}textbf{A}^T$ as $epsilon$ goes to 0? Obviously this reduces to $(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$, but I'm not sure where to go from here without specific values for A
$endgroup$
– W. MacTurk
Dec 6 '18 at 1:58











1












$begingroup$

You can show that $A^top A$ is positive semi-definite (specifically, that it has nonnegative eigenvalues $sigma_1^2, ldots, sigma_r^2, 0, ldots, 0$).



[It is not always invertible. Specifically, if some of its eigenvalues are zero, then it is not invertible.]



Knowing this fact about $A^top A$, can you explicitly write down the eigenvalues of $A^top A + epsilon I$? What values of $epsilon$ make this matrix invertible or non-invertible?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wouldn't the eigenvalues just be $sigma_{1}^{2}+epsilon , sigma_{2}^{2}+epsilon , ...sigma_{r}^{2}+epsilon$, with trailing eigenvalues of $epsilon$ if $textbf{A}^Ttextbf{A}$ was not invertible?
    $endgroup$
    – W. MacTurk
    Dec 7 '18 at 3:24
















1












$begingroup$

You can show that $A^top A$ is positive semi-definite (specifically, that it has nonnegative eigenvalues $sigma_1^2, ldots, sigma_r^2, 0, ldots, 0$).



[It is not always invertible. Specifically, if some of its eigenvalues are zero, then it is not invertible.]



Knowing this fact about $A^top A$, can you explicitly write down the eigenvalues of $A^top A + epsilon I$? What values of $epsilon$ make this matrix invertible or non-invertible?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wouldn't the eigenvalues just be $sigma_{1}^{2}+epsilon , sigma_{2}^{2}+epsilon , ...sigma_{r}^{2}+epsilon$, with trailing eigenvalues of $epsilon$ if $textbf{A}^Ttextbf{A}$ was not invertible?
    $endgroup$
    – W. MacTurk
    Dec 7 '18 at 3:24














1












1








1





$begingroup$

You can show that $A^top A$ is positive semi-definite (specifically, that it has nonnegative eigenvalues $sigma_1^2, ldots, sigma_r^2, 0, ldots, 0$).



[It is not always invertible. Specifically, if some of its eigenvalues are zero, then it is not invertible.]



Knowing this fact about $A^top A$, can you explicitly write down the eigenvalues of $A^top A + epsilon I$? What values of $epsilon$ make this matrix invertible or non-invertible?






share|cite|improve this answer









$endgroup$



You can show that $A^top A$ is positive semi-definite (specifically, that it has nonnegative eigenvalues $sigma_1^2, ldots, sigma_r^2, 0, ldots, 0$).



[It is not always invertible. Specifically, if some of its eigenvalues are zero, then it is not invertible.]



Knowing this fact about $A^top A$, can you explicitly write down the eigenvalues of $A^top A + epsilon I$? What values of $epsilon$ make this matrix invertible or non-invertible?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 21:45









angryavianangryavian

40.6k23380




40.6k23380












  • $begingroup$
    Wouldn't the eigenvalues just be $sigma_{1}^{2}+epsilon , sigma_{2}^{2}+epsilon , ...sigma_{r}^{2}+epsilon$, with trailing eigenvalues of $epsilon$ if $textbf{A}^Ttextbf{A}$ was not invertible?
    $endgroup$
    – W. MacTurk
    Dec 7 '18 at 3:24


















  • $begingroup$
    Wouldn't the eigenvalues just be $sigma_{1}^{2}+epsilon , sigma_{2}^{2}+epsilon , ...sigma_{r}^{2}+epsilon$, with trailing eigenvalues of $epsilon$ if $textbf{A}^Ttextbf{A}$ was not invertible?
    $endgroup$
    – W. MacTurk
    Dec 7 '18 at 3:24
















$begingroup$
Wouldn't the eigenvalues just be $sigma_{1}^{2}+epsilon , sigma_{2}^{2}+epsilon , ...sigma_{r}^{2}+epsilon$, with trailing eigenvalues of $epsilon$ if $textbf{A}^Ttextbf{A}$ was not invertible?
$endgroup$
– W. MacTurk
Dec 7 '18 at 3:24




$begingroup$
Wouldn't the eigenvalues just be $sigma_{1}^{2}+epsilon , sigma_{2}^{2}+epsilon , ...sigma_{r}^{2}+epsilon$, with trailing eigenvalues of $epsilon$ if $textbf{A}^Ttextbf{A}$ was not invertible?
$endgroup$
– W. MacTurk
Dec 7 '18 at 3:24


















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