Mixing
Basic question: How to solve the following $2 times 2$ (block-based) linear system analytically?
$begingroup$
Sorry for asking basic question, but I think I have some trouble in understanding the below part. Please excuse me for asking trivial question.
Say $2 times 2$ linear system is
begin{align}
begin{bmatrix}
H & A^T \
A & 0
end{bmatrix}
begin{bmatrix}
x \
y
end{bmatrix}
&=
begin{bmatrix}
b \
c
end{bmatrix}
end{align}
$H in mathbb{R}^{n times n}$ is a non-singular matrix, $A in mathbb{R}^{m times n}$, $b in mathbb{R}^{n times 1}$, $c in mathbb{R}^{m times 1} $, the unknowns $x in mathbb{R}^{n times 1} $, and $y in mathbb{R}^{m times 1} $.
Then solve for $x$ and $y$ based on
begin{align}
A H^{-1} A^T y &= -c + AH^{-1}b \
H x &= b - A^Ty .
end{align}
Question:
How to reach to those two system of equations?
linear-algebra
asked Jan 28 at 20:54
learninglearning
718
$endgroup$
add a comment |
$begingroup$
Sorry for asking basic question, but I think I have some trouble in understanding the below part. Please excuse me for asking trivial question.
Say $2 times 2$ linear system is
begin{align}
begin{bmatrix}
H & A^T \
A & 0
end{bmatrix}
begin{bmatrix}
x \
y
end{bmatrix}
&=
begin{bmatrix}
b \
c
end{bmatrix}
end{align}
$H in mathbb{R}^{n times n}$ is a non-singular matrix, $A in mathbb{R}^{m times n}$, $b in mathbb{R}^{n times 1}$, $c in mathbb{R}^{m times 1} $, the unknowns $x in mathbb{R}^{n times 1} $, and $y in mathbb{R}^{m times 1} $.
Then solve for $x$ and $y$ based on
begin{align}
A H^{-1} A^T y &= -c + AH^{-1}b \
H x &= b - A^Ty .
end{align}
Question:
How to reach to those two system of equations?
linear-algebra
asked Jan 28 at 20:54
learninglearning
718
$endgroup$
$begingroup$
Could you give us some context? Where did you encounter this? Do you understand that the system can be rewritten as $$ Hx + A^Ty = b\ Ax = c? $$
$endgroup$
– Omnomnomnom
Jan 28 at 21:13
$begingroup$
I Think I understand the matrix vector Products. Yes, I found this on page 12 of this slidekit see.stanford.edu/materials/lsocoee364a/11EqualityMin.pdf.
$endgroup$
– learning
Jan 28 at 21:16
add a comment |
$begingroup$
Sorry for asking basic question, but I think I have some trouble in understanding the below part. Please excuse me for asking trivial question.
Say $2 times 2$ linear system is
begin{align}
begin{bmatrix}
H & A^T \
A & 0
end{bmatrix}
begin{bmatrix}
x \
y
end{bmatrix}
&=
begin{bmatrix}
b \
c
end{bmatrix}
end{align}
$H in mathbb{R}^{n times n}$ is a non-singular matrix, $A in mathbb{R}^{m times n}$, $b in mathbb{R}^{n times 1}$, $c in mathbb{R}^{m times 1} $, the unknowns $x in mathbb{R}^{n times 1} $, and $y in mathbb{R}^{m times 1} $.
Then solve for $x$ and $y$ based on
begin{align}
A H^{-1} A^T y &= -c + AH^{-1}b \
H x &= b - A^Ty .
end{align}
Question:
How to reach to those two system of equations?
linear-algebra
asked Jan 28 at 20:54
learninglearning
718
$endgroup$
Sorry for asking basic question, but I think I have some trouble in understanding the below part. Please excuse me for asking trivial question.
Say $2 times 2$ linear system is
begin{align}
begin{bmatrix}
H & A^T \
A & 0
end{bmatrix}
begin{bmatrix}
x \
y
end{bmatrix}
&=
begin{bmatrix}
b \
c
end{bmatrix}
end{align}
$H in mathbb{R}^{n times n}$ is a non-singular matrix, $A in mathbb{R}^{m times n}$, $b in mathbb{R}^{n times 1}$, $c in mathbb{R}^{m times 1} $, the unknowns $x in mathbb{R}^{n times 1} $, and $y in mathbb{R}^{m times 1} $.
Then solve for $x$ and $y$ based on
begin{align}
A H^{-1} A^T y &= -c + AH^{-1}b \
H x &= b - A^Ty .
end{align}
Question:
How to reach to those two system of equations?
linear-algebra
linear-algebra
asked Jan 28 at 20:54
learninglearning
718
asked Jan 28 at 20:54
learninglearning
718
edited Jan 28 at 21:24
learning
asked Jan 28 at 20:54
learninglearning
718
asked Jan 28 at 20:54
learninglearning
718
asked Jan 28 at 20:54
learninglearning
718
718
$begingroup$
Could you give us some context? Where did you encounter this? Do you understand that the system can be rewritten as $$ Hx + A^Ty = b\ Ax = c? $$
$endgroup$
– Omnomnomnom
Jan 28 at 21:13
$begingroup$
I Think I understand the matrix vector Products. Yes, I found this on page 12 of this slidekit see.stanford.edu/materials/lsocoee364a/11EqualityMin.pdf.
$endgroup$
– learning
Jan 28 at 21:16
add a comment |
$begingroup$
Could you give us some context? Where did you encounter this? Do you understand that the system can be rewritten as $$ Hx + A^Ty = b\ Ax = c? $$
$endgroup$
– Omnomnomnom
Jan 28 at 21:13
$begingroup$
I Think I understand the matrix vector Products. Yes, I found this on page 12 of this slidekit see.stanford.edu/materials/lsocoee364a/11EqualityMin.pdf.
$endgroup$
– learning
Jan 28 at 21:16
$begingroup$
Could you give us some context? Where did you encounter this? Do you understand that the system can be rewritten as $$ Hx + A^Ty = b\ Ax = c? $$
$endgroup$
– Omnomnomnom
Jan 28 at 21:13
$begingroup$
Could you give us some context? Where did you encounter this? Do you understand that the system can be rewritten as $$ Hx + A^Ty = b\ Ax = c? $$
$endgroup$
– Omnomnomnom
Jan 28 at 21:13
$begingroup$
I Think I understand the matrix vector Products. Yes, I found this on page 12 of this slidekit see.stanford.edu/materials/lsocoee364a/11EqualityMin.pdf.
$endgroup$
– learning
Jan 28 at 21:16
$begingroup$
I Think I understand the matrix vector Products. Yes, I found this on page 12 of this slidekit see.stanford.edu/materials/lsocoee364a/11EqualityMin.pdf.
$endgroup$
– learning
Jan 28 at 21:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
One approach is to use block-wise row-operations. That is:
$$
left[begin{array}{cc|c}
H&A^T & b\A&0&c
end{array}right] to \
left[begin{array}{cc|c}
I& H^{-1}A^T & H^{-1}b\A&0&c
end{array}right] to \
left[begin{array}{cc|c}
I& H^{-1}A^T & H^{-1}b\
0&-AH^{-1}A^T & c - AH^{-1}b
end{array}right]
$$
which leaves us with the system of equations
$$
x + H^{-1}A^Ty = H^{-1}b\
-AHA^Ty = c - AH^{-1}b
$$
answered Jan 28 at 21:22
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
1
$begingroup$
I think this is what they mean by "elimination". The step of multiplying the first row by $H^{-1}$ is not necessary, strictly speaking
$endgroup$
– Omnomnomnom
Jan 28 at 21:23
$begingroup$
Thank you so much for your help
$endgroup$
– learning
Jan 28 at 21:24
add a comment |
$begingroup$
The system reads
$$
Hx + A^T y = b, \Ax = c
$$
If matrix $A$ was invertible, this would a trivial problem, since from the second equation we would have $x = A^{-1} c$ and the 1st equation would simply become $H A^{-1}c + A^{T}y = b Leftrightarrow y = A^{-T}(b-H A^{T}c)$. However, this is not the case... So we must manipulate the equations in such a way that we are able to isolate x and y. We are allowed to use $H^{-1}$, but not $A^{-1}$ or $A^{-T}$.
Taking the 1st equation, you get $x=H^{-1}(b-A^{T}y)$. Substituting x in the second equation, you get $A H^{-1}(b-A^{T}y) = c Leftrightarrow A H^{-1} A^{T} y = -c + A H^{-1} b $. Hope it helps!
answered Jan 28 at 21:31
PierreCarrePierreCarre
1,695212
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One approach is to use block-wise row-operations. That is:
$$
left[begin{array}{cc|c}
H&A^T & b\A&0&c
end{array}right] to \
left[begin{array}{cc|c}
I& H^{-1}A^T & H^{-1}b\A&0&c
end{array}right] to \
left[begin{array}{cc|c}
I& H^{-1}A^T & H^{-1}b\
0&-AH^{-1}A^T & c - AH^{-1}b
end{array}right]
$$
which leaves us with the system of equations
$$
x + H^{-1}A^Ty = H^{-1}b\
-AHA^Ty = c - AH^{-1}b
$$
answered Jan 28 at 21:22
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
1
$begingroup$
I think this is what they mean by "elimination". The step of multiplying the first row by $H^{-1}$ is not necessary, strictly speaking
$endgroup$
– Omnomnomnom
Jan 28 at 21:23
$begingroup$
Thank you so much for your help
$endgroup$
– learning
Jan 28 at 21:24
add a comment |
$begingroup$
One approach is to use block-wise row-operations. That is:
$$
left[begin{array}{cc|c}
H&A^T & b\A&0&c
end{array}right] to \
left[begin{array}{cc|c}
I& H^{-1}A^T & H^{-1}b\A&0&c
end{array}right] to \
left[begin{array}{cc|c}
I& H^{-1}A^T & H^{-1}b\
0&-AH^{-1}A^T & c - AH^{-1}b
end{array}right]
$$
which leaves us with the system of equations
$$
x + H^{-1}A^Ty = H^{-1}b\
-AHA^Ty = c - AH^{-1}b
$$
answered Jan 28 at 21:22
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
1
$begingroup$
I think this is what they mean by "elimination". The step of multiplying the first row by $H^{-1}$ is not necessary, strictly speaking
$endgroup$
– Omnomnomnom
Jan 28 at 21:23
$begingroup$
Thank you so much for your help
$endgroup$
– learning
Jan 28 at 21:24
add a comment |
$begingroup$
One approach is to use block-wise row-operations. That is:
$$
left[begin{array}{cc|c}
H&A^T & b\A&0&c
end{array}right] to \
left[begin{array}{cc|c}
I& H^{-1}A^T & H^{-1}b\A&0&c
end{array}right] to \
left[begin{array}{cc|c}
I& H^{-1}A^T & H^{-1}b\
0&-AH^{-1}A^T & c - AH^{-1}b
end{array}right]
$$
which leaves us with the system of equations
$$
x + H^{-1}A^Ty = H^{-1}b\
-AHA^Ty = c - AH^{-1}b
$$
answered Jan 28 at 21:22
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
One approach is to use block-wise row-operations. That is:
$$
left[begin{array}{cc|c}
H&A^T & b\A&0&c
end{array}right] to \
left[begin{array}{cc|c}
I& H^{-1}A^T & H^{-1}b\A&0&c
end{array}right] to \
left[begin{array}{cc|c}
I& H^{-1}A^T & H^{-1}b\
0&-AH^{-1}A^T & c - AH^{-1}b
end{array}right]
$$
which leaves us with the system of equations
$$
x + H^{-1}A^Ty = H^{-1}b\
-AHA^Ty = c - AH^{-1}b
$$
answered Jan 28 at 21:22
OmnomnomnomOmnomnomnom
129k792186
edited Jan 28 at 21:26
answered Jan 28 at 21:22
OmnomnomnomOmnomnomnom
129k792186
answered Jan 28 at 21:22
OmnomnomnomOmnomnomnom
129k792186
answered Jan 28 at 21:22
OmnomnomnomOmnomnomnom
129k792186
129k792186
1
$begingroup$
I think this is what they mean by "elimination". The step of multiplying the first row by $H^{-1}$ is not necessary, strictly speaking
$endgroup$
– Omnomnomnom
Jan 28 at 21:23
$begingroup$
Thank you so much for your help
$endgroup$
– learning
Jan 28 at 21:24
add a comment |
1
$begingroup$
I think this is what they mean by "elimination". The step of multiplying the first row by $H^{-1}$ is not necessary, strictly speaking
$endgroup$
– Omnomnomnom
Jan 28 at 21:23
$begingroup$
Thank you so much for your help
$endgroup$
– learning
Jan 28 at 21:24
1
1
$begingroup$
I think this is what they mean by "elimination". The step of multiplying the first row by $H^{-1}$ is not necessary, strictly speaking
$endgroup$
– Omnomnomnom
Jan 28 at 21:23
$begingroup$
I think this is what they mean by "elimination". The step of multiplying the first row by $H^{-1}$ is not necessary, strictly speaking
$endgroup$
– Omnomnomnom
Jan 28 at 21:23
$begingroup$
Thank you so much for your help
$endgroup$
– learning
Jan 28 at 21:24
$begingroup$
Thank you so much for your help
$endgroup$
– learning
Jan 28 at 21:24
add a comment |
$begingroup$
The system reads
$$
Hx + A^T y = b, \Ax = c
$$
If matrix $A$ was invertible, this would a trivial problem, since from the second equation we would have $x = A^{-1} c$ and the 1st equation would simply become $H A^{-1}c + A^{T}y = b Leftrightarrow y = A^{-T}(b-H A^{T}c)$. However, this is not the case... So we must manipulate the equations in such a way that we are able to isolate x and y. We are allowed to use $H^{-1}$, but not $A^{-1}$ or $A^{-T}$.
Taking the 1st equation, you get $x=H^{-1}(b-A^{T}y)$. Substituting x in the second equation, you get $A H^{-1}(b-A^{T}y) = c Leftrightarrow A H^{-1} A^{T} y = -c + A H^{-1} b $. Hope it helps!
answered Jan 28 at 21:31
PierreCarrePierreCarre
1,695212
$endgroup$
add a comment |
$begingroup$
The system reads
$$
Hx + A^T y = b, \Ax = c
$$
If matrix $A$ was invertible, this would a trivial problem, since from the second equation we would have $x = A^{-1} c$ and the 1st equation would simply become $H A^{-1}c + A^{T}y = b Leftrightarrow y = A^{-T}(b-H A^{T}c)$. However, this is not the case... So we must manipulate the equations in such a way that we are able to isolate x and y. We are allowed to use $H^{-1}$, but not $A^{-1}$ or $A^{-T}$.
Taking the 1st equation, you get $x=H^{-1}(b-A^{T}y)$. Substituting x in the second equation, you get $A H^{-1}(b-A^{T}y) = c Leftrightarrow A H^{-1} A^{T} y = -c + A H^{-1} b $. Hope it helps!
answered Jan 28 at 21:31
PierreCarrePierreCarre
1,695212
$endgroup$
add a comment |
$begingroup$
The system reads
$$
Hx + A^T y = b, \Ax = c
$$
If matrix $A$ was invertible, this would a trivial problem, since from the second equation we would have $x = A^{-1} c$ and the 1st equation would simply become $H A^{-1}c + A^{T}y = b Leftrightarrow y = A^{-T}(b-H A^{T}c)$. However, this is not the case... So we must manipulate the equations in such a way that we are able to isolate x and y. We are allowed to use $H^{-1}$, but not $A^{-1}$ or $A^{-T}$.
Taking the 1st equation, you get $x=H^{-1}(b-A^{T}y)$. Substituting x in the second equation, you get $A H^{-1}(b-A^{T}y) = c Leftrightarrow A H^{-1} A^{T} y = -c + A H^{-1} b $. Hope it helps!
answered Jan 28 at 21:31
PierreCarrePierreCarre
1,695212
$endgroup$
The system reads
$$
Hx + A^T y = b, \Ax = c
$$
If matrix $A$ was invertible, this would a trivial problem, since from the second equation we would have $x = A^{-1} c$ and the 1st equation would simply become $H A^{-1}c + A^{T}y = b Leftrightarrow y = A^{-T}(b-H A^{T}c)$. However, this is not the case... So we must manipulate the equations in such a way that we are able to isolate x and y. We are allowed to use $H^{-1}$, but not $A^{-1}$ or $A^{-T}$.
Taking the 1st equation, you get $x=H^{-1}(b-A^{T}y)$. Substituting x in the second equation, you get $A H^{-1}(b-A^{T}y) = c Leftrightarrow A H^{-1} A^{T} y = -c + A H^{-1} b $. Hope it helps!
answered Jan 28 at 21:31
PierreCarrePierreCarre
1,695212
answered Jan 28 at 21:31
PierreCarrePierreCarre
1,695212
answered Jan 28 at 21:31
PierreCarrePierreCarre
1,695212
answered Jan 28 at 21:31
PierreCarrePierreCarre
1,695212
1,695212
add a comment |
add a comment |
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$begingroup$
Could you give us some context? Where did you encounter this? Do you understand that the system can be rewritten as $$ Hx + A^Ty = b\ Ax = c? $$
$endgroup$
– Omnomnomnom
Jan 28 at 21:13
$begingroup$
I Think I understand the matrix vector Products. Yes, I found this on page 12 of this slidekit see.stanford.edu/materials/lsocoee364a/11EqualityMin.pdf.
$endgroup$
– learning
Jan 28 at 21:16