How to prove $|A^+|_2leq|A^{-1}_1|_2$ in the question?












1












$begingroup$


Suppose the $m×n$ matrix $A$ has the form $A=begin{pmatrix}
A_1 \
A_2
end{pmatrix}$ where $ A_1$ is a nonsingular matrix of dimension $n×n$ and $A_2$ is an arbitrary matrix of dimension $(m-n)×n$. Prove that $|A^+|_2leq|A^{-1}_1|_2$.



My idea of this problem:
In order to calculate the 2-norm, I calculate the eigenvalues of $(A^+)^*A^+$ and $(A^{-1}_1)^*A^{-1}_1$(which are $(AA^*)^{-1}$ and $(A_1A_1^*)^{-1}$ after simplifying). Now I do not know how to compare their eigenvalues.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $A_1$ is $n times n$ and $A_2$ is $(m-n) times n$ then would not A be $m times 2n$ the way you have defined it?
    $endgroup$
    – Wintermute
    Dec 18 '18 at 18:16
















1












$begingroup$


Suppose the $m×n$ matrix $A$ has the form $A=begin{pmatrix}
A_1 \
A_2
end{pmatrix}$ where $ A_1$ is a nonsingular matrix of dimension $n×n$ and $A_2$ is an arbitrary matrix of dimension $(m-n)×n$. Prove that $|A^+|_2leq|A^{-1}_1|_2$.



My idea of this problem:
In order to calculate the 2-norm, I calculate the eigenvalues of $(A^+)^*A^+$ and $(A^{-1}_1)^*A^{-1}_1$(which are $(AA^*)^{-1}$ and $(A_1A_1^*)^{-1}$ after simplifying). Now I do not know how to compare their eigenvalues.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $A_1$ is $n times n$ and $A_2$ is $(m-n) times n$ then would not A be $m times 2n$ the way you have defined it?
    $endgroup$
    – Wintermute
    Dec 18 '18 at 18:16














1












1








1





$begingroup$


Suppose the $m×n$ matrix $A$ has the form $A=begin{pmatrix}
A_1 \
A_2
end{pmatrix}$ where $ A_1$ is a nonsingular matrix of dimension $n×n$ and $A_2$ is an arbitrary matrix of dimension $(m-n)×n$. Prove that $|A^+|_2leq|A^{-1}_1|_2$.



My idea of this problem:
In order to calculate the 2-norm, I calculate the eigenvalues of $(A^+)^*A^+$ and $(A^{-1}_1)^*A^{-1}_1$(which are $(AA^*)^{-1}$ and $(A_1A_1^*)^{-1}$ after simplifying). Now I do not know how to compare their eigenvalues.






linear-algebra







share|cite|improve this question











$endgroup$




Suppose the $m×n$ matrix $A$ has the form $A=begin{pmatrix}
A_1 \
A_2
end{pmatrix}$ where $ A_1$ is a nonsingular matrix of dimension $n×n$ and $A_2$ is an arbitrary matrix of dimension $(m-n)×n$. Prove that $|A^+|_2leq|A^{-1}_1|_2$.



My idea of this problem:
In order to calculate the 2-norm, I calculate the eigenvalues of $(A^+)^*A^+$ and $(A^{-1}_1)^*A^{-1}_1$(which are $(AA^*)^{-1}$ and $(A_1A_1^*)^{-1}$ after simplifying). Now I do not know how to compare their eigenvalues.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 12:23







Kristy

















asked Dec 18 '18 at 1:04









KristyKristy

454




454












  • $begingroup$
    If $A_1$ is $n times n$ and $A_2$ is $(m-n) times n$ then would not A be $m times 2n$ the way you have defined it?
    $endgroup$
    – Wintermute
    Dec 18 '18 at 18:16


















  • $begingroup$
    If $A_1$ is $n times n$ and $A_2$ is $(m-n) times n$ then would not A be $m times 2n$ the way you have defined it?
    $endgroup$
    – Wintermute
    Dec 18 '18 at 18:16
















$begingroup$
If $A_1$ is $n times n$ and $A_2$ is $(m-n) times n$ then would not A be $m times 2n$ the way you have defined it?
$endgroup$
– Wintermute
Dec 18 '18 at 18:16




$begingroup$
If $A_1$ is $n times n$ and $A_2$ is $(m-n) times n$ then would not A be $m times 2n$ the way you have defined it?
$endgroup$
– Wintermute
Dec 18 '18 at 18:16










1 Answer
1






active

oldest

votes


















1












$begingroup$

Since the columns of $A$ are independent,



$A^+=(A_1^*A_1+A_2^*A_2)^{-1}[A_1^*,A_2^*]$ and, therefore



$A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$.



Moreover $A_1^{-1}{A_1^{-1}}^*=(A_1^*A_1)^{-1}$.



Since $0_n<A_1^*A_1leq A_1^*A_1+A_2^*A_2$,



$lambda_{min}(A_1^*A_1)leq lambda_{min}(A_1^*A_1+A_2^*A_2)$ and



$lambda_{max}((A_1^*A_1)^{-1})geq lambda_{max}((A_1^*A_1+A_2^*A_2)^{-1})$ and we are done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why $A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$?
    $endgroup$
    – Kristy
    Dec 19 '18 at 1:00










  • $begingroup$
    Write the product.
    $endgroup$
    – loup blanc
    Dec 19 '18 at 8:21











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Since the columns of $A$ are independent,



$A^+=(A_1^*A_1+A_2^*A_2)^{-1}[A_1^*,A_2^*]$ and, therefore



$A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$.



Moreover $A_1^{-1}{A_1^{-1}}^*=(A_1^*A_1)^{-1}$.



Since $0_n<A_1^*A_1leq A_1^*A_1+A_2^*A_2$,



$lambda_{min}(A_1^*A_1)leq lambda_{min}(A_1^*A_1+A_2^*A_2)$ and



$lambda_{max}((A_1^*A_1)^{-1})geq lambda_{max}((A_1^*A_1+A_2^*A_2)^{-1})$ and we are done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why $A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$?
    $endgroup$
    – Kristy
    Dec 19 '18 at 1:00










  • $begingroup$
    Write the product.
    $endgroup$
    – loup blanc
    Dec 19 '18 at 8:21
















1












$begingroup$

Since the columns of $A$ are independent,



$A^+=(A_1^*A_1+A_2^*A_2)^{-1}[A_1^*,A_2^*]$ and, therefore



$A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$.



Moreover $A_1^{-1}{A_1^{-1}}^*=(A_1^*A_1)^{-1}$.



Since $0_n<A_1^*A_1leq A_1^*A_1+A_2^*A_2$,



$lambda_{min}(A_1^*A_1)leq lambda_{min}(A_1^*A_1+A_2^*A_2)$ and



$lambda_{max}((A_1^*A_1)^{-1})geq lambda_{max}((A_1^*A_1+A_2^*A_2)^{-1})$ and we are done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why $A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$?
    $endgroup$
    – Kristy
    Dec 19 '18 at 1:00










  • $begingroup$
    Write the product.
    $endgroup$
    – loup blanc
    Dec 19 '18 at 8:21














1












1








1





$begingroup$

Since the columns of $A$ are independent,



$A^+=(A_1^*A_1+A_2^*A_2)^{-1}[A_1^*,A_2^*]$ and, therefore



$A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$.



Moreover $A_1^{-1}{A_1^{-1}}^*=(A_1^*A_1)^{-1}$.



Since $0_n<A_1^*A_1leq A_1^*A_1+A_2^*A_2$,



$lambda_{min}(A_1^*A_1)leq lambda_{min}(A_1^*A_1+A_2^*A_2)$ and



$lambda_{max}((A_1^*A_1)^{-1})geq lambda_{max}((A_1^*A_1+A_2^*A_2)^{-1})$ and we are done.






share|cite|improve this answer









$endgroup$



Since the columns of $A$ are independent,



$A^+=(A_1^*A_1+A_2^*A_2)^{-1}[A_1^*,A_2^*]$ and, therefore



$A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$.



Moreover $A_1^{-1}{A_1^{-1}}^*=(A_1^*A_1)^{-1}$.



Since $0_n<A_1^*A_1leq A_1^*A_1+A_2^*A_2$,



$lambda_{min}(A_1^*A_1)leq lambda_{min}(A_1^*A_1+A_2^*A_2)$ and



$lambda_{max}((A_1^*A_1)^{-1})geq lambda_{max}((A_1^*A_1+A_2^*A_2)^{-1})$ and we are done.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 18 '18 at 18:07









loup blancloup blanc

24.1k21851




24.1k21851












  • $begingroup$
    why $A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$?
    $endgroup$
    – Kristy
    Dec 19 '18 at 1:00










  • $begingroup$
    Write the product.
    $endgroup$
    – loup blanc
    Dec 19 '18 at 8:21


















  • $begingroup$
    why $A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$?
    $endgroup$
    – Kristy
    Dec 19 '18 at 1:00










  • $begingroup$
    Write the product.
    $endgroup$
    – loup blanc
    Dec 19 '18 at 8:21
















$begingroup$
why $A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$?
$endgroup$
– Kristy
Dec 19 '18 at 1:00




$begingroup$
why $A^+{A^+}^*=(A_1^*A_1+A_2^*A_2)^{-1}$?
$endgroup$
– Kristy
Dec 19 '18 at 1:00












$begingroup$
Write the product.
$endgroup$
– loup blanc
Dec 19 '18 at 8:21




$begingroup$
Write the product.
$endgroup$
– loup blanc
Dec 19 '18 at 8:21


















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