Proving det AB = det A det B starting from the generalized eigenvalue definition?
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I've been following the book Linear Algebra Done Right, and it's a good book, but I don't like how the subject of determinant is approached. So I tried to deduce the formula for the determinant on my own, but I hit a roadblock, as I need the property I asked for in the title, or something similar.
How would I go about doing this in a motivated way?
linear-algebra determinant
asked Sep 8 at 12:32
Robly18
637
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up vote
2
down vote
favorite
1
I've been following the book Linear Algebra Done Right, and it's a good book, but I don't like how the subject of determinant is approached. So I tried to deduce the formula for the determinant on my own, but I hit a roadblock, as I need the property I asked for in the title, or something similar.
How would I go about doing this in a motivated way?
linear-algebra determinant
asked Sep 8 at 12:32
Robly18
637
Small nitpick: there's no such thing as generalised eigenvalues, only eigenvalues, and (sometimes generalised) eigenvectors.
– Theo Bendit
Sep 8 at 12:52
So, to be clear, you're reading the book, which defines determinants as the product of eigenvalues to the power of their multiplicities. What about the book's approach did you not like? Am I right in thinking that you still wish to use this definition of determinant?
– Theo Bendit
Sep 8 at 12:55
@Theo You're right about the generalized eigenvector thing, my bad. Yes, I would like to keep this definition. What I did not like is how the explicit formula (sum over the permutations etc) was developed.
– Robly18
Sep 8 at 13:08
1
There is no easy way to derive the formula for $detleft(ABright)$ from the eigenvalue "definition". The eigenvalues of $AB$ usually have nothing to do with those of $A$ and of $B$. For example, it is easy to find two matrices $A$ and $B$ whose eigenvalues are rational while those of $AB$ are not. (Just take a generic matrix with irrational eigenvalues, and consider its LU decomposition.)
– darij grinberg
Sep 8 at 13:11
2
@Robly18 It's tricky to answer this without the book in front of me, let alone out-do Axler's approach. I quite like the eigenvalue definition too, but it does leave a gap between defining the determinant and actually using the determinant. I think maybe the most intuitive way to get from this definition to $det(AB) = det(A)det(B)$ is using measure. Show that the image of a unit hypercube under $A$ has Lebesgue measure $det(A)$. This connects most immediately with the geometry of eigenvectors. But, it's still a fair slog to get there.
– Theo Bendit
Sep 8 at 13:15
add a comment |Â
up vote
2
down vote
favorite
1
up vote
2
down vote
favorite
1
1
I've been following the book Linear Algebra Done Right, and it's a good book, but I don't like how the subject of determinant is approached. So I tried to deduce the formula for the determinant on my own, but I hit a roadblock, as I need the property I asked for in the title, or something similar.
How would I go about doing this in a motivated way?
linear-algebra determinant
asked Sep 8 at 12:32
Robly18
637
I've been following the book Linear Algebra Done Right, and it's a good book, but I don't like how the subject of determinant is approached. So I tried to deduce the formula for the determinant on my own, but I hit a roadblock, as I need the property I asked for in the title, or something similar.
How would I go about doing this in a motivated way?
linear-algebra determinant
linear-algebra determinant
asked Sep 8 at 12:32
Robly18
637
asked Sep 8 at 12:32
Robly18
637
asked Sep 8 at 12:32
Robly18
637
asked Sep 8 at 12:32
Robly18
637
asked Sep 8 at 12:32
Robly18
637
637
Small nitpick: there's no such thing as generalised eigenvalues, only eigenvalues, and (sometimes generalised) eigenvectors.
– Theo Bendit
Sep 8 at 12:52
So, to be clear, you're reading the book, which defines determinants as the product of eigenvalues to the power of their multiplicities. What about the book's approach did you not like? Am I right in thinking that you still wish to use this definition of determinant?
– Theo Bendit
Sep 8 at 12:55
@Theo You're right about the generalized eigenvector thing, my bad. Yes, I would like to keep this definition. What I did not like is how the explicit formula (sum over the permutations etc) was developed.
– Robly18
Sep 8 at 13:08
1
There is no easy way to derive the formula for $detleft(ABright)$ from the eigenvalue "definition". The eigenvalues of $AB$ usually have nothing to do with those of $A$ and of $B$. For example, it is easy to find two matrices $A$ and $B$ whose eigenvalues are rational while those of $AB$ are not. (Just take a generic matrix with irrational eigenvalues, and consider its LU decomposition.)
– darij grinberg
Sep 8 at 13:11
2
@Robly18 It's tricky to answer this without the book in front of me, let alone out-do Axler's approach. I quite like the eigenvalue definition too, but it does leave a gap between defining the determinant and actually using the determinant. I think maybe the most intuitive way to get from this definition to $det(AB) = det(A)det(B)$ is using measure. Show that the image of a unit hypercube under $A$ has Lebesgue measure $det(A)$. This connects most immediately with the geometry of eigenvectors. But, it's still a fair slog to get there.
– Theo Bendit
Sep 8 at 13:15
add a comment |Â
Small nitpick: there's no such thing as generalised eigenvalues, only eigenvalues, and (sometimes generalised) eigenvectors.
– Theo Bendit
Sep 8 at 12:52
So, to be clear, you're reading the book, which defines determinants as the product of eigenvalues to the power of their multiplicities. What about the book's approach did you not like? Am I right in thinking that you still wish to use this definition of determinant?
– Theo Bendit
Sep 8 at 12:55
@Theo You're right about the generalized eigenvector thing, my bad. Yes, I would like to keep this definition. What I did not like is how the explicit formula (sum over the permutations etc) was developed.
– Robly18
Sep 8 at 13:08
1
There is no easy way to derive the formula for $detleft(ABright)$ from the eigenvalue "definition". The eigenvalues of $AB$ usually have nothing to do with those of $A$ and of $B$. For example, it is easy to find two matrices $A$ and $B$ whose eigenvalues are rational while those of $AB$ are not. (Just take a generic matrix with irrational eigenvalues, and consider its LU decomposition.)
– darij grinberg
Sep 8 at 13:11
2
@Robly18 It's tricky to answer this without the book in front of me, let alone out-do Axler's approach. I quite like the eigenvalue definition too, but it does leave a gap between defining the determinant and actually using the determinant. I think maybe the most intuitive way to get from this definition to $det(AB) = det(A)det(B)$ is using measure. Show that the image of a unit hypercube under $A$ has Lebesgue measure $det(A)$. This connects most immediately with the geometry of eigenvectors. But, it's still a fair slog to get there.
– Theo Bendit
Sep 8 at 13:15
Small nitpick: there's no such thing as generalised eigenvalues, only eigenvalues, and (sometimes generalised) eigenvectors.
– Theo Bendit
Sep 8 at 12:52
Small nitpick: there's no such thing as generalised eigenvalues, only eigenvalues, and (sometimes generalised) eigenvectors.
– Theo Bendit
Sep 8 at 12:52
So, to be clear, you're reading the book, which defines determinants as the product of eigenvalues to the power of their multiplicities. What about the book's approach did you not like? Am I right in thinking that you still wish to use this definition of determinant?
– Theo Bendit
Sep 8 at 12:55
So, to be clear, you're reading the book, which defines determinants as the product of eigenvalues to the power of their multiplicities. What about the book's approach did you not like? Am I right in thinking that you still wish to use this definition of determinant?
– Theo Bendit
Sep 8 at 12:55
@Theo You're right about the generalized eigenvector thing, my bad. Yes, I would like to keep this definition. What I did not like is how the explicit formula (sum over the permutations etc) was developed.
– Robly18
Sep 8 at 13:08
@Theo You're right about the generalized eigenvector thing, my bad. Yes, I would like to keep this definition. What I did not like is how the explicit formula (sum over the permutations etc) was developed.
– Robly18
Sep 8 at 13:08
1
1
There is no easy way to derive the formula for $detleft(ABright)$ from the eigenvalue "definition". The eigenvalues of $AB$ usually have nothing to do with those of $A$ and of $B$. For example, it is easy to find two matrices $A$ and $B$ whose eigenvalues are rational while those of $AB$ are not. (Just take a generic matrix with irrational eigenvalues, and consider its LU decomposition.)
– darij grinberg
Sep 8 at 13:11
There is no easy way to derive the formula for $detleft(ABright)$ from the eigenvalue "definition". The eigenvalues of $AB$ usually have nothing to do with those of $A$ and of $B$. For example, it is easy to find two matrices $A$ and $B$ whose eigenvalues are rational while those of $AB$ are not. (Just take a generic matrix with irrational eigenvalues, and consider its LU decomposition.)
– darij grinberg
Sep 8 at 13:11
2
2
@Robly18 It's tricky to answer this without the book in front of me, let alone out-do Axler's approach. I quite like the eigenvalue definition too, but it does leave a gap between defining the determinant and actually using the determinant. I think maybe the most intuitive way to get from this definition to $det(AB) = det(A)det(B)$ is using measure. Show that the image of a unit hypercube under $A$ has Lebesgue measure $det(A)$. This connects most immediately with the geometry of eigenvectors. But, it's still a fair slog to get there.
– Theo Bendit
Sep 8 at 13:15
@Robly18 It's tricky to answer this without the book in front of me, let alone out-do Axler's approach. I quite like the eigenvalue definition too, but it does leave a gap between defining the determinant and actually using the determinant. I think maybe the most intuitive way to get from this definition to $det(AB) = det(A)det(B)$ is using measure. Show that the image of a unit hypercube under $A$ has Lebesgue measure $det(A)$. This connects most immediately with the geometry of eigenvectors. But, it's still a fair slog to get there.
– Theo Bendit
Sep 8 at 13:15
add a comment |Â
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Small nitpick: there's no such thing as generalised eigenvalues, only eigenvalues, and (sometimes generalised) eigenvectors.
– Theo Bendit
Sep 8 at 12:52
So, to be clear, you're reading the book, which defines determinants as the product of eigenvalues to the power of their multiplicities. What about the book's approach did you not like? Am I right in thinking that you still wish to use this definition of determinant?
– Theo Bendit
Sep 8 at 12:55
@Theo You're right about the generalized eigenvector thing, my bad. Yes, I would like to keep this definition. What I did not like is how the explicit formula (sum over the permutations etc) was developed.
– Robly18
Sep 8 at 13:08
1
There is no easy way to derive the formula for $detleft(ABright)$ from the eigenvalue "definition". The eigenvalues of $AB$ usually have nothing to do with those of $A$ and of $B$. For example, it is easy to find two matrices $A$ and $B$ whose eigenvalues are rational while those of $AB$ are not. (Just take a generic matrix with irrational eigenvalues, and consider its LU decomposition.)
– darij grinberg
Sep 8 at 13:11
2
@Robly18 It's tricky to answer this without the book in front of me, let alone out-do Axler's approach. I quite like the eigenvalue definition too, but it does leave a gap between defining the determinant and actually using the determinant. I think maybe the most intuitive way to get from this definition to $det(AB) = det(A)det(B)$ is using measure. Show that the image of a unit hypercube under $A$ has Lebesgue measure $det(A)$. This connects most immediately with the geometry of eigenvectors. But, it's still a fair slog to get there.
– Theo Bendit
Sep 8 at 13:15