Proving nonsingularity of this block matrix
$begingroup$
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}$$
I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!
linear-algebra matrices matrix-calculus
asked Dec 11 '18 at 15:07
Adam WarlockAdam Warlock
1016
$endgroup$
add a comment |
$begingroup$
I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?
Let $A,B,C,D in R^{n times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then
$$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}$$
I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!
linear-algebra matrices matrix-calculus
asked Dec 11 '18 at 15:07
Adam WarlockAdam Warlock
1016
$endgroup$
$begingroup$
Your alleged inverse is not quite right. Several typos.
$endgroup$
– Robert Israel
Dec 11 '18 at 15:22
$begingroup$
I adjusted it, typos have been fixed!
$endgroup$
– Adam Warlock
Dec 11 '18 at 15:32
add a comment |
$begingroup$
I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?
Let $A,B,C,D in R^{n times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then
$$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}$$
I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!
linear-algebra matrices matrix-calculus
asked Dec 11 '18 at 15:07
Adam WarlockAdam Warlock
1016
$endgroup$
I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?
Let $A,B,C,D in R^{n times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then
$$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}$$
I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!
linear-algebra matrices matrix-calculus
linear-algebra matrices matrix-calculus
asked Dec 11 '18 at 15:07
Adam WarlockAdam Warlock
1016
asked Dec 11 '18 at 15:07
Adam WarlockAdam Warlock
1016
edited Dec 11 '18 at 15:31
Adam Warlock
asked Dec 11 '18 at 15:07
Adam WarlockAdam Warlock
1016
asked Dec 11 '18 at 15:07
Adam WarlockAdam Warlock
1016
asked Dec 11 '18 at 15:07
Adam WarlockAdam Warlock
1016
1016
$begingroup$
Your alleged inverse is not quite right. Several typos.
$endgroup$
– Robert Israel
Dec 11 '18 at 15:22
$begingroup$
I adjusted it, typos have been fixed!
$endgroup$
– Adam Warlock
Dec 11 '18 at 15:32
add a comment |
$begingroup$
Your alleged inverse is not quite right. Several typos.
$endgroup$
– Robert Israel
Dec 11 '18 at 15:22
$begingroup$
I adjusted it, typos have been fixed!
$endgroup$
– Adam Warlock
Dec 11 '18 at 15:32
$begingroup$
Your alleged inverse is not quite right. Several typos.
$endgroup$
– Robert Israel
Dec 11 '18 at 15:22
$begingroup$
Your alleged inverse is not quite right. Several typos.
$endgroup$
– Robert Israel
Dec 11 '18 at 15:22
$begingroup$
I adjusted it, typos have been fixed!
$endgroup$
– Adam Warlock
Dec 11 '18 at 15:32
$begingroup$
I adjusted it, typos have been fixed!
$endgroup$
– Adam Warlock
Dec 11 '18 at 15:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ 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]$$
So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.
answered Dec 11 '18 at 15:24
Robert IsraelRobert Israel
321k23211465
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ 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]$$
So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.
answered Dec 11 '18 at 15:24
Robert IsraelRobert Israel
321k23211465
$endgroup$
add a comment |
$begingroup$
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ 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]$$
So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.
answered Dec 11 '18 at 15:24
Robert IsraelRobert Israel
321k23211465
$endgroup$
add a comment |
$begingroup$
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ 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]$$
So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.
answered Dec 11 '18 at 15:24
Robert IsraelRobert Israel
321k23211465
$endgroup$
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ 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 $left[matrix{I & 0cr 0 & I}right]$.
answered Dec 11 '18 at 15:24
Robert IsraelRobert Israel
321k23211465
answered Dec 11 '18 at 15:24
Robert IsraelRobert Israel
321k23211465
answered Dec 11 '18 at 15:24
Robert IsraelRobert Israel
321k23211465
answered Dec 11 '18 at 15:24
Robert IsraelRobert Israel
321k23211465
321k23211465
add a comment |
add a comment |
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$begingroup$
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