Converse of statements about the rank of adjugate matrices
$begingroup$
Let the adjugate of a matrix be defined as the transpose of the cofactor matrix, denoted $A^{*}$. (Also termed the Classical Adjoint)
It can be proven that for any $ntimes n$ matrix $A$,
1) if rank($A$)= $n$ then rank(adj($A$)) = $n$
2) if rank($A$) = $n-1$ then rank(adj($A$)) = $1$
2) if rank($A$) < $n-1$ then rank(adj($A$)) = $0$
Suppose we were given an arbitrary $ntimes n$ matrix $A$. If we know that rank((adj($A$)) = $1$, does it necessarily imply that the rank of $A$ is $n-1$? For example, suppose $A$ has an unknown entry but it is given that rank(adj($A$)) =$1$, is it a valid approach to conclude that the rank of $A$ is $n-1$ and proceed to determine the unknown value by solving $det(A)=0$?
Similarly for the other two statements. Are they single direction statements or are they really if and only if statements?
linear-algebra matrices
asked Dec 30 '18 at 15:08
NetUser5y62NetUser5y62
525215
$endgroup$
add a comment |
$begingroup$
Let the adjugate of a matrix be defined as the transpose of the cofactor matrix, denoted $A^{*}$. (Also termed the Classical Adjoint)
It can be proven that for any $ntimes n$ matrix $A$,
1) if rank($A$)= $n$ then rank(adj($A$)) = $n$
2) if rank($A$) = $n-1$ then rank(adj($A$)) = $1$
2) if rank($A$) < $n-1$ then rank(adj($A$)) = $0$
Suppose we were given an arbitrary $ntimes n$ matrix $A$. If we know that rank((adj($A$)) = $1$, does it necessarily imply that the rank of $A$ is $n-1$? For example, suppose $A$ has an unknown entry but it is given that rank(adj($A$)) =$1$, is it a valid approach to conclude that the rank of $A$ is $n-1$ and proceed to determine the unknown value by solving $det(A)=0$?
Similarly for the other two statements. Are they single direction statements or are they really if and only if statements?
linear-algebra matrices
asked Dec 30 '18 at 15:08
NetUser5y62NetUser5y62
525215
$endgroup$
$begingroup$
I'm not familiar with the statement you posted. What is your definition of adjoint matrix? Either $operatorname{rank} A = n$, $operatorname{rank} A = n-1$, or $operatorname{rank} A < n-1$. Assuming your statement is correct, if $operatorname{rank}operatorname{adj} A = 1$ and $operatorname{rank} A neq n-1$ you have a contradiction so $operatorname{rank}(A)$ must be $n-1$.
$endgroup$
– tch
Dec 30 '18 at 15:23
$begingroup$
@TylerChen : thank you for the reply, an adjoint matrix is the conjugate transpose of a matrix. I updated the post with some queries.
$endgroup$
– NetUser5y62
Dec 30 '18 at 15:35
$begingroup$
How do you use that definition of $A$ is not invertible? In any case, I don't think that is the definition of conjugate transpose.
$endgroup$
– tch
Dec 30 '18 at 15:53
$begingroup$
@TylerChen thank you for correcting that, I didn’t realize that was just a property thats applies when $A$ is invertible. I have removed it from the definition.
$endgroup$
– NetUser5y62
Dec 30 '18 at 16:02
add a comment |
$begingroup$
Let the adjugate of a matrix be defined as the transpose of the cofactor matrix, denoted $A^{*}$. (Also termed the Classical Adjoint)
It can be proven that for any $ntimes n$ matrix $A$,
1) if rank($A$)= $n$ then rank(adj($A$)) = $n$
2) if rank($A$) = $n-1$ then rank(adj($A$)) = $1$
2) if rank($A$) < $n-1$ then rank(adj($A$)) = $0$
Suppose we were given an arbitrary $ntimes n$ matrix $A$. If we know that rank((adj($A$)) = $1$, does it necessarily imply that the rank of $A$ is $n-1$? For example, suppose $A$ has an unknown entry but it is given that rank(adj($A$)) =$1$, is it a valid approach to conclude that the rank of $A$ is $n-1$ and proceed to determine the unknown value by solving $det(A)=0$?
Similarly for the other two statements. Are they single direction statements or are they really if and only if statements?
linear-algebra matrices
asked Dec 30 '18 at 15:08
NetUser5y62NetUser5y62
525215
$endgroup$
Let the adjugate of a matrix be defined as the transpose of the cofactor matrix, denoted $A^{*}$. (Also termed the Classical Adjoint)
It can be proven that for any $ntimes n$ matrix $A$,
1) if rank($A$)= $n$ then rank(adj($A$)) = $n$
2) if rank($A$) = $n-1$ then rank(adj($A$)) = $1$
2) if rank($A$) < $n-1$ then rank(adj($A$)) = $0$
Suppose we were given an arbitrary $ntimes n$ matrix $A$. If we know that rank((adj($A$)) = $1$, does it necessarily imply that the rank of $A$ is $n-1$? For example, suppose $A$ has an unknown entry but it is given that rank(adj($A$)) =$1$, is it a valid approach to conclude that the rank of $A$ is $n-1$ and proceed to determine the unknown value by solving $det(A)=0$?
Similarly for the other two statements. Are they single direction statements or are they really if and only if statements?
linear-algebra matrices
linear-algebra matrices
asked Dec 30 '18 at 15:08
NetUser5y62NetUser5y62
525215
asked Dec 30 '18 at 15:08
NetUser5y62NetUser5y62
525215
edited Dec 31 '18 at 4:08
NetUser5y62
asked Dec 30 '18 at 15:08
NetUser5y62NetUser5y62
525215
asked Dec 30 '18 at 15:08
NetUser5y62NetUser5y62
525215
asked Dec 30 '18 at 15:08
NetUser5y62NetUser5y62
525215
525215
$begingroup$
I'm not familiar with the statement you posted. What is your definition of adjoint matrix? Either $operatorname{rank} A = n$, $operatorname{rank} A = n-1$, or $operatorname{rank} A < n-1$. Assuming your statement is correct, if $operatorname{rank}operatorname{adj} A = 1$ and $operatorname{rank} A neq n-1$ you have a contradiction so $operatorname{rank}(A)$ must be $n-1$.
$endgroup$
– tch
Dec 30 '18 at 15:23
$begingroup$
@TylerChen : thank you for the reply, an adjoint matrix is the conjugate transpose of a matrix. I updated the post with some queries.
$endgroup$
– NetUser5y62
Dec 30 '18 at 15:35
$begingroup$
How do you use that definition of $A$ is not invertible? In any case, I don't think that is the definition of conjugate transpose.
$endgroup$
– tch
Dec 30 '18 at 15:53
$begingroup$
@TylerChen thank you for correcting that, I didn’t realize that was just a property thats applies when $A$ is invertible. I have removed it from the definition.
$endgroup$
– NetUser5y62
Dec 30 '18 at 16:02
add a comment |
$begingroup$
I'm not familiar with the statement you posted. What is your definition of adjoint matrix? Either $operatorname{rank} A = n$, $operatorname{rank} A = n-1$, or $operatorname{rank} A < n-1$. Assuming your statement is correct, if $operatorname{rank}operatorname{adj} A = 1$ and $operatorname{rank} A neq n-1$ you have a contradiction so $operatorname{rank}(A)$ must be $n-1$.
$endgroup$
– tch
Dec 30 '18 at 15:23
$begingroup$
@TylerChen : thank you for the reply, an adjoint matrix is the conjugate transpose of a matrix. I updated the post with some queries.
$endgroup$
– NetUser5y62
Dec 30 '18 at 15:35
$begingroup$
How do you use that definition of $A$ is not invertible? In any case, I don't think that is the definition of conjugate transpose.
$endgroup$
– tch
Dec 30 '18 at 15:53
$begingroup$
@TylerChen thank you for correcting that, I didn’t realize that was just a property thats applies when $A$ is invertible. I have removed it from the definition.
$endgroup$
– NetUser5y62
Dec 30 '18 at 16:02
$begingroup$
I'm not familiar with the statement you posted. What is your definition of adjoint matrix? Either $operatorname{rank} A = n$, $operatorname{rank} A = n-1$, or $operatorname{rank} A < n-1$. Assuming your statement is correct, if $operatorname{rank}operatorname{adj} A = 1$ and $operatorname{rank} A neq n-1$ you have a contradiction so $operatorname{rank}(A)$ must be $n-1$.
$endgroup$
– tch
Dec 30 '18 at 15:23
$begingroup$
I'm not familiar with the statement you posted. What is your definition of adjoint matrix? Either $operatorname{rank} A = n$, $operatorname{rank} A = n-1$, or $operatorname{rank} A < n-1$. Assuming your statement is correct, if $operatorname{rank}operatorname{adj} A = 1$ and $operatorname{rank} A neq n-1$ you have a contradiction so $operatorname{rank}(A)$ must be $n-1$.
$endgroup$
– tch
Dec 30 '18 at 15:23
$begingroup$
@TylerChen : thank you for the reply, an adjoint matrix is the conjugate transpose of a matrix. I updated the post with some queries.
$endgroup$
– NetUser5y62
Dec 30 '18 at 15:35
$begingroup$
@TylerChen : thank you for the reply, an adjoint matrix is the conjugate transpose of a matrix. I updated the post with some queries.
$endgroup$
– NetUser5y62
Dec 30 '18 at 15:35
$begingroup$
How do you use that definition of $A$ is not invertible? In any case, I don't think that is the definition of conjugate transpose.
$endgroup$
– tch
Dec 30 '18 at 15:53
$begingroup$
How do you use that definition of $A$ is not invertible? In any case, I don't think that is the definition of conjugate transpose.
$endgroup$
– tch
Dec 30 '18 at 15:53
$begingroup$
@TylerChen thank you for correcting that, I didn’t realize that was just a property thats applies when $A$ is invertible. I have removed it from the definition.
$endgroup$
– NetUser5y62
Dec 30 '18 at 16:02
$begingroup$
@TylerChen thank you for correcting that, I didn’t realize that was just a property thats applies when $A$ is invertible. I have removed it from the definition.
$endgroup$
– NetUser5y62
Dec 30 '18 at 16:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
1)Yes. If $mathrm{rank}(mathrm{adj}(A))=n$ then $mathrm{adj}(A)$ is invertible, so $A=mathrm{det}(A)mathrm{adj}(A)^{-1}$ is invertible, because it can not be zero.
2)Yes. If $mathrm{rank}(mathrm{adj}(A))=1$ then $A$ is not invertible so $mathrm{rank}(A)leq n-1$ but by 3) we can not have $mathrm{rank}(A)< n-1$, since it would imply $mathrm{adj}(A)=0$. Thus $mathrm{rank}(A)= n-1$.
3)Yes. If $mathrm{adj}(A)=0$ then all $(n−1)×(n−1)$ minors of $A$ are zero, hence $mathrm{rank}(A)leq 2$
answered Dec 30 '18 at 15:44
moutheticsmouthetics
52137
$endgroup$
add a comment |
$begingroup$
If by adjoint of $A$ you mean adjoint operator, i.e. $operatorname{adj}(A) = A^*$ then your claimed properties are not true. For instance, any Hermetian matrix satisfies $A=A^*$ so the rank of $A$ is the rank of $A^*$.
More generally, for any matrix $A$, the conjugate transpose $A^*$ always has the same rank.
As an explicit example, consider the $3times 3$ matrix:
$$
A = begin{bmatrix}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 0
end{bmatrix}
$$
This is clearly rank $2$, but the conjugate transpose is also rank 2.
That said, if there exists some definition of "adjoint" satisfying the 3 properties you stated, then it is true that $operatorname{rank}operatorname{adj}(A)=1$ implies $operatorname{rank}(A) = n-1$. To see this, note that if $operatorname{rank}(A)$ did not equal $n-1$ then $operatorname{rank}operatorname{adj}(A)neq1$
answered Dec 30 '18 at 16:29
tchtch
833310
$endgroup$
$begingroup$
$mathrm{adj}(A)$ is the transpose of the cofactor matrix of $A$ which make it different from $A^*$ the transpose of the conjugate.
$endgroup$
– mouthetics
Dec 31 '18 at 15:15
add a comment |
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2 Answers
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2 Answers
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$begingroup$
1)Yes. If $mathrm{rank}(mathrm{adj}(A))=n$ then $mathrm{adj}(A)$ is invertible, so $A=mathrm{det}(A)mathrm{adj}(A)^{-1}$ is invertible, because it can not be zero.
2)Yes. If $mathrm{rank}(mathrm{adj}(A))=1$ then $A$ is not invertible so $mathrm{rank}(A)leq n-1$ but by 3) we can not have $mathrm{rank}(A)< n-1$, since it would imply $mathrm{adj}(A)=0$. Thus $mathrm{rank}(A)= n-1$.
3)Yes. If $mathrm{adj}(A)=0$ then all $(n−1)×(n−1)$ minors of $A$ are zero, hence $mathrm{rank}(A)leq 2$
answered Dec 30 '18 at 15:44
moutheticsmouthetics
52137
$endgroup$
add a comment |
$begingroup$
1)Yes. If $mathrm{rank}(mathrm{adj}(A))=n$ then $mathrm{adj}(A)$ is invertible, so $A=mathrm{det}(A)mathrm{adj}(A)^{-1}$ is invertible, because it can not be zero.
2)Yes. If $mathrm{rank}(mathrm{adj}(A))=1$ then $A$ is not invertible so $mathrm{rank}(A)leq n-1$ but by 3) we can not have $mathrm{rank}(A)< n-1$, since it would imply $mathrm{adj}(A)=0$. Thus $mathrm{rank}(A)= n-1$.
3)Yes. If $mathrm{adj}(A)=0$ then all $(n−1)×(n−1)$ minors of $A$ are zero, hence $mathrm{rank}(A)leq 2$
answered Dec 30 '18 at 15:44
moutheticsmouthetics
52137
$endgroup$
add a comment |
$begingroup$
1)Yes. If $mathrm{rank}(mathrm{adj}(A))=n$ then $mathrm{adj}(A)$ is invertible, so $A=mathrm{det}(A)mathrm{adj}(A)^{-1}$ is invertible, because it can not be zero.
2)Yes. If $mathrm{rank}(mathrm{adj}(A))=1$ then $A$ is not invertible so $mathrm{rank}(A)leq n-1$ but by 3) we can not have $mathrm{rank}(A)< n-1$, since it would imply $mathrm{adj}(A)=0$. Thus $mathrm{rank}(A)= n-1$.
3)Yes. If $mathrm{adj}(A)=0$ then all $(n−1)×(n−1)$ minors of $A$ are zero, hence $mathrm{rank}(A)leq 2$
answered Dec 30 '18 at 15:44
moutheticsmouthetics
52137
$endgroup$
1)Yes. If $mathrm{rank}(mathrm{adj}(A))=n$ then $mathrm{adj}(A)$ is invertible, so $A=mathrm{det}(A)mathrm{adj}(A)^{-1}$ is invertible, because it can not be zero.
2)Yes. If $mathrm{rank}(mathrm{adj}(A))=1$ then $A$ is not invertible so $mathrm{rank}(A)leq n-1$ but by 3) we can not have $mathrm{rank}(A)< n-1$, since it would imply $mathrm{adj}(A)=0$. Thus $mathrm{rank}(A)= n-1$.
3)Yes. If $mathrm{adj}(A)=0$ then all $(n−1)×(n−1)$ minors of $A$ are zero, hence $mathrm{rank}(A)leq 2$
answered Dec 30 '18 at 15:44
moutheticsmouthetics
52137
edited Dec 30 '18 at 16:10
answered Dec 30 '18 at 15:44
moutheticsmouthetics
52137
answered Dec 30 '18 at 15:44
moutheticsmouthetics
52137
answered Dec 30 '18 at 15:44
moutheticsmouthetics
52137
52137
add a comment |
add a comment |
$begingroup$
If by adjoint of $A$ you mean adjoint operator, i.e. $operatorname{adj}(A) = A^*$ then your claimed properties are not true. For instance, any Hermetian matrix satisfies $A=A^*$ so the rank of $A$ is the rank of $A^*$.
More generally, for any matrix $A$, the conjugate transpose $A^*$ always has the same rank.
As an explicit example, consider the $3times 3$ matrix:
$$
A = begin{bmatrix}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 0
end{bmatrix}
$$
This is clearly rank $2$, but the conjugate transpose is also rank 2.
That said, if there exists some definition of "adjoint" satisfying the 3 properties you stated, then it is true that $operatorname{rank}operatorname{adj}(A)=1$ implies $operatorname{rank}(A) = n-1$. To see this, note that if $operatorname{rank}(A)$ did not equal $n-1$ then $operatorname{rank}operatorname{adj}(A)neq1$
answered Dec 30 '18 at 16:29
tchtch
833310
$endgroup$
$begingroup$
$mathrm{adj}(A)$ is the transpose of the cofactor matrix of $A$ which make it different from $A^*$ the transpose of the conjugate.
$endgroup$
– mouthetics
Dec 31 '18 at 15:15
add a comment |
$begingroup$
If by adjoint of $A$ you mean adjoint operator, i.e. $operatorname{adj}(A) = A^*$ then your claimed properties are not true. For instance, any Hermetian matrix satisfies $A=A^*$ so the rank of $A$ is the rank of $A^*$.
More generally, for any matrix $A$, the conjugate transpose $A^*$ always has the same rank.
As an explicit example, consider the $3times 3$ matrix:
$$
A = begin{bmatrix}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 0
end{bmatrix}
$$
This is clearly rank $2$, but the conjugate transpose is also rank 2.
That said, if there exists some definition of "adjoint" satisfying the 3 properties you stated, then it is true that $operatorname{rank}operatorname{adj}(A)=1$ implies $operatorname{rank}(A) = n-1$. To see this, note that if $operatorname{rank}(A)$ did not equal $n-1$ then $operatorname{rank}operatorname{adj}(A)neq1$
answered Dec 30 '18 at 16:29
tchtch
833310
$endgroup$
$begingroup$
$mathrm{adj}(A)$ is the transpose of the cofactor matrix of $A$ which make it different from $A^*$ the transpose of the conjugate.
$endgroup$
– mouthetics
Dec 31 '18 at 15:15
add a comment |
$begingroup$
If by adjoint of $A$ you mean adjoint operator, i.e. $operatorname{adj}(A) = A^*$ then your claimed properties are not true. For instance, any Hermetian matrix satisfies $A=A^*$ so the rank of $A$ is the rank of $A^*$.
More generally, for any matrix $A$, the conjugate transpose $A^*$ always has the same rank.
As an explicit example, consider the $3times 3$ matrix:
$$
A = begin{bmatrix}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 0
end{bmatrix}
$$
This is clearly rank $2$, but the conjugate transpose is also rank 2.
That said, if there exists some definition of "adjoint" satisfying the 3 properties you stated, then it is true that $operatorname{rank}operatorname{adj}(A)=1$ implies $operatorname{rank}(A) = n-1$. To see this, note that if $operatorname{rank}(A)$ did not equal $n-1$ then $operatorname{rank}operatorname{adj}(A)neq1$
answered Dec 30 '18 at 16:29
tchtch
833310
$endgroup$
If by adjoint of $A$ you mean adjoint operator, i.e. $operatorname{adj}(A) = A^*$ then your claimed properties are not true. For instance, any Hermetian matrix satisfies $A=A^*$ so the rank of $A$ is the rank of $A^*$.
More generally, for any matrix $A$, the conjugate transpose $A^*$ always has the same rank.
As an explicit example, consider the $3times 3$ matrix:
$$
A = begin{bmatrix}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 0
end{bmatrix}
$$
This is clearly rank $2$, but the conjugate transpose is also rank 2.
That said, if there exists some definition of "adjoint" satisfying the 3 properties you stated, then it is true that $operatorname{rank}operatorname{adj}(A)=1$ implies $operatorname{rank}(A) = n-1$. To see this, note that if $operatorname{rank}(A)$ did not equal $n-1$ then $operatorname{rank}operatorname{adj}(A)neq1$
answered Dec 30 '18 at 16:29
tchtch
833310
edited Dec 30 '18 at 16:41
answered Dec 30 '18 at 16:29
tchtch
833310
answered Dec 30 '18 at 16:29
tchtch
833310
answered Dec 30 '18 at 16:29
tchtch
833310
833310
$begingroup$
$mathrm{adj}(A)$ is the transpose of the cofactor matrix of $A$ which make it different from $A^*$ the transpose of the conjugate.
$endgroup$
– mouthetics
Dec 31 '18 at 15:15
add a comment |
$begingroup$
$mathrm{adj}(A)$ is the transpose of the cofactor matrix of $A$ which make it different from $A^*$ the transpose of the conjugate.
$endgroup$
– mouthetics
Dec 31 '18 at 15:15
$begingroup$
$mathrm{adj}(A)$ is the transpose of the cofactor matrix of $A$ which make it different from $A^*$ the transpose of the conjugate.
$endgroup$
– mouthetics
Dec 31 '18 at 15:15
$begingroup$
$mathrm{adj}(A)$ is the transpose of the cofactor matrix of $A$ which make it different from $A^*$ the transpose of the conjugate.
$endgroup$
– mouthetics
Dec 31 '18 at 15:15
add a comment |
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$begingroup$
I'm not familiar with the statement you posted. What is your definition of adjoint matrix? Either $operatorname{rank} A = n$, $operatorname{rank} A = n-1$, or $operatorname{rank} A < n-1$. Assuming your statement is correct, if $operatorname{rank}operatorname{adj} A = 1$ and $operatorname{rank} A neq n-1$ you have a contradiction so $operatorname{rank}(A)$ must be $n-1$.
$endgroup$
– tch
Dec 30 '18 at 15:23
$begingroup$
@TylerChen : thank you for the reply, an adjoint matrix is the conjugate transpose of a matrix. I updated the post with some queries.
$endgroup$
– NetUser5y62
Dec 30 '18 at 15:35
$begingroup$
How do you use that definition of $A$ is not invertible? In any case, I don't think that is the definition of conjugate transpose.
$endgroup$
– tch
Dec 30 '18 at 15:53
$begingroup$
@TylerChen thank you for correcting that, I didn’t realize that was just a property thats applies when $A$ is invertible. I have removed it from the definition.
$endgroup$
– NetUser5y62
Dec 30 '18 at 16:02