Mixing
$det(EB)=det(E)det(B)$?
$begingroup$
I was doing a proof from my book and I had a question regarding a step.
Proof:
$Ito E$ (Row switch)
$-det I=det E$
$det E=-1$
$det(EB)=det(-1cdot B)
=mathbf{-1det B}$
In the last step of the proof(in bold), how was $-1$ taken out of the determinant?
Isn't $det(ncdot B)$ not equal to $ndet(B)$ where $n$ is a number?
linear-algebra proof-explanation
edited Jan 13 at 6:41
Thomas Shelby
3,1571524
asked Jan 13 at 5:29
JohnySmith12JohnySmith12
413
$endgroup$
add a comment |
$begingroup$
I was doing a proof from my book and I had a question regarding a step.
Proof:
$Ito E$ (Row switch)
$-det I=det E$
$det E=-1$
$det(EB)=det(-1cdot B)
=mathbf{-1det B}$
In the last step of the proof(in bold), how was $-1$ taken out of the determinant?
Isn't $det(ncdot B)$ not equal to $ndet(B)$ where $n$ is a number?
linear-algebra proof-explanation
edited Jan 13 at 6:41
Thomas Shelby
3,1571524
asked Jan 13 at 5:29
JohnySmith12JohnySmith12
413
$endgroup$
add a comment |
$begingroup$
I was doing a proof from my book and I had a question regarding a step.
Proof:
$Ito E$ (Row switch)
$-det I=det E$
$det E=-1$
$det(EB)=det(-1cdot B)
=mathbf{-1det B}$
In the last step of the proof(in bold), how was $-1$ taken out of the determinant?
Isn't $det(ncdot B)$ not equal to $ndet(B)$ where $n$ is a number?
linear-algebra proof-explanation
edited Jan 13 at 6:41
Thomas Shelby
3,1571524
asked Jan 13 at 5:29
JohnySmith12JohnySmith12
413
$endgroup$
I was doing a proof from my book and I had a question regarding a step.
Proof:
$Ito E$ (Row switch)
$-det I=det E$
$det E=-1$
$det(EB)=det(-1cdot B)
=mathbf{-1det B}$
In the last step of the proof(in bold), how was $-1$ taken out of the determinant?
Isn't $det(ncdot B)$ not equal to $ndet(B)$ where $n$ is a number?
linear-algebra proof-explanation
linear-algebra proof-explanation
edited Jan 13 at 6:41
Thomas Shelby
3,1571524
asked Jan 13 at 5:29
JohnySmith12JohnySmith12
413
edited Jan 13 at 6:41
Thomas Shelby
3,1571524
asked Jan 13 at 5:29
JohnySmith12JohnySmith12
413
edited Jan 13 at 6:41
Thomas Shelby
3,1571524
edited Jan 13 at 6:41
Thomas Shelby
3,1571524
edited Jan 13 at 6:41
Thomas Shelby
3,1571524
3,1571524
asked Jan 13 at 5:29
JohnySmith12JohnySmith12
413
asked Jan 13 at 5:29
JohnySmith12JohnySmith12
413
asked Jan 13 at 5:29
JohnySmith12JohnySmith12
413
413
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I can't really follow the working properly as presented, but you're right to be sceptical of the last step. In general, $det(kM) = k^n det(M)$, where $M$ is an $n times n$ matrix. If $n$ is odd, (e.g. a $3 times 3$ matrix), then this would be valid, as $(-1)^n = -1$ when $n$ is odd, but my suspicion is that the last step is a typo.
answered Jan 13 at 5:46
Theo BenditTheo Bendit
18.1k12152
$endgroup$
add a comment |
$begingroup$
|EB|=|-1B| =-1|B|
the middle part is not correct. It must be:
$$|EB|=|E|cdot |B|=-1cdot |B|,$$
because $E$ is an elementary matrix, not a scalar.
answered Jan 13 at 6:17
farruhotafarruhota
20.2k2738
$endgroup$
add a comment |
$begingroup$
I don't think $det (EB) = det (-B)$ is correct, because $E$ only swaps two rows, not the multiplication by $-1$. Specifically the first $=$ on the "last step" is not correct, and the boldface part is not correct either.
answered Jan 13 at 6:04
xbhxbh
6,1351522
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I can't really follow the working properly as presented, but you're right to be sceptical of the last step. In general, $det(kM) = k^n det(M)$, where $M$ is an $n times n$ matrix. If $n$ is odd, (e.g. a $3 times 3$ matrix), then this would be valid, as $(-1)^n = -1$ when $n$ is odd, but my suspicion is that the last step is a typo.
answered Jan 13 at 5:46
Theo BenditTheo Bendit
18.1k12152
$endgroup$
add a comment |
$begingroup$
I can't really follow the working properly as presented, but you're right to be sceptical of the last step. In general, $det(kM) = k^n det(M)$, where $M$ is an $n times n$ matrix. If $n$ is odd, (e.g. a $3 times 3$ matrix), then this would be valid, as $(-1)^n = -1$ when $n$ is odd, but my suspicion is that the last step is a typo.
answered Jan 13 at 5:46
Theo BenditTheo Bendit
18.1k12152
$endgroup$
add a comment |
$begingroup$
I can't really follow the working properly as presented, but you're right to be sceptical of the last step. In general, $det(kM) = k^n det(M)$, where $M$ is an $n times n$ matrix. If $n$ is odd, (e.g. a $3 times 3$ matrix), then this would be valid, as $(-1)^n = -1$ when $n$ is odd, but my suspicion is that the last step is a typo.
answered Jan 13 at 5:46
Theo BenditTheo Bendit
18.1k12152
$endgroup$
I can't really follow the working properly as presented, but you're right to be sceptical of the last step. In general, $det(kM) = k^n det(M)$, where $M$ is an $n times n$ matrix. If $n$ is odd, (e.g. a $3 times 3$ matrix), then this would be valid, as $(-1)^n = -1$ when $n$ is odd, but my suspicion is that the last step is a typo.
answered Jan 13 at 5:46
Theo BenditTheo Bendit
18.1k12152
answered Jan 13 at 5:46
Theo BenditTheo Bendit
18.1k12152
answered Jan 13 at 5:46
Theo BenditTheo Bendit
18.1k12152
answered Jan 13 at 5:46
Theo BenditTheo Bendit
18.1k12152
18.1k12152
add a comment |
add a comment |
$begingroup$
|EB|=|-1B| =-1|B|
the middle part is not correct. It must be:
$$|EB|=|E|cdot |B|=-1cdot |B|,$$
because $E$ is an elementary matrix, not a scalar.
answered Jan 13 at 6:17
farruhotafarruhota
20.2k2738
$endgroup$
add a comment |
$begingroup$
|EB|=|-1B| =-1|B|
the middle part is not correct. It must be:
$$|EB|=|E|cdot |B|=-1cdot |B|,$$
because $E$ is an elementary matrix, not a scalar.
answered Jan 13 at 6:17
farruhotafarruhota
20.2k2738
$endgroup$
add a comment |
$begingroup$
|EB|=|-1B| =-1|B|
the middle part is not correct. It must be:
$$|EB|=|E|cdot |B|=-1cdot |B|,$$
because $E$ is an elementary matrix, not a scalar.
answered Jan 13 at 6:17
farruhotafarruhota
20.2k2738
$endgroup$
|EB|=|-1B| =-1|B|
the middle part is not correct. It must be:
$$|EB|=|E|cdot |B|=-1cdot |B|,$$
because $E$ is an elementary matrix, not a scalar.
answered Jan 13 at 6:17
farruhotafarruhota
20.2k2738
answered Jan 13 at 6:17
farruhotafarruhota
20.2k2738
answered Jan 13 at 6:17
farruhotafarruhota
20.2k2738
answered Jan 13 at 6:17
farruhotafarruhota
20.2k2738
20.2k2738
add a comment |
add a comment |
$begingroup$
I don't think $det (EB) = det (-B)$ is correct, because $E$ only swaps two rows, not the multiplication by $-1$. Specifically the first $=$ on the "last step" is not correct, and the boldface part is not correct either.
answered Jan 13 at 6:04
xbhxbh
6,1351522
$endgroup$
add a comment |
$begingroup$
I don't think $det (EB) = det (-B)$ is correct, because $E$ only swaps two rows, not the multiplication by $-1$. Specifically the first $=$ on the "last step" is not correct, and the boldface part is not correct either.
answered Jan 13 at 6:04
xbhxbh
6,1351522
$endgroup$
add a comment |
$begingroup$
I don't think $det (EB) = det (-B)$ is correct, because $E$ only swaps two rows, not the multiplication by $-1$. Specifically the first $=$ on the "last step" is not correct, and the boldface part is not correct either.
answered Jan 13 at 6:04
xbhxbh
6,1351522
$endgroup$
I don't think $det (EB) = det (-B)$ is correct, because $E$ only swaps two rows, not the multiplication by $-1$. Specifically the first $=$ on the "last step" is not correct, and the boldface part is not correct either.
answered Jan 13 at 6:04
xbhxbh
6,1351522
answered Jan 13 at 6:04
xbhxbh
6,1351522
answered Jan 13 at 6:04
xbhxbh
6,1351522
answered Jan 13 at 6:04
xbhxbh
6,1351522
6,1351522
add a comment |
add a comment |
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