Mixing
Why do the properties of determinants (used to calculate determinants from multiple matrices) apply not only...
$begingroup$
A set of rules in my textbook is as follows:
a. If $A$ has a zero row (column), then $det A = 0$
b. If $B$ is obtained by interchanging two rows (columns) of $A$, then $det B = -det A$
c. If $A$ has two identical rows (columns), then $det A = 0$
d. If $B$ is obtained by multiplying a row (column) of $A$ by $k$, then $det B = kcdotdet A$
e. If $A$, $B$, and $C$ are identical except that the $i$-th row (column) of $C$ is the sum of the $i$-th rows (columns) of $A$ and $B$, then $det C = det B + det A$
f. If $B$ is obtained by adding a multiple of one row (column) of $A$ to another row (column), then $det B = det A$
I don't understand why "column" is in parentheses after every instance of row. Is that the same as saying, for example with clause a: "if $A$ has a zero row or a zero column, then $det A = 0$"? As in, it's saying that column and row can be used interchangably in the statement, as the statement holds true either way?
If the above interpretation is correct, then how? Why would these statements that apply to rows also apply to columns. The only situation I could see it applying is if the matrix is symmetrical, but the question doesn't specify that, it only says that the matrix is square.
Any help is appreciated.
linear-algebra matrices determinant
edited Jan 1 at 20:54
amWhy
192k28225439
asked Jan 1 at 20:41
Nest Doberman Nest Doberman
311
$endgroup$
add a comment |
$begingroup$
A set of rules in my textbook is as follows:
a. If $A$ has a zero row (column), then $det A = 0$
b. If $B$ is obtained by interchanging two rows (columns) of $A$, then $det B = -det A$
c. If $A$ has two identical rows (columns), then $det A = 0$
d. If $B$ is obtained by multiplying a row (column) of $A$ by $k$, then $det B = kcdotdet A$
e. If $A$, $B$, and $C$ are identical except that the $i$-th row (column) of $C$ is the sum of the $i$-th rows (columns) of $A$ and $B$, then $det C = det B + det A$
f. If $B$ is obtained by adding a multiple of one row (column) of $A$ to another row (column), then $det B = det A$
I don't understand why "column" is in parentheses after every instance of row. Is that the same as saying, for example with clause a: "if $A$ has a zero row or a zero column, then $det A = 0$"? As in, it's saying that column and row can be used interchangably in the statement, as the statement holds true either way?
If the above interpretation is correct, then how? Why would these statements that apply to rows also apply to columns. The only situation I could see it applying is if the matrix is symmetrical, but the question doesn't specify that, it only says that the matrix is square.
Any help is appreciated.
linear-algebra matrices determinant
edited Jan 1 at 20:54
amWhy
192k28225439
asked Jan 1 at 20:41
Nest Doberman Nest Doberman
311
$endgroup$
$begingroup$
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
$endgroup$
– timtfj
Jan 1 at 22:40
$begingroup$
$det(M^T)=det(M)$.
$endgroup$
– amd
Jan 2 at 4:41
add a comment |
$begingroup$
A set of rules in my textbook is as follows:
a. If $A$ has a zero row (column), then $det A = 0$
b. If $B$ is obtained by interchanging two rows (columns) of $A$, then $det B = -det A$
c. If $A$ has two identical rows (columns), then $det A = 0$
d. If $B$ is obtained by multiplying a row (column) of $A$ by $k$, then $det B = kcdotdet A$
e. If $A$, $B$, and $C$ are identical except that the $i$-th row (column) of $C$ is the sum of the $i$-th rows (columns) of $A$ and $B$, then $det C = det B + det A$
f. If $B$ is obtained by adding a multiple of one row (column) of $A$ to another row (column), then $det B = det A$
I don't understand why "column" is in parentheses after every instance of row. Is that the same as saying, for example with clause a: "if $A$ has a zero row or a zero column, then $det A = 0$"? As in, it's saying that column and row can be used interchangably in the statement, as the statement holds true either way?
If the above interpretation is correct, then how? Why would these statements that apply to rows also apply to columns. The only situation I could see it applying is if the matrix is symmetrical, but the question doesn't specify that, it only says that the matrix is square.
Any help is appreciated.
linear-algebra matrices determinant
edited Jan 1 at 20:54
amWhy
192k28225439
asked Jan 1 at 20:41
Nest Doberman Nest Doberman
311
$endgroup$
A set of rules in my textbook is as follows:
a. If $A$ has a zero row (column), then $det A = 0$
b. If $B$ is obtained by interchanging two rows (columns) of $A$, then $det B = -det A$
c. If $A$ has two identical rows (columns), then $det A = 0$
d. If $B$ is obtained by multiplying a row (column) of $A$ by $k$, then $det B = kcdotdet A$
e. If $A$, $B$, and $C$ are identical except that the $i$-th row (column) of $C$ is the sum of the $i$-th rows (columns) of $A$ and $B$, then $det C = det B + det A$
f. If $B$ is obtained by adding a multiple of one row (column) of $A$ to another row (column), then $det B = det A$
I don't understand why "column" is in parentheses after every instance of row. Is that the same as saying, for example with clause a: "if $A$ has a zero row or a zero column, then $det A = 0$"? As in, it's saying that column and row can be used interchangably in the statement, as the statement holds true either way?
If the above interpretation is correct, then how? Why would these statements that apply to rows also apply to columns. The only situation I could see it applying is if the matrix is symmetrical, but the question doesn't specify that, it only says that the matrix is square.
Any help is appreciated.
linear-algebra matrices determinant
linear-algebra matrices determinant
edited Jan 1 at 20:54
amWhy
192k28225439
asked Jan 1 at 20:41
Nest Doberman Nest Doberman
311
edited Jan 1 at 20:54
amWhy
192k28225439
asked Jan 1 at 20:41
Nest Doberman Nest Doberman
311
edited Jan 1 at 20:54
amWhy
192k28225439
edited Jan 1 at 20:54
amWhy
192k28225439
edited Jan 1 at 20:54
amWhy
192k28225439
192k28225439
asked Jan 1 at 20:41
Nest Doberman Nest Doberman
311
asked Jan 1 at 20:41
Nest Doberman Nest Doberman
311
asked Jan 1 at 20:41
Nest Doberman Nest Doberman
311
311
$begingroup$
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
$endgroup$
– timtfj
Jan 1 at 22:40
$begingroup$
$det(M^T)=det(M)$.
$endgroup$
– amd
Jan 2 at 4:41
add a comment |
$begingroup$
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
$endgroup$
– timtfj
Jan 1 at 22:40
$begingroup$
$det(M^T)=det(M)$.
$endgroup$
– amd
Jan 2 at 4:41
$begingroup$
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
$endgroup$
– timtfj
Jan 1 at 22:40
$begingroup$
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
$endgroup$
– timtfj
Jan 1 at 22:40
$begingroup$
$det(M^T)=det(M)$.
$endgroup$
– amd
Jan 2 at 4:41
$begingroup$
$det(M^T)=det(M)$.
$endgroup$
– amd
Jan 2 at 4:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The trick is that $det A^T=det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $det A^T=0$. You can check all the other copy-paste-to-column results in that way.
answered Jan 1 at 20:43
J.G.J.G.
23.6k22338
$endgroup$
4
$begingroup$
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
$endgroup$
– Eric Wofsey
Jan 1 at 20:59
$begingroup$
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
$endgroup$
– J.G.
Jan 1 at 21:08
$begingroup$
@EricWofsey: I definitely prefer the definition by permutations, because it immediately gives all the properties of determinants. I don't know why so many people even bother with the cofactor definition, which is really quite useless in my opinion.
$endgroup$
– user21820
Jan 2 at 5:34
add a comment |
$begingroup$
That's because your textbook isn't really explaining determinants at all to give you any intuition. It's just listing a bunch of their properties.
The key thing to remember is that $det A$ is the volumetric scaling factor of a matrix $A$ (i.e., $A$ scales the unit cube's volume by $det A$).
If you've seen eigenvalues, this means that $det A$ is the product of $A$'s eigenvalues. (This can be a little mindblowing the first time you realize it, if you've already learned linear algebra for some time.)
If you start from this, then it's far easier to see why rows and columns don't make a difference:
The unit cube is represented by its orthogonal edges via $I$, the identity matrix.
Right-multiplying by $I$ (which is an operation linearly combining $A$'s columns) gives the same result ($A$) as as left-multiplying by $I$ (which is an operation linearly combining $A$'s rows)
The fact that we have the same output solid regardless of whether we operate on rows or columns of $A$ means that the scaling factor is the same, and hence the determinant is the same.
answered Jan 2 at 2:10
MehrdadMehrdad
6,62463778
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The trick is that $det A^T=det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $det A^T=0$. You can check all the other copy-paste-to-column results in that way.
answered Jan 1 at 20:43
J.G.J.G.
23.6k22338
$endgroup$
4
$begingroup$
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
$endgroup$
– Eric Wofsey
Jan 1 at 20:59
$begingroup$
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
$endgroup$
– J.G.
Jan 1 at 21:08
$begingroup$
@EricWofsey: I definitely prefer the definition by permutations, because it immediately gives all the properties of determinants. I don't know why so many people even bother with the cofactor definition, which is really quite useless in my opinion.
$endgroup$
– user21820
Jan 2 at 5:34
add a comment |
$begingroup$
The trick is that $det A^T=det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $det A^T=0$. You can check all the other copy-paste-to-column results in that way.
answered Jan 1 at 20:43
J.G.J.G.
23.6k22338
$endgroup$
4
$begingroup$
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
$endgroup$
– Eric Wofsey
Jan 1 at 20:59
$begingroup$
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
$endgroup$
– J.G.
Jan 1 at 21:08
$begingroup$
@EricWofsey: I definitely prefer the definition by permutations, because it immediately gives all the properties of determinants. I don't know why so many people even bother with the cofactor definition, which is really quite useless in my opinion.
$endgroup$
– user21820
Jan 2 at 5:34
add a comment |
$begingroup$
The trick is that $det A^T=det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $det A^T=0$. You can check all the other copy-paste-to-column results in that way.
answered Jan 1 at 20:43
J.G.J.G.
23.6k22338
$endgroup$
The trick is that $det A^T=det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $det A^T=0$. You can check all the other copy-paste-to-column results in that way.
answered Jan 1 at 20:43
J.G.J.G.
23.6k22338
answered Jan 1 at 20:43
J.G.J.G.
23.6k22338
answered Jan 1 at 20:43
J.G.J.G.
23.6k22338
answered Jan 1 at 20:43
J.G.J.G.
23.6k22338
23.6k22338
4
$begingroup$
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
$endgroup$
– Eric Wofsey
Jan 1 at 20:59
$begingroup$
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
$endgroup$
– J.G.
Jan 1 at 21:08
$begingroup$
@EricWofsey: I definitely prefer the definition by permutations, because it immediately gives all the properties of determinants. I don't know why so many people even bother with the cofactor definition, which is really quite useless in my opinion.
$endgroup$
– user21820
Jan 2 at 5:34
add a comment |
4
$begingroup$
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
$endgroup$
– Eric Wofsey
Jan 1 at 20:59
$begingroup$
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
$endgroup$
– J.G.
Jan 1 at 21:08
$begingroup$
@EricWofsey: I definitely prefer the definition by permutations, because it immediately gives all the properties of determinants. I don't know why so many people even bother with the cofactor definition, which is really quite useless in my opinion.
$endgroup$
– user21820
Jan 2 at 5:34
4
4
$begingroup$
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
$endgroup$
– Eric Wofsey
Jan 1 at 20:59
$begingroup$
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
$endgroup$
– Eric Wofsey
Jan 1 at 20:59
$begingroup$
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
$endgroup$
– J.G.
Jan 1 at 21:08
$begingroup$
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
$endgroup$
– J.G.
Jan 1 at 21:08
$begingroup$
@EricWofsey: I definitely prefer the definition by permutations, because it immediately gives all the properties of determinants. I don't know why so many people even bother with the cofactor definition, which is really quite useless in my opinion.
$endgroup$
– user21820
Jan 2 at 5:34
$begingroup$
@EricWofsey: I definitely prefer the definition by permutations, because it immediately gives all the properties of determinants. I don't know why so many people even bother with the cofactor definition, which is really quite useless in my opinion.
$endgroup$
– user21820
Jan 2 at 5:34
add a comment |
$begingroup$
That's because your textbook isn't really explaining determinants at all to give you any intuition. It's just listing a bunch of their properties.
The key thing to remember is that $det A$ is the volumetric scaling factor of a matrix $A$ (i.e., $A$ scales the unit cube's volume by $det A$).
If you've seen eigenvalues, this means that $det A$ is the product of $A$'s eigenvalues. (This can be a little mindblowing the first time you realize it, if you've already learned linear algebra for some time.)
If you start from this, then it's far easier to see why rows and columns don't make a difference:
The unit cube is represented by its orthogonal edges via $I$, the identity matrix.
Right-multiplying by $I$ (which is an operation linearly combining $A$'s columns) gives the same result ($A$) as as left-multiplying by $I$ (which is an operation linearly combining $A$'s rows)
The fact that we have the same output solid regardless of whether we operate on rows or columns of $A$ means that the scaling factor is the same, and hence the determinant is the same.
answered Jan 2 at 2:10
MehrdadMehrdad
6,62463778
$endgroup$
add a comment |
$begingroup$
That's because your textbook isn't really explaining determinants at all to give you any intuition. It's just listing a bunch of their properties.
The key thing to remember is that $det A$ is the volumetric scaling factor of a matrix $A$ (i.e., $A$ scales the unit cube's volume by $det A$).
If you've seen eigenvalues, this means that $det A$ is the product of $A$'s eigenvalues. (This can be a little mindblowing the first time you realize it, if you've already learned linear algebra for some time.)
If you start from this, then it's far easier to see why rows and columns don't make a difference:
The unit cube is represented by its orthogonal edges via $I$, the identity matrix.
Right-multiplying by $I$ (which is an operation linearly combining $A$'s columns) gives the same result ($A$) as as left-multiplying by $I$ (which is an operation linearly combining $A$'s rows)
The fact that we have the same output solid regardless of whether we operate on rows or columns of $A$ means that the scaling factor is the same, and hence the determinant is the same.
answered Jan 2 at 2:10
MehrdadMehrdad
6,62463778
$endgroup$
add a comment |
$begingroup$
That's because your textbook isn't really explaining determinants at all to give you any intuition. It's just listing a bunch of their properties.
The key thing to remember is that $det A$ is the volumetric scaling factor of a matrix $A$ (i.e., $A$ scales the unit cube's volume by $det A$).
If you've seen eigenvalues, this means that $det A$ is the product of $A$'s eigenvalues. (This can be a little mindblowing the first time you realize it, if you've already learned linear algebra for some time.)
If you start from this, then it's far easier to see why rows and columns don't make a difference:
The unit cube is represented by its orthogonal edges via $I$, the identity matrix.
Right-multiplying by $I$ (which is an operation linearly combining $A$'s columns) gives the same result ($A$) as as left-multiplying by $I$ (which is an operation linearly combining $A$'s rows)
The fact that we have the same output solid regardless of whether we operate on rows or columns of $A$ means that the scaling factor is the same, and hence the determinant is the same.
answered Jan 2 at 2:10
MehrdadMehrdad
6,62463778
$endgroup$
That's because your textbook isn't really explaining determinants at all to give you any intuition. It's just listing a bunch of their properties.
The key thing to remember is that $det A$ is the volumetric scaling factor of a matrix $A$ (i.e., $A$ scales the unit cube's volume by $det A$).
If you've seen eigenvalues, this means that $det A$ is the product of $A$'s eigenvalues. (This can be a little mindblowing the first time you realize it, if you've already learned linear algebra for some time.)
If you start from this, then it's far easier to see why rows and columns don't make a difference:
The unit cube is represented by its orthogonal edges via $I$, the identity matrix.
Right-multiplying by $I$ (which is an operation linearly combining $A$'s columns) gives the same result ($A$) as as left-multiplying by $I$ (which is an operation linearly combining $A$'s rows)
The fact that we have the same output solid regardless of whether we operate on rows or columns of $A$ means that the scaling factor is the same, and hence the determinant is the same.
answered Jan 2 at 2:10
MehrdadMehrdad
6,62463778
answered Jan 2 at 2:10
MehrdadMehrdad
6,62463778
answered Jan 2 at 2:10
MehrdadMehrdad
6,62463778
answered Jan 2 at 2:10
MehrdadMehrdad
6,62463778
6,62463778
add a comment |
add a comment |
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$begingroup$
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
$endgroup$
– timtfj
Jan 1 at 22:40
$begingroup$
$det(M^T)=det(M)$.
$endgroup$
– amd
Jan 2 at 4:41