Proving nonsingularity of this block matrix
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}$$
I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!
linear-algebra matrices matrix-calculus
asked Dec 11 '18 at 15:07
Adam Warlock
1016
add a comment |
I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?
Let $A,B,C,D in R^{n times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then
$$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}$$
I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!
linear-algebra matrices matrix-calculus
asked Dec 11 '18 at 15:07
Adam Warlock
1016
Your alleged inverse is not quite right. Several typos.
– Robert Israel
Dec 11 '18 at 15:22
I adjusted it, typos have been fixed!
– Adam Warlock
Dec 11 '18 at 15:32
add a comment |
I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?
Let $A,B,C,D in R^{n times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then
$$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}$$
I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!
linear-algebra matrices matrix-calculus
asked Dec 11 '18 at 15:07
Adam Warlock
1016
I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?
Let $A,B,C,D in R^{n times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then
$$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}$$
I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!
linear-algebra matrices matrix-calculus
linear-algebra matrices matrix-calculus
asked Dec 11 '18 at 15:07
Adam Warlock
1016
asked Dec 11 '18 at 15:07
Adam Warlock
1016
edited Dec 11 '18 at 15:31
asked Dec 11 '18 at 15:07
Adam Warlock
1016
asked Dec 11 '18 at 15:07
Adam Warlock
1016
asked Dec 11 '18 at 15:07
Adam Warlock
1016
1016
Your alleged inverse is not quite right. Several typos.
– Robert Israel
Dec 11 '18 at 15:22
I adjusted it, typos have been fixed!
– Adam Warlock
Dec 11 '18 at 15:32
add a comment |
Your alleged inverse is not quite right. Several typos.
– Robert Israel
Dec 11 '18 at 15:22
I adjusted it, typos have been fixed!
– Adam Warlock
Dec 11 '18 at 15:32
Your alleged inverse is not quite right. Several typos.
– Robert Israel
Dec 11 '18 at 15:22
Your alleged inverse is not quite right. Several typos.
– Robert Israel
Dec 11 '18 at 15:22
I adjusted it, typos have been fixed!
– Adam Warlock
Dec 11 '18 at 15:32
I adjusted it, typos have been fixed!
– Adam Warlock
Dec 11 '18 at 15:32
add a comment |
1 Answer
1
active
oldest
votes
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ 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]$$
So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.
answered Dec 11 '18 at 15:24
Robert Israel
318k23208457
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ 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]$$
So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.
answered Dec 11 '18 at 15:24
Robert Israel
318k23208457
add a comment |
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ 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]$$
So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.
answered Dec 11 '18 at 15:24
Robert Israel
318k23208457
add a comment |
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ 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]$$
So multiply $left[ matrix{A & Bcr C & Dcr} right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $left[matrix{I & 0cr 0 & I}right]$.
answered Dec 11 '18 at 15:24
Robert Israel
318k23208457
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't
commute. Thus
$$ 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 Israel
318k23208457
answered Dec 11 '18 at 15:24
Robert Israel
318k23208457
answered Dec 11 '18 at 15:24
Robert Israel
318k23208457
answered Dec 11 '18 at 15:24
Robert Israel
318k23208457
318k23208457
add a comment |
add a comment |
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Your alleged inverse is not quite right. Several typos.
– Robert Israel
Dec 11 '18 at 15:22
I adjusted it, typos have been fixed!
– Adam Warlock
Dec 11 '18 at 15:32