Is this block matrix invertible? [closed]
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Suppose that $A$ is full column rank matrix. Define
$$
L=begin{pmatrix}
A&0\
0&A^{T}
end{pmatrix}.
$$
Can we prove/disprove that $L$ is invertible?
linear-algebra
closed as off-topic by amWhy, KReiser, Cesareo, Saad, Brahadeesh Nov 28 '18 at 5:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, KReiser, Cesareo, Saad, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Suppose that $A$ is full column rank matrix. Define
$$
L=begin{pmatrix}
A&0\
0&A^{T}
end{pmatrix}.
$$
Can we prove/disprove that $L$ is invertible?
linear-algebra
closed as off-topic by amWhy, KReiser, Cesareo, Saad, Brahadeesh Nov 28 '18 at 5:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, KReiser, Cesareo, Saad, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
3
You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
– darij grinberg
Nov 27 '18 at 21:10
add a comment |
Suppose that $A$ is full column rank matrix. Define
$$
L=begin{pmatrix}
A&0\
0&A^{T}
end{pmatrix}.
$$
Can we prove/disprove that $L$ is invertible?
linear-algebra
Suppose that $A$ is full column rank matrix. Define
$$
L=begin{pmatrix}
A&0\
0&A^{T}
end{pmatrix}.
$$
Can we prove/disprove that $L$ is invertible?
linear-algebra
linear-algebra
asked Nov 27 '18 at 21:01
Math_Math_
113
113
closed as off-topic by amWhy, KReiser, Cesareo, Saad, Brahadeesh Nov 28 '18 at 5:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, KReiser, Cesareo, Saad, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, KReiser, Cesareo, Saad, Brahadeesh Nov 28 '18 at 5:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, KReiser, Cesareo, Saad, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
3
You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
– darij grinberg
Nov 27 '18 at 21:10
add a comment |
3
You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
– darij grinberg
Nov 27 '18 at 21:10
3
3
You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
– darij grinberg
Nov 27 '18 at 21:10
You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
– darij grinberg
Nov 27 '18 at 21:10
add a comment |
3 Answers
3
active
oldest
votes
This is true only when $A$ is square.
Take
$$
A = begin{pmatrix}
1 \
0 end{pmatrix}; qquad
L = begin{pmatrix}
1&0&0 \
0&0&0 \
0&1&0 end{pmatrix}.$$
Our matrix $L$ is certainly not invertible.
add a comment |
You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
begin{bmatrix}
A^{-1} & 0 \
0 & (A^{T})^{-1}
end{bmatrix}
is the inverse of the original matrix.
Here $A$ is not square matrix.
– Math_
Nov 27 '18 at 21:15
add a comment |
Based on your comments, we have that $A$ is an $n times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m times n$ matrix.
A quick aside:
Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$det begin{pmatrix} X & 0 \ Y & Z end{pmatrix} = det(X) det(Z)$$
(ripped from the wikipedia page on the determinant)
So in our case, the block matrix $begin{pmatrix} A & 0 \ 0 & A^T end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.
For example, for a $3 times 2$ matrix, $$left ( begin{array}{c c c | c c}
a & b & 0 & 0 & 0 \
c & d & 0 & 0 & 0 \
e & f & 0 & 0 & 0 \
hline
0 & 0 & a & c & e \
0 & 0 & b & d & f \
end{array} right )$$
However since we appended a column of zeros, the determinant of the upper square block is $0$, which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is true only when $A$ is square.
Take
$$
A = begin{pmatrix}
1 \
0 end{pmatrix}; qquad
L = begin{pmatrix}
1&0&0 \
0&0&0 \
0&1&0 end{pmatrix}.$$
Our matrix $L$ is certainly not invertible.
add a comment |
This is true only when $A$ is square.
Take
$$
A = begin{pmatrix}
1 \
0 end{pmatrix}; qquad
L = begin{pmatrix}
1&0&0 \
0&0&0 \
0&1&0 end{pmatrix}.$$
Our matrix $L$ is certainly not invertible.
add a comment |
This is true only when $A$ is square.
Take
$$
A = begin{pmatrix}
1 \
0 end{pmatrix}; qquad
L = begin{pmatrix}
1&0&0 \
0&0&0 \
0&1&0 end{pmatrix}.$$
Our matrix $L$ is certainly not invertible.
This is true only when $A$ is square.
Take
$$
A = begin{pmatrix}
1 \
0 end{pmatrix}; qquad
L = begin{pmatrix}
1&0&0 \
0&0&0 \
0&1&0 end{pmatrix}.$$
Our matrix $L$ is certainly not invertible.
answered Nov 27 '18 at 22:03
Santana AftonSantana Afton
2,5752629
2,5752629
add a comment |
add a comment |
You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
begin{bmatrix}
A^{-1} & 0 \
0 & (A^{T})^{-1}
end{bmatrix}
is the inverse of the original matrix.
Here $A$ is not square matrix.
– Math_
Nov 27 '18 at 21:15
add a comment |
You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
begin{bmatrix}
A^{-1} & 0 \
0 & (A^{T})^{-1}
end{bmatrix}
is the inverse of the original matrix.
Here $A$ is not square matrix.
– Math_
Nov 27 '18 at 21:15
add a comment |
You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
begin{bmatrix}
A^{-1} & 0 \
0 & (A^{T})^{-1}
end{bmatrix}
is the inverse of the original matrix.
You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
begin{bmatrix}
A^{-1} & 0 \
0 & (A^{T})^{-1}
end{bmatrix}
is the inverse of the original matrix.
answered Nov 27 '18 at 21:13
mathnoobmathnoob
1,792422
1,792422
Here $A$ is not square matrix.
– Math_
Nov 27 '18 at 21:15
add a comment |
Here $A$ is not square matrix.
– Math_
Nov 27 '18 at 21:15
Here $A$ is not square matrix.
– Math_
Nov 27 '18 at 21:15
Here $A$ is not square matrix.
– Math_
Nov 27 '18 at 21:15
add a comment |
Based on your comments, we have that $A$ is an $n times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m times n$ matrix.
A quick aside:
Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$det begin{pmatrix} X & 0 \ Y & Z end{pmatrix} = det(X) det(Z)$$
(ripped from the wikipedia page on the determinant)
So in our case, the block matrix $begin{pmatrix} A & 0 \ 0 & A^T end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.
For example, for a $3 times 2$ matrix, $$left ( begin{array}{c c c | c c}
a & b & 0 & 0 & 0 \
c & d & 0 & 0 & 0 \
e & f & 0 & 0 & 0 \
hline
0 & 0 & a & c & e \
0 & 0 & b & d & f \
end{array} right )$$
However since we appended a column of zeros, the determinant of the upper square block is $0$, which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.
add a comment |
Based on your comments, we have that $A$ is an $n times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m times n$ matrix.
A quick aside:
Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$det begin{pmatrix} X & 0 \ Y & Z end{pmatrix} = det(X) det(Z)$$
(ripped from the wikipedia page on the determinant)
So in our case, the block matrix $begin{pmatrix} A & 0 \ 0 & A^T end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.
For example, for a $3 times 2$ matrix, $$left ( begin{array}{c c c | c c}
a & b & 0 & 0 & 0 \
c & d & 0 & 0 & 0 \
e & f & 0 & 0 & 0 \
hline
0 & 0 & a & c & e \
0 & 0 & b & d & f \
end{array} right )$$
However since we appended a column of zeros, the determinant of the upper square block is $0$, which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.
add a comment |
Based on your comments, we have that $A$ is an $n times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m times n$ matrix.
A quick aside:
Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$det begin{pmatrix} X & 0 \ Y & Z end{pmatrix} = det(X) det(Z)$$
(ripped from the wikipedia page on the determinant)
So in our case, the block matrix $begin{pmatrix} A & 0 \ 0 & A^T end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.
For example, for a $3 times 2$ matrix, $$left ( begin{array}{c c c | c c}
a & b & 0 & 0 & 0 \
c & d & 0 & 0 & 0 \
e & f & 0 & 0 & 0 \
hline
0 & 0 & a & c & e \
0 & 0 & b & d & f \
end{array} right )$$
However since we appended a column of zeros, the determinant of the upper square block is $0$, which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.
Based on your comments, we have that $A$ is an $n times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m times n$ matrix.
A quick aside:
Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$det begin{pmatrix} X & 0 \ Y & Z end{pmatrix} = det(X) det(Z)$$
(ripped from the wikipedia page on the determinant)
So in our case, the block matrix $begin{pmatrix} A & 0 \ 0 & A^T end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.
For example, for a $3 times 2$ matrix, $$left ( begin{array}{c c c | c c}
a & b & 0 & 0 & 0 \
c & d & 0 & 0 & 0 \
e & f & 0 & 0 & 0 \
hline
0 & 0 & a & c & e \
0 & 0 & b & d & f \
end{array} right )$$
However since we appended a column of zeros, the determinant of the upper square block is $0$, which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.
answered Nov 27 '18 at 21:32
JoeJoe
64119
64119
add a comment |
add a comment |
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3
You can prove it when $A$ is square (by constructing the inverse) and disprove it otherwise.
– darij grinberg
Nov 27 '18 at 21:10