Special property of matrix $Ainmathbb{R}^{n,n}$ and it's determinant
$begingroup$
How to prove for matrix $Ainmathbb{R}^{n,n}$ that if in places where $x$ columns crosses with $y$ rows are placed $0$, then we can be sure, that
$$
det A = 0qquad qquad text{if }quad x+y > n
$$
.
matrices determinant
asked Dec 7 '18 at 8:19
avan1235avan1235
3217
$endgroup$
add a comment |
$begingroup$
How to prove for matrix $Ainmathbb{R}^{n,n}$ that if in places where $x$ columns crosses with $y$ rows are placed $0$, then we can be sure, that
$$
det A = 0qquad qquad text{if }quad x+y > n
$$
.
matrices determinant
asked Dec 7 '18 at 8:19
avan1235avan1235
3217
$endgroup$
$begingroup$
what are $x$ and $y$?
$endgroup$
– Lau
Dec 7 '18 at 8:27
$begingroup$
@Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
$endgroup$
– avan1235
Dec 7 '18 at 8:29
add a comment |
$begingroup$
How to prove for matrix $Ainmathbb{R}^{n,n}$ that if in places where $x$ columns crosses with $y$ rows are placed $0$, then we can be sure, that
$$
det A = 0qquad qquad text{if }quad x+y > n
$$
.
matrices determinant
asked Dec 7 '18 at 8:19
avan1235avan1235
3217
$endgroup$
How to prove for matrix $Ainmathbb{R}^{n,n}$ that if in places where $x$ columns crosses with $y$ rows are placed $0$, then we can be sure, that
$$
det A = 0qquad qquad text{if }quad x+y > n
$$
.
matrices determinant
matrices determinant
asked Dec 7 '18 at 8:19
avan1235avan1235
3217
asked Dec 7 '18 at 8:19
avan1235avan1235
3217
asked Dec 7 '18 at 8:19
avan1235avan1235
3217
asked Dec 7 '18 at 8:19
avan1235avan1235
3217
asked Dec 7 '18 at 8:19
avan1235avan1235
3217
3217
$begingroup$
what are $x$ and $y$?
$endgroup$
– Lau
Dec 7 '18 at 8:27
$begingroup$
@Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
$endgroup$
– avan1235
Dec 7 '18 at 8:29
add a comment |
$begingroup$
what are $x$ and $y$?
$endgroup$
– Lau
Dec 7 '18 at 8:27
$begingroup$
@Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
$endgroup$
– avan1235
Dec 7 '18 at 8:29
$begingroup$
what are $x$ and $y$?
$endgroup$
– Lau
Dec 7 '18 at 8:27
$begingroup$
what are $x$ and $y$?
$endgroup$
– Lau
Dec 7 '18 at 8:27
$begingroup$
@Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
$endgroup$
– avan1235
Dec 7 '18 at 8:29
$begingroup$
@Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
$endgroup$
– avan1235
Dec 7 '18 at 8:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $A=[a_{ij}]_n=begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\a_{21}&a_{22}&a_{23}&...&a_{2n}\vdots&vdots&vdots&&vdots\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}end{bmatrix}_n$
Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:
$A'=begin{bmatrix}0_{ytimes x}&B_{ytimes(n-x)}\C_{(n-y)times x}&D_{(n-y)times(n-x)}end{bmatrix}_n$
where $0_{ytimes x}$ is the zero matrix of indicated order.
Also note that $det A'=(-1)^kcdotdet A$ or $|det A'|=|det A|$
Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $begin{bmatrix}0&Bend{bmatrix}_{ytimes n}$ are linearly dependent. This means $det A'=0impliesdet A=0$.
answered Dec 7 '18 at 8:57
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
$begingroup$
If I rightly understand your question, I have a solution:
Consider
$$det A = det
begin{pmatrix}
O& B\
C& D
end{pmatrix}
$$
where $O$ is a $xtimes y$ zero matrix, $B$ is a $(n-x)times y$ matrix, $C$ is a $xtimes(n-y)$ matrix and $D$ is a $(n-x)times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $det{A}=0$.
In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).
answered Dec 7 '18 at 8:52
LauLau
517315
$endgroup$
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
$begingroup$
Let $A=[a_{ij}]_n=begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\a_{21}&a_{22}&a_{23}&...&a_{2n}\vdots&vdots&vdots&&vdots\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}end{bmatrix}_n$
Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:
$A'=begin{bmatrix}0_{ytimes x}&B_{ytimes(n-x)}\C_{(n-y)times x}&D_{(n-y)times(n-x)}end{bmatrix}_n$
where $0_{ytimes x}$ is the zero matrix of indicated order.
Also note that $det A'=(-1)^kcdotdet A$ or $|det A'|=|det A|$
Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $begin{bmatrix}0&Bend{bmatrix}_{ytimes n}$ are linearly dependent. This means $det A'=0impliesdet A=0$.
answered Dec 7 '18 at 8:57
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
$begingroup$
Let $A=[a_{ij}]_n=begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\a_{21}&a_{22}&a_{23}&...&a_{2n}\vdots&vdots&vdots&&vdots\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}end{bmatrix}_n$
Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:
$A'=begin{bmatrix}0_{ytimes x}&B_{ytimes(n-x)}\C_{(n-y)times x}&D_{(n-y)times(n-x)}end{bmatrix}_n$
where $0_{ytimes x}$ is the zero matrix of indicated order.
Also note that $det A'=(-1)^kcdotdet A$ or $|det A'|=|det A|$
Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $begin{bmatrix}0&Bend{bmatrix}_{ytimes n}$ are linearly dependent. This means $det A'=0impliesdet A=0$.
answered Dec 7 '18 at 8:57
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
$begingroup$
Let $A=[a_{ij}]_n=begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\a_{21}&a_{22}&a_{23}&...&a_{2n}\vdots&vdots&vdots&&vdots\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}end{bmatrix}_n$
Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:
$A'=begin{bmatrix}0_{ytimes x}&B_{ytimes(n-x)}\C_{(n-y)times x}&D_{(n-y)times(n-x)}end{bmatrix}_n$
where $0_{ytimes x}$ is the zero matrix of indicated order.
Also note that $det A'=(-1)^kcdotdet A$ or $|det A'|=|det A|$
Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $begin{bmatrix}0&Bend{bmatrix}_{ytimes n}$ are linearly dependent. This means $det A'=0impliesdet A=0$.
answered Dec 7 '18 at 8:57
Shubham JohriShubham Johri
5,172717
$endgroup$
Let $A=[a_{ij}]_n=begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\a_{21}&a_{22}&a_{23}&...&a_{2n}\vdots&vdots&vdots&&vdots\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}end{bmatrix}_n$
Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:
$A'=begin{bmatrix}0_{ytimes x}&B_{ytimes(n-x)}\C_{(n-y)times x}&D_{(n-y)times(n-x)}end{bmatrix}_n$
where $0_{ytimes x}$ is the zero matrix of indicated order.
Also note that $det A'=(-1)^kcdotdet A$ or $|det A'|=|det A|$
Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $begin{bmatrix}0&Bend{bmatrix}_{ytimes n}$ are linearly dependent. This means $det A'=0impliesdet A=0$.
answered Dec 7 '18 at 8:57
Shubham JohriShubham Johri
5,172717
edited Dec 7 '18 at 9:54
answered Dec 7 '18 at 8:57
Shubham JohriShubham Johri
5,172717
answered Dec 7 '18 at 8:57
Shubham JohriShubham Johri
5,172717
answered Dec 7 '18 at 8:57
Shubham JohriShubham Johri
5,172717
5,172717
add a comment |
add a comment |
$begingroup$
If I rightly understand your question, I have a solution:
Consider
$$det A = det
begin{pmatrix}
O& B\
C& D
end{pmatrix}
$$
where $O$ is a $xtimes y$ zero matrix, $B$ is a $(n-x)times y$ matrix, $C$ is a $xtimes(n-y)$ matrix and $D$ is a $(n-x)times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $det{A}=0$.
In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).
answered Dec 7 '18 at 8:52
LauLau
517315
$endgroup$
add a comment |
$begingroup$
If I rightly understand your question, I have a solution:
Consider
$$det A = det
begin{pmatrix}
O& B\
C& D
end{pmatrix}
$$
where $O$ is a $xtimes y$ zero matrix, $B$ is a $(n-x)times y$ matrix, $C$ is a $xtimes(n-y)$ matrix and $D$ is a $(n-x)times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $det{A}=0$.
In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).
answered Dec 7 '18 at 8:52
LauLau
517315
$endgroup$
add a comment |
$begingroup$
If I rightly understand your question, I have a solution:
Consider
$$det A = det
begin{pmatrix}
O& B\
C& D
end{pmatrix}
$$
where $O$ is a $xtimes y$ zero matrix, $B$ is a $(n-x)times y$ matrix, $C$ is a $xtimes(n-y)$ matrix and $D$ is a $(n-x)times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $det{A}=0$.
In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).
answered Dec 7 '18 at 8:52
LauLau
517315
$endgroup$
If I rightly understand your question, I have a solution:
Consider
$$det A = det
begin{pmatrix}
O& B\
C& D
end{pmatrix}
$$
where $O$ is a $xtimes y$ zero matrix, $B$ is a $(n-x)times y$ matrix, $C$ is a $xtimes(n-y)$ matrix and $D$ is a $(n-x)times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $det{A}=0$.
In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).
answered Dec 7 '18 at 8:52
LauLau
517315
answered Dec 7 '18 at 8:52
LauLau
517315
answered Dec 7 '18 at 8:52
LauLau
517315
answered Dec 7 '18 at 8:52
LauLau
517315
517315
add a comment |
add a comment |
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$begingroup$
what are $x$ and $y$?
$endgroup$
– Lau
Dec 7 '18 at 8:27
$begingroup$
@Lau $x$ is a number of columns and $y$ is number of rows of which "crosses places" are placed zeros.
$endgroup$
– avan1235
Dec 7 '18 at 8:29