Proving nonsingularity of this block matrix












0












$begingroup$


I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?



Let $A,B,C,D in R^{n times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then
$$begin{bmatrix}
A & B\[0.3em]
C & D\[0.3em]
end{bmatrix}^{-1} = begin{bmatrix}
(A-BD^{-1}C)^{-1} & (C-DB^{-1}A)^{-1}\[0.3em]
-D^{-1}C(A-BD^{-1}C)^{-1} & D^{-1}-D^{-1}C(C-DB^{-1}A)^{-1}\[0.3em]
end{bmatrix}$$



I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!






linear-algebra matrices matrix-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your alleged inverse is not quite right. Several typos.
    $endgroup$
    – Robert Israel
    Dec 11 '18 at 15:22












  • $begingroup$
    I adjusted it, typos have been fixed!
    $endgroup$
    – Adam Warlock
    Dec 11 '18 at 15:32
















0












$begingroup$


I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?



Let $A,B,C,D in R^{n times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then
$$begin{bmatrix}
A & B\[0.3em]
C & D\[0.3em]
end{bmatrix}^{-1} = begin{bmatrix}
(A-BD^{-1}C)^{-1} & (C-DB^{-1}A)^{-1}\[0.3em]
-D^{-1}C(A-BD^{-1}C)^{-1} & D^{-1}-D^{-1}C(C-DB^{-1}A)^{-1}\[0.3em]
end{bmatrix}$$



I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!






linear-algebra matrices matrix-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your alleged inverse is not quite right. Several typos.
    $endgroup$
    – Robert Israel
    Dec 11 '18 at 15:22












  • $begingroup$
    I adjusted it, typos have been fixed!
    $endgroup$
    – Adam Warlock
    Dec 11 '18 at 15:32














0












0








0





$begingroup$


I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?



Let $A,B,C,D in R^{n times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then
$$begin{bmatrix}
A & B\[0.3em]
C & D\[0.3em]
end{bmatrix}^{-1} = begin{bmatrix}
(A-BD^{-1}C)^{-1} & (C-DB^{-1}A)^{-1}\[0.3em]
-D^{-1}C(A-BD^{-1}C)^{-1} & D^{-1}-D^{-1}C(C-DB^{-1}A)^{-1}\[0.3em]
end{bmatrix}$$



I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!






linear-algebra matrices matrix-calculus







share|cite|improve this question











$endgroup$




I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?



Let $A,B,C,D in R^{n times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then
$$begin{bmatrix}
A & B\[0.3em]
C & D\[0.3em]
end{bmatrix}^{-1} = begin{bmatrix}
(A-BD^{-1}C)^{-1} & (C-DB^{-1}A)^{-1}\[0.3em]
-D^{-1}C(A-BD^{-1}C)^{-1} & D^{-1}-D^{-1}C(C-DB^{-1}A)^{-1}\[0.3em]
end{bmatrix}$$



I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!






linear-algebra matrices matrix-calculus




linear-algebra matrices matrix-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 15:31







Adam Warlock

















asked Dec 11 '18 at 15:07









Adam WarlockAdam Warlock

1016




1016












  • $begingroup$
    Your alleged inverse is not quite right. Several typos.
    $endgroup$
    – Robert Israel
    Dec 11 '18 at 15:22












  • $begingroup$
    I adjusted it, typos have been fixed!
    $endgroup$
    – Adam Warlock
    Dec 11 '18 at 15:32


















  • $begingroup$
    Your alleged inverse is not quite right. Several typos.
    $endgroup$
    – Robert Israel
    Dec 11 '18 at 15:22












  • $begingroup$
    I adjusted it, typos have been fixed!
    $endgroup$
    – Adam Warlock
    Dec 11 '18 at 15:32
















$begingroup$
Your alleged inverse is not quite right. Several typos.
$endgroup$
– Robert Israel
Dec 11 '18 at 15:22






$begingroup$
Your alleged inverse is not quite right. Several typos.
$endgroup$
– Robert Israel
Dec 11 '18 at 15:22














$begingroup$
I adjusted it, typos have been fixed!
$endgroup$
– Adam Warlock
Dec 11 '18 at 15:32




$begingroup$
I adjusted it, typos have been fixed!
$endgroup$
– Adam Warlock
Dec 11 '18 at 15:32










1 Answer
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$begingroup$

Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ left[ matrix{A & Bcr C & Dcr} right] left[ matrix{E & Fcr G & Hcr} right] = left[ matrix{AE+BG & AF + BHcr CE +DG & CF + DHcr} right]$$



So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.






share|cite|improve this answer









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    $begingroup$

    Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
    commute. Thus
    $$ left[ matrix{A & Bcr C & Dcr} right] left[ matrix{E & Fcr G & Hcr} right] = left[ matrix{AE+BG & AF + BHcr CE +DG & CF + DHcr} right]$$



    So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
      commute. Thus
      $$ left[ matrix{A & Bcr C & Dcr} right] left[ matrix{E & Fcr G & Hcr} right] = left[ matrix{AE+BG & AF + BHcr CE +DG & CF + DHcr} right]$$



      So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
        commute. Thus
        $$ left[ matrix{A & Bcr C & Dcr} right] left[ matrix{E & Fcr G & Hcr} right] = left[ matrix{AE+BG & AF + BHcr CE +DG & CF + DHcr} right]$$



        So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.






        share|cite|improve this answer









        $endgroup$



        Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
        commute. Thus
        $$ left[ matrix{A & Bcr C & Dcr} right] left[ matrix{E & Fcr G & Hcr} right] = left[ matrix{AE+BG & AF + BHcr CE +DG & CF + DHcr} right]$$



        So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 15:24









        Robert IsraelRobert Israel

        321k23211465




        321k23211465






























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