Which of the following statements about determinants are correct?
$begingroup$
Which of the following statements about determinants are correct?
$det(A^2)>0$, for all invertible matrices $A$
$det(A+A^{-1})=det(A)+dfrac{1}{det(A)}$, for all invertible matrices $A$
$det(vv^T)>0$, for all column vectors $v ≠ 0$
$det(AB^T)=det((A^T)B)$, for all square matrices $A$ and $B$
Which of the statements are correct? I do not feel secure about which of them that is true.
My answer:
My calculations have given me that $2$ and $4$ are true. Am I correct?
matrices symmetric-matrices
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|
show 6 more comments
$begingroup$
Which of the following statements about determinants are correct?
$det(A^2)>0$, for all invertible matrices $A$
$det(A+A^{-1})=det(A)+dfrac{1}{det(A)}$, for all invertible matrices $A$
$det(vv^T)>0$, for all column vectors $v ≠ 0$
$det(AB^T)=det((A^T)B)$, for all square matrices $A$ and $B$
Which of the statements are correct? I do not feel secure about which of them that is true.
My answer:
My calculations have given me that $2$ and $4$ are true. Am I correct?
matrices symmetric-matrices
$endgroup$
1
$begingroup$
kindly edit your post to include your attempts using mathjax.
$endgroup$
– Siong Thye Goh
Dec 19 '18 at 6:54
$begingroup$
Okey. Sorry I forgot
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 6:55
$begingroup$
What do you know about determinants?
$endgroup$
– bubba
Dec 19 '18 at 6:58
1
$begingroup$
Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
$endgroup$
– bubba
Dec 19 '18 at 6:59
$begingroup$
Sorry I have calculate totally wrong
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:00
|
show 6 more comments
$begingroup$
Which of the following statements about determinants are correct?
$det(A^2)>0$, for all invertible matrices $A$
$det(A+A^{-1})=det(A)+dfrac{1}{det(A)}$, for all invertible matrices $A$
$det(vv^T)>0$, for all column vectors $v ≠ 0$
$det(AB^T)=det((A^T)B)$, for all square matrices $A$ and $B$
Which of the statements are correct? I do not feel secure about which of them that is true.
My answer:
My calculations have given me that $2$ and $4$ are true. Am I correct?
matrices symmetric-matrices
$endgroup$
Which of the following statements about determinants are correct?
$det(A^2)>0$, for all invertible matrices $A$
$det(A+A^{-1})=det(A)+dfrac{1}{det(A)}$, for all invertible matrices $A$
$det(vv^T)>0$, for all column vectors $v ≠ 0$
$det(AB^T)=det((A^T)B)$, for all square matrices $A$ and $B$
Which of the statements are correct? I do not feel secure about which of them that is true.
My answer:
My calculations have given me that $2$ and $4$ are true. Am I correct?
matrices symmetric-matrices
matrices symmetric-matrices
edited Dec 19 '18 at 7:12
Henno Brandsma
114k348124
114k348124
asked Dec 19 '18 at 6:53
Jacob AndreassonJacob Andreasson
35
35
1
$begingroup$
kindly edit your post to include your attempts using mathjax.
$endgroup$
– Siong Thye Goh
Dec 19 '18 at 6:54
$begingroup$
Okey. Sorry I forgot
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 6:55
$begingroup$
What do you know about determinants?
$endgroup$
– bubba
Dec 19 '18 at 6:58
1
$begingroup$
Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
$endgroup$
– bubba
Dec 19 '18 at 6:59
$begingroup$
Sorry I have calculate totally wrong
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:00
|
show 6 more comments
1
$begingroup$
kindly edit your post to include your attempts using mathjax.
$endgroup$
– Siong Thye Goh
Dec 19 '18 at 6:54
$begingroup$
Okey. Sorry I forgot
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 6:55
$begingroup$
What do you know about determinants?
$endgroup$
– bubba
Dec 19 '18 at 6:58
1
$begingroup$
Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
$endgroup$
– bubba
Dec 19 '18 at 6:59
$begingroup$
Sorry I have calculate totally wrong
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:00
1
1
$begingroup$
kindly edit your post to include your attempts using mathjax.
$endgroup$
– Siong Thye Goh
Dec 19 '18 at 6:54
$begingroup$
kindly edit your post to include your attempts using mathjax.
$endgroup$
– Siong Thye Goh
Dec 19 '18 at 6:54
$begingroup$
Okey. Sorry I forgot
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 6:55
$begingroup$
Okey. Sorry I forgot
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 6:55
$begingroup$
What do you know about determinants?
$endgroup$
– bubba
Dec 19 '18 at 6:58
$begingroup$
What do you know about determinants?
$endgroup$
– bubba
Dec 19 '18 at 6:58
1
1
$begingroup$
Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
$endgroup$
– bubba
Dec 19 '18 at 6:59
$begingroup$
Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
$endgroup$
– bubba
Dec 19 '18 at 6:59
$begingroup$
Sorry I have calculate totally wrong
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:00
$begingroup$
Sorry I have calculate totally wrong
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:00
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
- $det (A^2)=det A cdot det A=(det A)^2>0$
$det (I+I)=det (2I)=2^n .det I=2^n$ whereas $det I+frac{1}{det I}=2$
$vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $det vv^T=0$
- $det (AB^T)=det A cdot det B^T=det A cdot det B=det A^T.det B=det(A^TB)$
$endgroup$
$begingroup$
so the only one that is true is 4?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:27
$begingroup$
it is A*B$^T$ not (AB)$^T$
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:30
$begingroup$
see my answer again. use $det A=det A^T$
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:33
$begingroup$
1 and 4 is true then?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:36
$begingroup$
yes...................
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:36
add a comment |
$begingroup$
For 1, use that $det(AB)= det(A)det(B)$ to find that $det(A^2) = det(A)^2 > 0$.
For 2, take $A = I $.
For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.
For 4, use that $det(A^T) = det(A)$, so $det(AB^T) = det((AB^T)^T) = det(BA^T) = det(A^TB)$.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
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$begingroup$
- $det (A^2)=det A cdot det A=(det A)^2>0$
$det (I+I)=det (2I)=2^n .det I=2^n$ whereas $det I+frac{1}{det I}=2$
$vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $det vv^T=0$
- $det (AB^T)=det A cdot det B^T=det A cdot det B=det A^T.det B=det(A^TB)$
$endgroup$
$begingroup$
so the only one that is true is 4?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:27
$begingroup$
it is A*B$^T$ not (AB)$^T$
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:30
$begingroup$
see my answer again. use $det A=det A^T$
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:33
$begingroup$
1 and 4 is true then?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:36
$begingroup$
yes...................
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:36
add a comment |
$begingroup$
- $det (A^2)=det A cdot det A=(det A)^2>0$
$det (I+I)=det (2I)=2^n .det I=2^n$ whereas $det I+frac{1}{det I}=2$
$vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $det vv^T=0$
- $det (AB^T)=det A cdot det B^T=det A cdot det B=det A^T.det B=det(A^TB)$
$endgroup$
$begingroup$
so the only one that is true is 4?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:27
$begingroup$
it is A*B$^T$ not (AB)$^T$
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:30
$begingroup$
see my answer again. use $det A=det A^T$
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:33
$begingroup$
1 and 4 is true then?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:36
$begingroup$
yes...................
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:36
add a comment |
$begingroup$
- $det (A^2)=det A cdot det A=(det A)^2>0$
$det (I+I)=det (2I)=2^n .det I=2^n$ whereas $det I+frac{1}{det I}=2$
$vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $det vv^T=0$
- $det (AB^T)=det A cdot det B^T=det A cdot det B=det A^T.det B=det(A^TB)$
$endgroup$
- $det (A^2)=det A cdot det A=(det A)^2>0$
$det (I+I)=det (2I)=2^n .det I=2^n$ whereas $det I+frac{1}{det I}=2$
$vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $det vv^T=0$
- $det (AB^T)=det A cdot det B^T=det A cdot det B=det A^T.det B=det(A^TB)$
edited Dec 19 '18 at 7:32
answered Dec 19 '18 at 7:25
Chinnapparaj RChinnapparaj R
5,8682928
5,8682928
$begingroup$
so the only one that is true is 4?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:27
$begingroup$
it is A*B$^T$ not (AB)$^T$
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:30
$begingroup$
see my answer again. use $det A=det A^T$
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:33
$begingroup$
1 and 4 is true then?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:36
$begingroup$
yes...................
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:36
add a comment |
$begingroup$
so the only one that is true is 4?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:27
$begingroup$
it is A*B$^T$ not (AB)$^T$
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:30
$begingroup$
see my answer again. use $det A=det A^T$
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:33
$begingroup$
1 and 4 is true then?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:36
$begingroup$
yes...................
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:36
$begingroup$
so the only one that is true is 4?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:27
$begingroup$
so the only one that is true is 4?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:27
$begingroup$
it is A*B$^T$ not (AB)$^T$
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:30
$begingroup$
it is A*B$^T$ not (AB)$^T$
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:30
$begingroup$
see my answer again. use $det A=det A^T$
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:33
$begingroup$
see my answer again. use $det A=det A^T$
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:33
$begingroup$
1 and 4 is true then?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:36
$begingroup$
1 and 4 is true then?
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:36
$begingroup$
yes...................
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:36
$begingroup$
yes...................
$endgroup$
– Chinnapparaj R
Dec 19 '18 at 7:36
add a comment |
$begingroup$
For 1, use that $det(AB)= det(A)det(B)$ to find that $det(A^2) = det(A)^2 > 0$.
For 2, take $A = I $.
For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.
For 4, use that $det(A^T) = det(A)$, so $det(AB^T) = det((AB^T)^T) = det(BA^T) = det(A^TB)$.
$endgroup$
add a comment |
$begingroup$
For 1, use that $det(AB)= det(A)det(B)$ to find that $det(A^2) = det(A)^2 > 0$.
For 2, take $A = I $.
For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.
For 4, use that $det(A^T) = det(A)$, so $det(AB^T) = det((AB^T)^T) = det(BA^T) = det(A^TB)$.
$endgroup$
add a comment |
$begingroup$
For 1, use that $det(AB)= det(A)det(B)$ to find that $det(A^2) = det(A)^2 > 0$.
For 2, take $A = I $.
For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.
For 4, use that $det(A^T) = det(A)$, so $det(AB^T) = det((AB^T)^T) = det(BA^T) = det(A^TB)$.
$endgroup$
For 1, use that $det(AB)= det(A)det(B)$ to find that $det(A^2) = det(A)^2 > 0$.
For 2, take $A = I $.
For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.
For 4, use that $det(A^T) = det(A)$, so $det(AB^T) = det((AB^T)^T) = det(BA^T) = det(A^TB)$.
answered Dec 19 '18 at 7:31
Anthony TerAnthony Ter
37116
37116
add a comment |
add a comment |
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1
$begingroup$
kindly edit your post to include your attempts using mathjax.
$endgroup$
– Siong Thye Goh
Dec 19 '18 at 6:54
$begingroup$
Okey. Sorry I forgot
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 6:55
$begingroup$
What do you know about determinants?
$endgroup$
– bubba
Dec 19 '18 at 6:58
1
$begingroup$
Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
$endgroup$
– bubba
Dec 19 '18 at 6:59
$begingroup$
Sorry I have calculate totally wrong
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:00