How to inverse a block diagonal matrix?
$begingroup$
Given a matrix $$x = begin{bmatrix}
40 & 0 & 0 & 0\
0 & 80 & 100 & 0 \
0 & 40 & 120 & 0 \
0 & 0 & 0 & 60end{bmatrix}$$
How to find the inverse of that matrix?
What I know: $det(x) = ac-bd$,
inverse of a 2x2 matrix: $$x^{-1} = frac{1}{det(x)}cdot begin{bmatrix} d &-b\ -c &aend{bmatrix}.$$
There is a lot of content online; however none of them has a specific numerical example.
matrices numerical-linear-algebra matrix-decomposition
edited Dec 3 '18 at 13:12
amWhy
1
asked Dec 3 '18 at 12:22
JayDoughJayDough
12
$endgroup$
add a comment |
$begingroup$
Given a matrix $$x = begin{bmatrix}
40 & 0 & 0 & 0\
0 & 80 & 100 & 0 \
0 & 40 & 120 & 0 \
0 & 0 & 0 & 60end{bmatrix}$$
How to find the inverse of that matrix?
What I know: $det(x) = ac-bd$,
inverse of a 2x2 matrix: $$x^{-1} = frac{1}{det(x)}cdot begin{bmatrix} d &-b\ -c &aend{bmatrix}.$$
There is a lot of content online; however none of them has a specific numerical example.
matrices numerical-linear-algebra matrix-decomposition
edited Dec 3 '18 at 13:12
amWhy
1
asked Dec 3 '18 at 12:22
JayDoughJayDough
12
$endgroup$
add a comment |
$begingroup$
Given a matrix $$x = begin{bmatrix}
40 & 0 & 0 & 0\
0 & 80 & 100 & 0 \
0 & 40 & 120 & 0 \
0 & 0 & 0 & 60end{bmatrix}$$
How to find the inverse of that matrix?
What I know: $det(x) = ac-bd$,
inverse of a 2x2 matrix: $$x^{-1} = frac{1}{det(x)}cdot begin{bmatrix} d &-b\ -c &aend{bmatrix}.$$
There is a lot of content online; however none of them has a specific numerical example.
matrices numerical-linear-algebra matrix-decomposition
edited Dec 3 '18 at 13:12
amWhy
1
asked Dec 3 '18 at 12:22
JayDoughJayDough
12
$endgroup$
Given a matrix $$x = begin{bmatrix}
40 & 0 & 0 & 0\
0 & 80 & 100 & 0 \
0 & 40 & 120 & 0 \
0 & 0 & 0 & 60end{bmatrix}$$
How to find the inverse of that matrix?
What I know: $det(x) = ac-bd$,
inverse of a 2x2 matrix: $$x^{-1} = frac{1}{det(x)}cdot begin{bmatrix} d &-b\ -c &aend{bmatrix}.$$
There is a lot of content online; however none of them has a specific numerical example.
matrices numerical-linear-algebra matrix-decomposition
matrices numerical-linear-algebra matrix-decomposition
edited Dec 3 '18 at 13:12
amWhy
1
asked Dec 3 '18 at 12:22
JayDoughJayDough
12
edited Dec 3 '18 at 13:12
amWhy
1
asked Dec 3 '18 at 12:22
JayDoughJayDough
12
edited Dec 3 '18 at 13:12
amWhy
1
edited Dec 3 '18 at 13:12
amWhy
1
edited Dec 3 '18 at 13:12
amWhy
1
1
asked Dec 3 '18 at 12:22
JayDoughJayDough
12
asked Dec 3 '18 at 12:22
JayDoughJayDough
12
asked Dec 3 '18 at 12:22
JayDoughJayDough
12
12
add a comment |
add a comment |
2 Answers
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oldest
votes
$begingroup$
Block diagonal matrices can be inverted block by block. See also [*].
In your example:
$$begin{bmatrix} 40 & 0 & 0 & 0 \ 0 & 80 & 100 & 0 \ 0 & 40 & 120 & 0 \ 0 & 0 & 0 & 60end{bmatrix}^{-1}
= begin{bmatrix} [40]^{-1} & begin{matrix} 0 quad & 0quad end{matrix} & 0 \ begin{matrix} 0 \ 0 end{matrix} & begin{bmatrix} 80 & 100 \ 40 & 120 end{bmatrix}^{-1} & begin{matrix} 0 \ 0 end{matrix} \
0 & begin{matrix} 0quad & 0quad end{matrix} & [60]^{-1}
end{bmatrix}.
$$
Can you take it from here?
answered Dec 3 '18 at 12:39
FlorianFlorian
1,3661721
$endgroup$
add a comment |
$begingroup$
Guide:
If you have a matrix of the form of
$$diag(D_1, D_2, D_3),$$
where each block is invertible, then its inverse is $$diag(D_1^{-1},D_2^{-1}, D_3^{-1}).$$
You should verify this.
In your question $D_2$ is $2$ by $2$ and the other two blocks are scalar.
answered Dec 3 '18 at 12:39
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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$begingroup$
Block diagonal matrices can be inverted block by block. See also [*].
In your example:
$$begin{bmatrix} 40 & 0 & 0 & 0 \ 0 & 80 & 100 & 0 \ 0 & 40 & 120 & 0 \ 0 & 0 & 0 & 60end{bmatrix}^{-1}
= begin{bmatrix} [40]^{-1} & begin{matrix} 0 quad & 0quad end{matrix} & 0 \ begin{matrix} 0 \ 0 end{matrix} & begin{bmatrix} 80 & 100 \ 40 & 120 end{bmatrix}^{-1} & begin{matrix} 0 \ 0 end{matrix} \
0 & begin{matrix} 0quad & 0quad end{matrix} & [60]^{-1}
end{bmatrix}.
$$
Can you take it from here?
answered Dec 3 '18 at 12:39
FlorianFlorian
1,3661721
$endgroup$
add a comment |
$begingroup$
Block diagonal matrices can be inverted block by block. See also [*].
In your example:
$$begin{bmatrix} 40 & 0 & 0 & 0 \ 0 & 80 & 100 & 0 \ 0 & 40 & 120 & 0 \ 0 & 0 & 0 & 60end{bmatrix}^{-1}
= begin{bmatrix} [40]^{-1} & begin{matrix} 0 quad & 0quad end{matrix} & 0 \ begin{matrix} 0 \ 0 end{matrix} & begin{bmatrix} 80 & 100 \ 40 & 120 end{bmatrix}^{-1} & begin{matrix} 0 \ 0 end{matrix} \
0 & begin{matrix} 0quad & 0quad end{matrix} & [60]^{-1}
end{bmatrix}.
$$
Can you take it from here?
answered Dec 3 '18 at 12:39
FlorianFlorian
1,3661721
$endgroup$
add a comment |
$begingroup$
Block diagonal matrices can be inverted block by block. See also [*].
In your example:
$$begin{bmatrix} 40 & 0 & 0 & 0 \ 0 & 80 & 100 & 0 \ 0 & 40 & 120 & 0 \ 0 & 0 & 0 & 60end{bmatrix}^{-1}
= begin{bmatrix} [40]^{-1} & begin{matrix} 0 quad & 0quad end{matrix} & 0 \ begin{matrix} 0 \ 0 end{matrix} & begin{bmatrix} 80 & 100 \ 40 & 120 end{bmatrix}^{-1} & begin{matrix} 0 \ 0 end{matrix} \
0 & begin{matrix} 0quad & 0quad end{matrix} & [60]^{-1}
end{bmatrix}.
$$
Can you take it from here?
answered Dec 3 '18 at 12:39
FlorianFlorian
1,3661721
$endgroup$
Block diagonal matrices can be inverted block by block. See also [*].
In your example:
$$begin{bmatrix} 40 & 0 & 0 & 0 \ 0 & 80 & 100 & 0 \ 0 & 40 & 120 & 0 \ 0 & 0 & 0 & 60end{bmatrix}^{-1}
= begin{bmatrix} [40]^{-1} & begin{matrix} 0 quad & 0quad end{matrix} & 0 \ begin{matrix} 0 \ 0 end{matrix} & begin{bmatrix} 80 & 100 \ 40 & 120 end{bmatrix}^{-1} & begin{matrix} 0 \ 0 end{matrix} \
0 & begin{matrix} 0quad & 0quad end{matrix} & [60]^{-1}
end{bmatrix}.
$$
Can you take it from here?
answered Dec 3 '18 at 12:39
FlorianFlorian
1,3661721
edited Dec 3 '18 at 13:22
answered Dec 3 '18 at 12:39
FlorianFlorian
1,3661721
answered Dec 3 '18 at 12:39
FlorianFlorian
1,3661721
answered Dec 3 '18 at 12:39
FlorianFlorian
1,3661721
1,3661721
add a comment |
add a comment |
$begingroup$
Guide:
If you have a matrix of the form of
$$diag(D_1, D_2, D_3),$$
where each block is invertible, then its inverse is $$diag(D_1^{-1},D_2^{-1}, D_3^{-1}).$$
You should verify this.
In your question $D_2$ is $2$ by $2$ and the other two blocks are scalar.
answered Dec 3 '18 at 12:39
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
Guide:
If you have a matrix of the form of
$$diag(D_1, D_2, D_3),$$
where each block is invertible, then its inverse is $$diag(D_1^{-1},D_2^{-1}, D_3^{-1}).$$
You should verify this.
In your question $D_2$ is $2$ by $2$ and the other two blocks are scalar.
answered Dec 3 '18 at 12:39
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
Guide:
If you have a matrix of the form of
$$diag(D_1, D_2, D_3),$$
where each block is invertible, then its inverse is $$diag(D_1^{-1},D_2^{-1}, D_3^{-1}).$$
You should verify this.
In your question $D_2$ is $2$ by $2$ and the other two blocks are scalar.
answered Dec 3 '18 at 12:39
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
Guide:
If you have a matrix of the form of
$$diag(D_1, D_2, D_3),$$
where each block is invertible, then its inverse is $$diag(D_1^{-1},D_2^{-1}, D_3^{-1}).$$
You should verify this.
In your question $D_2$ is $2$ by $2$ and the other two blocks are scalar.
answered Dec 3 '18 at 12:39
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 3 '18 at 12:39
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 3 '18 at 12:39
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 3 '18 at 12:39
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
add a comment |
add a comment |
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