Mixing
Determinant of matrix times elementary matrix
I have been reading Linear Algebra Done Wrong and came across this Lemma:
Lemma 3.6. For a square matrix A and an elementary matrix E (of the
same size)
det(AE) = (det A)(det E)
Proof: The proof can be done just by direct checking: determinants of
special matrices are easy to compute; right multiplication by an elementary matrix is a column operation, and effect of column operations on the
determinant is well known.
This can look like a lucky coincidence, that the determinants of elementary matrices agree with the corresponding column operations, but it is not
a coincidence at all.
Namely, for a column operation the corresponding elementary matrix
can be obtained from the identity matrix I by this column operation. So, its
determinant is 1 (determinant of I) times the effect of the column operation.
And that is all! It may be hard to realize at first, but the above paragraph is a complete and rigorous proof of the lemma!
I am having a very hard time understanding this proof any help would be great, thanks!
linear-algebra determinant
asked Jan 1 at 2:24
jake walshjake walsh
808
add a comment |
I have been reading Linear Algebra Done Wrong and came across this Lemma:
Lemma 3.6. For a square matrix A and an elementary matrix E (of the
same size)
det(AE) = (det A)(det E)
Proof: The proof can be done just by direct checking: determinants of
special matrices are easy to compute; right multiplication by an elementary matrix is a column operation, and effect of column operations on the
determinant is well known.
This can look like a lucky coincidence, that the determinants of elementary matrices agree with the corresponding column operations, but it is not
a coincidence at all.
Namely, for a column operation the corresponding elementary matrix
can be obtained from the identity matrix I by this column operation. So, its
determinant is 1 (determinant of I) times the effect of the column operation.
And that is all! It may be hard to realize at first, but the above paragraph is a complete and rigorous proof of the lemma!
I am having a very hard time understanding this proof any help would be great, thanks!
linear-algebra determinant
asked Jan 1 at 2:24
jake walshjake walsh
808
add a comment |
I have been reading Linear Algebra Done Wrong and came across this Lemma:
Lemma 3.6. For a square matrix A and an elementary matrix E (of the
same size)
det(AE) = (det A)(det E)
Proof: The proof can be done just by direct checking: determinants of
special matrices are easy to compute; right multiplication by an elementary matrix is a column operation, and effect of column operations on the
determinant is well known.
This can look like a lucky coincidence, that the determinants of elementary matrices agree with the corresponding column operations, but it is not
a coincidence at all.
Namely, for a column operation the corresponding elementary matrix
can be obtained from the identity matrix I by this column operation. So, its
determinant is 1 (determinant of I) times the effect of the column operation.
And that is all! It may be hard to realize at first, but the above paragraph is a complete and rigorous proof of the lemma!
I am having a very hard time understanding this proof any help would be great, thanks!
linear-algebra determinant
asked Jan 1 at 2:24
jake walshjake walsh
808
I have been reading Linear Algebra Done Wrong and came across this Lemma:
Lemma 3.6. For a square matrix A and an elementary matrix E (of the
same size)
det(AE) = (det A)(det E)
Proof: The proof can be done just by direct checking: determinants of
special matrices are easy to compute; right multiplication by an elementary matrix is a column operation, and effect of column operations on the
determinant is well known.
This can look like a lucky coincidence, that the determinants of elementary matrices agree with the corresponding column operations, but it is not
a coincidence at all.
Namely, for a column operation the corresponding elementary matrix
can be obtained from the identity matrix I by this column operation. So, its
determinant is 1 (determinant of I) times the effect of the column operation.
And that is all! It may be hard to realize at first, but the above paragraph is a complete and rigorous proof of the lemma!
I am having a very hard time understanding this proof any help would be great, thanks!
linear-algebra determinant
linear-algebra determinant
asked Jan 1 at 2:24
jake walshjake walsh
808
asked Jan 1 at 2:24
jake walshjake walsh
808
asked Jan 1 at 2:24
jake walshjake walsh
808
asked Jan 1 at 2:24
jake walshjake walsh
808
asked Jan 1 at 2:24
jake walshjake walsh
808
808
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
OK, so this is a weird proof, but there's really only two parts of this proof that actually matter here. First:
[...] right multiplication by an elementary matrix is a column operation, and effect of column operations on the determinant is well known.
What this means is that when you multiply $A$ by $E$ on the right, you are doing a column operation on $A$, whether that be switching two columns, doing a column addition, or multiplying a column by a scalar. All of these operations will multiply the determinant by some known number, $k$ (i.e. switching two columns multiplies determinant by $k=-1$, column addition multiplies determinant by $k=1$, and multiplying a column by the scalar $c$ multiplies determinant by $k=c$). Thus, for any matrix $A$, the determinant of $AE$ will be $k$ times the original determinant of $A$. We can write this in an equation as follows:
$$det AE=k(det A)$$
Second:
[...] for a column operation the corresponding elementary matrix can be obtained from the identity matrix I by this column operation. So, its determinant is 1 (determinant of I) times the effect of the column operation.
Now, this is really confusing at first, but it can be understood in terms of our $det AE=k(det A)$ above. See, this equation works for any matrix $A$, which means we could also substitute the identity matrix $I$ for $A$ into this equation. Therefore, we get:
$$det IE=k(det I)rightarrow det E=k$$
(Note that this uses the fact that $IE=E$ and $det I=1$.)
Therefore, we now know the value of $k$ is $det E$. Thus, we can substitute that back into our original equation $det AE=k(det A)$ to get:
$$det AE=(det E)(det A)$$
Since scalar multiplication is commutative, we can switch the right side around to get the final lemma:
$$det AE=(det A)(det E)$$
answered Jan 1 at 2:38
Noble MushtakNoble Mushtak
15.2k1735
Brilliant thanks very much!
– jake walsh
Jan 1 at 2:41
add a comment |
The text defines the determinant in terms of column operation and there is an elementary matrix for each column operation. They spend a lot of time building the determinant based on column operations and the payoff is that this lemma is essentially true by construction.
answered Jan 1 at 2:42
CyclotomicFieldCyclotomicField
2,1921313
add a comment |
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2 Answers
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2 Answers
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OK, so this is a weird proof, but there's really only two parts of this proof that actually matter here. First:
[...] right multiplication by an elementary matrix is a column operation, and effect of column operations on the determinant is well known.
What this means is that when you multiply $A$ by $E$ on the right, you are doing a column operation on $A$, whether that be switching two columns, doing a column addition, or multiplying a column by a scalar. All of these operations will multiply the determinant by some known number, $k$ (i.e. switching two columns multiplies determinant by $k=-1$, column addition multiplies determinant by $k=1$, and multiplying a column by the scalar $c$ multiplies determinant by $k=c$). Thus, for any matrix $A$, the determinant of $AE$ will be $k$ times the original determinant of $A$. We can write this in an equation as follows:
$$det AE=k(det A)$$
Second:
[...] for a column operation the corresponding elementary matrix can be obtained from the identity matrix I by this column operation. So, its determinant is 1 (determinant of I) times the effect of the column operation.
Now, this is really confusing at first, but it can be understood in terms of our $det AE=k(det A)$ above. See, this equation works for any matrix $A$, which means we could also substitute the identity matrix $I$ for $A$ into this equation. Therefore, we get:
$$det IE=k(det I)rightarrow det E=k$$
(Note that this uses the fact that $IE=E$ and $det I=1$.)
Therefore, we now know the value of $k$ is $det E$. Thus, we can substitute that back into our original equation $det AE=k(det A)$ to get:
$$det AE=(det E)(det A)$$
Since scalar multiplication is commutative, we can switch the right side around to get the final lemma:
$$det AE=(det A)(det E)$$
answered Jan 1 at 2:38
Noble MushtakNoble Mushtak
15.2k1735
Brilliant thanks very much!
– jake walsh
Jan 1 at 2:41
add a comment |
OK, so this is a weird proof, but there's really only two parts of this proof that actually matter here. First:
[...] right multiplication by an elementary matrix is a column operation, and effect of column operations on the determinant is well known.
What this means is that when you multiply $A$ by $E$ on the right, you are doing a column operation on $A$, whether that be switching two columns, doing a column addition, or multiplying a column by a scalar. All of these operations will multiply the determinant by some known number, $k$ (i.e. switching two columns multiplies determinant by $k=-1$, column addition multiplies determinant by $k=1$, and multiplying a column by the scalar $c$ multiplies determinant by $k=c$). Thus, for any matrix $A$, the determinant of $AE$ will be $k$ times the original determinant of $A$. We can write this in an equation as follows:
$$det AE=k(det A)$$
Second:
[...] for a column operation the corresponding elementary matrix can be obtained from the identity matrix I by this column operation. So, its determinant is 1 (determinant of I) times the effect of the column operation.
Now, this is really confusing at first, but it can be understood in terms of our $det AE=k(det A)$ above. See, this equation works for any matrix $A$, which means we could also substitute the identity matrix $I$ for $A$ into this equation. Therefore, we get:
$$det IE=k(det I)rightarrow det E=k$$
(Note that this uses the fact that $IE=E$ and $det I=1$.)
Therefore, we now know the value of $k$ is $det E$. Thus, we can substitute that back into our original equation $det AE=k(det A)$ to get:
$$det AE=(det E)(det A)$$
Since scalar multiplication is commutative, we can switch the right side around to get the final lemma:
$$det AE=(det A)(det E)$$
answered Jan 1 at 2:38
Noble MushtakNoble Mushtak
15.2k1735
Brilliant thanks very much!
– jake walsh
Jan 1 at 2:41
add a comment |
OK, so this is a weird proof, but there's really only two parts of this proof that actually matter here. First:
[...] right multiplication by an elementary matrix is a column operation, and effect of column operations on the determinant is well known.
What this means is that when you multiply $A$ by $E$ on the right, you are doing a column operation on $A$, whether that be switching two columns, doing a column addition, or multiplying a column by a scalar. All of these operations will multiply the determinant by some known number, $k$ (i.e. switching two columns multiplies determinant by $k=-1$, column addition multiplies determinant by $k=1$, and multiplying a column by the scalar $c$ multiplies determinant by $k=c$). Thus, for any matrix $A$, the determinant of $AE$ will be $k$ times the original determinant of $A$. We can write this in an equation as follows:
$$det AE=k(det A)$$
Second:
[...] for a column operation the corresponding elementary matrix can be obtained from the identity matrix I by this column operation. So, its determinant is 1 (determinant of I) times the effect of the column operation.
Now, this is really confusing at first, but it can be understood in terms of our $det AE=k(det A)$ above. See, this equation works for any matrix $A$, which means we could also substitute the identity matrix $I$ for $A$ into this equation. Therefore, we get:
$$det IE=k(det I)rightarrow det E=k$$
(Note that this uses the fact that $IE=E$ and $det I=1$.)
Therefore, we now know the value of $k$ is $det E$. Thus, we can substitute that back into our original equation $det AE=k(det A)$ to get:
$$det AE=(det E)(det A)$$
Since scalar multiplication is commutative, we can switch the right side around to get the final lemma:
$$det AE=(det A)(det E)$$
answered Jan 1 at 2:38
Noble MushtakNoble Mushtak
15.2k1735
OK, so this is a weird proof, but there's really only two parts of this proof that actually matter here. First:
[...] right multiplication by an elementary matrix is a column operation, and effect of column operations on the determinant is well known.
What this means is that when you multiply $A$ by $E$ on the right, you are doing a column operation on $A$, whether that be switching two columns, doing a column addition, or multiplying a column by a scalar. All of these operations will multiply the determinant by some known number, $k$ (i.e. switching two columns multiplies determinant by $k=-1$, column addition multiplies determinant by $k=1$, and multiplying a column by the scalar $c$ multiplies determinant by $k=c$). Thus, for any matrix $A$, the determinant of $AE$ will be $k$ times the original determinant of $A$. We can write this in an equation as follows:
$$det AE=k(det A)$$
Second:
[...] for a column operation the corresponding elementary matrix can be obtained from the identity matrix I by this column operation. So, its determinant is 1 (determinant of I) times the effect of the column operation.
Now, this is really confusing at first, but it can be understood in terms of our $det AE=k(det A)$ above. See, this equation works for any matrix $A$, which means we could also substitute the identity matrix $I$ for $A$ into this equation. Therefore, we get:
$$det IE=k(det I)rightarrow det E=k$$
(Note that this uses the fact that $IE=E$ and $det I=1$.)
Therefore, we now know the value of $k$ is $det E$. Thus, we can substitute that back into our original equation $det AE=k(det A)$ to get:
$$det AE=(det E)(det A)$$
Since scalar multiplication is commutative, we can switch the right side around to get the final lemma:
$$det AE=(det A)(det E)$$
answered Jan 1 at 2:38
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 1 at 2:38
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 1 at 2:38
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 1 at 2:38
Noble MushtakNoble Mushtak
15.2k1735
15.2k1735
Brilliant thanks very much!
– jake walsh
Jan 1 at 2:41
add a comment |
Brilliant thanks very much!
– jake walsh
Jan 1 at 2:41
Brilliant thanks very much!
– jake walsh
Jan 1 at 2:41
Brilliant thanks very much!
– jake walsh
Jan 1 at 2:41
add a comment |
The text defines the determinant in terms of column operation and there is an elementary matrix for each column operation. They spend a lot of time building the determinant based on column operations and the payoff is that this lemma is essentially true by construction.
answered Jan 1 at 2:42
CyclotomicFieldCyclotomicField
2,1921313
add a comment |
The text defines the determinant in terms of column operation and there is an elementary matrix for each column operation. They spend a lot of time building the determinant based on column operations and the payoff is that this lemma is essentially true by construction.
answered Jan 1 at 2:42
CyclotomicFieldCyclotomicField
2,1921313
add a comment |
The text defines the determinant in terms of column operation and there is an elementary matrix for each column operation. They spend a lot of time building the determinant based on column operations and the payoff is that this lemma is essentially true by construction.
answered Jan 1 at 2:42
CyclotomicFieldCyclotomicField
2,1921313
The text defines the determinant in terms of column operation and there is an elementary matrix for each column operation. They spend a lot of time building the determinant based on column operations and the payoff is that this lemma is essentially true by construction.
answered Jan 1 at 2:42
CyclotomicFieldCyclotomicField
2,1921313
answered Jan 1 at 2:42
CyclotomicFieldCyclotomicField
2,1921313
answered Jan 1 at 2:42
CyclotomicFieldCyclotomicField
2,1921313
answered Jan 1 at 2:42
CyclotomicFieldCyclotomicField
2,1921313
2,1921313
add a comment |
add a comment |
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