How do I solve this? Find determinant of a matrix
$begingroup$
I have a matrix with n rows and n columns: $A=(a_{j,k})$ is defined:
$$(a_{j,k})=begin{cases}
begin{split}
2j (jneq k) \ 0(j=k)
end{split} end{cases}$$
And I need to find $det(A)$. I think that I need to use gauss elimination but I have no idea how to do this here. can anyone help please? thanks.
linear-algebra determinant
asked Dec 23 '18 at 16:37
OmerOmer
3969
$endgroup$
add a comment |
$begingroup$
I have a matrix with n rows and n columns: $A=(a_{j,k})$ is defined:
$$(a_{j,k})=begin{cases}
begin{split}
2j (jneq k) \ 0(j=k)
end{split} end{cases}$$
And I need to find $det(A)$. I think that I need to use gauss elimination but I have no idea how to do this here. can anyone help please? thanks.
linear-algebra determinant
asked Dec 23 '18 at 16:37
OmerOmer
3969
$endgroup$
1
$begingroup$
You may start from factoring out $2j$ from each column in the determinant.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:05
$begingroup$
@A.Γ. sorry, I did not write the question well. I needed to write a(j,k) insted of a(k,j) but this is still the same, so now i need to factor 2j from each row. I have done that, and i get a messy matrix with much 1's, i understood that on the diagnole there are 0's and the other are 1's, but how can i calculate the determinant of that matrix?
$endgroup$
– Omer
Dec 23 '18 at 17:21
$begingroup$
If you add the identity matrix to the result, you will get a very nice matrix.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:22
$begingroup$
@A.Γ. Yes, I see, I will get a matrix of 1's, and its determinant of this is 0. but, there is no formula for det(A+B) as an expression of detA and detB, nothing that I was tought at least... so how does that help me?
$endgroup$
– Omer
Dec 23 '18 at 17:29
add a comment |
$begingroup$
I have a matrix with n rows and n columns: $A=(a_{j,k})$ is defined:
$$(a_{j,k})=begin{cases}
begin{split}
2j (jneq k) \ 0(j=k)
end{split} end{cases}$$
And I need to find $det(A)$. I think that I need to use gauss elimination but I have no idea how to do this here. can anyone help please? thanks.
linear-algebra determinant
asked Dec 23 '18 at 16:37
OmerOmer
3969
$endgroup$
I have a matrix with n rows and n columns: $A=(a_{j,k})$ is defined:
$$(a_{j,k})=begin{cases}
begin{split}
2j (jneq k) \ 0(j=k)
end{split} end{cases}$$
And I need to find $det(A)$. I think that I need to use gauss elimination but I have no idea how to do this here. can anyone help please? thanks.
linear-algebra determinant
linear-algebra determinant
asked Dec 23 '18 at 16:37
OmerOmer
3969
asked Dec 23 '18 at 16:37
OmerOmer
3969
edited Dec 23 '18 at 17:21
Omer
asked Dec 23 '18 at 16:37
OmerOmer
3969
asked Dec 23 '18 at 16:37
OmerOmer
3969
asked Dec 23 '18 at 16:37
OmerOmer
3969
3969
1
$begingroup$
You may start from factoring out $2j$ from each column in the determinant.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:05
$begingroup$
@A.Γ. sorry, I did not write the question well. I needed to write a(j,k) insted of a(k,j) but this is still the same, so now i need to factor 2j from each row. I have done that, and i get a messy matrix with much 1's, i understood that on the diagnole there are 0's and the other are 1's, but how can i calculate the determinant of that matrix?
$endgroup$
– Omer
Dec 23 '18 at 17:21
$begingroup$
If you add the identity matrix to the result, you will get a very nice matrix.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:22
$begingroup$
@A.Γ. Yes, I see, I will get a matrix of 1's, and its determinant of this is 0. but, there is no formula for det(A+B) as an expression of detA and detB, nothing that I was tought at least... so how does that help me?
$endgroup$
– Omer
Dec 23 '18 at 17:29
add a comment |
1
$begingroup$
You may start from factoring out $2j$ from each column in the determinant.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:05
$begingroup$
@A.Γ. sorry, I did not write the question well. I needed to write a(j,k) insted of a(k,j) but this is still the same, so now i need to factor 2j from each row. I have done that, and i get a messy matrix with much 1's, i understood that on the diagnole there are 0's and the other are 1's, but how can i calculate the determinant of that matrix?
$endgroup$
– Omer
Dec 23 '18 at 17:21
$begingroup$
If you add the identity matrix to the result, you will get a very nice matrix.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:22
$begingroup$
@A.Γ. Yes, I see, I will get a matrix of 1's, and its determinant of this is 0. but, there is no formula for det(A+B) as an expression of detA and detB, nothing that I was tought at least... so how does that help me?
$endgroup$
– Omer
Dec 23 '18 at 17:29
1
1
$begingroup$
You may start from factoring out $2j$ from each column in the determinant.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:05
$begingroup$
You may start from factoring out $2j$ from each column in the determinant.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:05
$begingroup$
@A.Γ. sorry, I did not write the question well. I needed to write a(j,k) insted of a(k,j) but this is still the same, so now i need to factor 2j from each row. I have done that, and i get a messy matrix with much 1's, i understood that on the diagnole there are 0's and the other are 1's, but how can i calculate the determinant of that matrix?
$endgroup$
– Omer
Dec 23 '18 at 17:21
$begingroup$
@A.Γ. sorry, I did not write the question well. I needed to write a(j,k) insted of a(k,j) but this is still the same, so now i need to factor 2j from each row. I have done that, and i get a messy matrix with much 1's, i understood that on the diagnole there are 0's and the other are 1's, but how can i calculate the determinant of that matrix?
$endgroup$
– Omer
Dec 23 '18 at 17:21
$begingroup$
If you add the identity matrix to the result, you will get a very nice matrix.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:22
$begingroup$
If you add the identity matrix to the result, you will get a very nice matrix.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:22
$begingroup$
@A.Γ. Yes, I see, I will get a matrix of 1's, and its determinant of this is 0. but, there is no formula for det(A+B) as an expression of detA and detB, nothing that I was tought at least... so how does that help me?
$endgroup$
– Omer
Dec 23 '18 at 17:29
$begingroup$
@A.Γ. Yes, I see, I will get a matrix of 1's, and its determinant of this is 0. but, there is no formula for det(A+B) as an expression of detA and detB, nothing that I was tought at least... so how does that help me?
$endgroup$
– Omer
Dec 23 '18 at 17:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
After factoring out $2j$ you will get the matrix
$$
begin{bmatrix}1\1\vdots\1end{bmatrix}begin{bmatrix}1 &1&ldots&1end{bmatrix}-I=ee^T-I.
$$
Now use the Sylvester determinant formula
$$
det(ee^T-I)=(-1)^ndet(I-ee^T)=(-1)^n(1-e^Te)=(-1)^n(1-n).
$$
answered Dec 23 '18 at 17:46
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
Taking out the $2j$ factors as suggested leaves $E-I$, with $I$ the identity matrix and $E$ all $1s$. If $a$ has shape $ntimes n$ the eigenvalues of $E$ are $0,,n$, of respective multiplicity $n-1,,1$. Then $$det (E-I)=(-1)^{n-1}(n-1),,det a=-n!(n-1)(-2)^n.$$
answered Dec 23 '18 at 17:33
J.G.J.G.
29.4k22846
$endgroup$
$begingroup$
sorry for keep asking stupid questions, but i still did not understand the last part of det(E-I). Is there a way to show this using gauss elimination
$endgroup$
– Omer
Dec 23 '18 at 17:42
$begingroup$
@Omer Which part is unclear: how I get the eigenvalues' multiplicities, how I get the determinant from there, or the scaling to compute $det a$? If it's the first part, the trick is that $Ev=0$ only places one constraint on an eigenvector, so the eigenspace has dimension $n-1$, while any $lambda=n$ eigenvector is parallel to the all-$1$s vector.
$endgroup$
– J.G.
Dec 23 '18 at 17:45
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
After factoring out $2j$ you will get the matrix
$$
begin{bmatrix}1\1\vdots\1end{bmatrix}begin{bmatrix}1 &1&ldots&1end{bmatrix}-I=ee^T-I.
$$
Now use the Sylvester determinant formula
$$
det(ee^T-I)=(-1)^ndet(I-ee^T)=(-1)^n(1-e^Te)=(-1)^n(1-n).
$$
answered Dec 23 '18 at 17:46
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
After factoring out $2j$ you will get the matrix
$$
begin{bmatrix}1\1\vdots\1end{bmatrix}begin{bmatrix}1 &1&ldots&1end{bmatrix}-I=ee^T-I.
$$
Now use the Sylvester determinant formula
$$
det(ee^T-I)=(-1)^ndet(I-ee^T)=(-1)^n(1-e^Te)=(-1)^n(1-n).
$$
answered Dec 23 '18 at 17:46
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
After factoring out $2j$ you will get the matrix
$$
begin{bmatrix}1\1\vdots\1end{bmatrix}begin{bmatrix}1 &1&ldots&1end{bmatrix}-I=ee^T-I.
$$
Now use the Sylvester determinant formula
$$
det(ee^T-I)=(-1)^ndet(I-ee^T)=(-1)^n(1-e^Te)=(-1)^n(1-n).
$$
answered Dec 23 '18 at 17:46
A.Γ.A.Γ.
22.8k32656
$endgroup$
After factoring out $2j$ you will get the matrix
$$
begin{bmatrix}1\1\vdots\1end{bmatrix}begin{bmatrix}1 &1&ldots&1end{bmatrix}-I=ee^T-I.
$$
Now use the Sylvester determinant formula
$$
det(ee^T-I)=(-1)^ndet(I-ee^T)=(-1)^n(1-e^Te)=(-1)^n(1-n).
$$
answered Dec 23 '18 at 17:46
A.Γ.A.Γ.
22.8k32656
edited Dec 23 '18 at 17:53
answered Dec 23 '18 at 17:46
A.Γ.A.Γ.
22.8k32656
answered Dec 23 '18 at 17:46
A.Γ.A.Γ.
22.8k32656
answered Dec 23 '18 at 17:46
A.Γ.A.Γ.
22.8k32656
22.8k32656
add a comment |
add a comment |
$begingroup$
Taking out the $2j$ factors as suggested leaves $E-I$, with $I$ the identity matrix and $E$ all $1s$. If $a$ has shape $ntimes n$ the eigenvalues of $E$ are $0,,n$, of respective multiplicity $n-1,,1$. Then $$det (E-I)=(-1)^{n-1}(n-1),,det a=-n!(n-1)(-2)^n.$$
answered Dec 23 '18 at 17:33
J.G.J.G.
29.4k22846
$endgroup$
$begingroup$
sorry for keep asking stupid questions, but i still did not understand the last part of det(E-I). Is there a way to show this using gauss elimination
$endgroup$
– Omer
Dec 23 '18 at 17:42
$begingroup$
@Omer Which part is unclear: how I get the eigenvalues' multiplicities, how I get the determinant from there, or the scaling to compute $det a$? If it's the first part, the trick is that $Ev=0$ only places one constraint on an eigenvector, so the eigenspace has dimension $n-1$, while any $lambda=n$ eigenvector is parallel to the all-$1$s vector.
$endgroup$
– J.G.
Dec 23 '18 at 17:45
add a comment |
$begingroup$
Taking out the $2j$ factors as suggested leaves $E-I$, with $I$ the identity matrix and $E$ all $1s$. If $a$ has shape $ntimes n$ the eigenvalues of $E$ are $0,,n$, of respective multiplicity $n-1,,1$. Then $$det (E-I)=(-1)^{n-1}(n-1),,det a=-n!(n-1)(-2)^n.$$
answered Dec 23 '18 at 17:33
J.G.J.G.
29.4k22846
$endgroup$
$begingroup$
sorry for keep asking stupid questions, but i still did not understand the last part of det(E-I). Is there a way to show this using gauss elimination
$endgroup$
– Omer
Dec 23 '18 at 17:42
$begingroup$
@Omer Which part is unclear: how I get the eigenvalues' multiplicities, how I get the determinant from there, or the scaling to compute $det a$? If it's the first part, the trick is that $Ev=0$ only places one constraint on an eigenvector, so the eigenspace has dimension $n-1$, while any $lambda=n$ eigenvector is parallel to the all-$1$s vector.
$endgroup$
– J.G.
Dec 23 '18 at 17:45
add a comment |
$begingroup$
Taking out the $2j$ factors as suggested leaves $E-I$, with $I$ the identity matrix and $E$ all $1s$. If $a$ has shape $ntimes n$ the eigenvalues of $E$ are $0,,n$, of respective multiplicity $n-1,,1$. Then $$det (E-I)=(-1)^{n-1}(n-1),,det a=-n!(n-1)(-2)^n.$$
answered Dec 23 '18 at 17:33
J.G.J.G.
29.4k22846
$endgroup$
Taking out the $2j$ factors as suggested leaves $E-I$, with $I$ the identity matrix and $E$ all $1s$. If $a$ has shape $ntimes n$ the eigenvalues of $E$ are $0,,n$, of respective multiplicity $n-1,,1$. Then $$det (E-I)=(-1)^{n-1}(n-1),,det a=-n!(n-1)(-2)^n.$$
answered Dec 23 '18 at 17:33
J.G.J.G.
29.4k22846
answered Dec 23 '18 at 17:33
J.G.J.G.
29.4k22846
answered Dec 23 '18 at 17:33
J.G.J.G.
29.4k22846
answered Dec 23 '18 at 17:33
J.G.J.G.
29.4k22846
29.4k22846
$begingroup$
sorry for keep asking stupid questions, but i still did not understand the last part of det(E-I). Is there a way to show this using gauss elimination
$endgroup$
– Omer
Dec 23 '18 at 17:42
$begingroup$
@Omer Which part is unclear: how I get the eigenvalues' multiplicities, how I get the determinant from there, or the scaling to compute $det a$? If it's the first part, the trick is that $Ev=0$ only places one constraint on an eigenvector, so the eigenspace has dimension $n-1$, while any $lambda=n$ eigenvector is parallel to the all-$1$s vector.
$endgroup$
– J.G.
Dec 23 '18 at 17:45
add a comment |
$begingroup$
sorry for keep asking stupid questions, but i still did not understand the last part of det(E-I). Is there a way to show this using gauss elimination
$endgroup$
– Omer
Dec 23 '18 at 17:42
$begingroup$
@Omer Which part is unclear: how I get the eigenvalues' multiplicities, how I get the determinant from there, or the scaling to compute $det a$? If it's the first part, the trick is that $Ev=0$ only places one constraint on an eigenvector, so the eigenspace has dimension $n-1$, while any $lambda=n$ eigenvector is parallel to the all-$1$s vector.
$endgroup$
– J.G.
Dec 23 '18 at 17:45
$begingroup$
sorry for keep asking stupid questions, but i still did not understand the last part of det(E-I). Is there a way to show this using gauss elimination
$endgroup$
– Omer
Dec 23 '18 at 17:42
$begingroup$
sorry for keep asking stupid questions, but i still did not understand the last part of det(E-I). Is there a way to show this using gauss elimination
$endgroup$
– Omer
Dec 23 '18 at 17:42
$begingroup$
@Omer Which part is unclear: how I get the eigenvalues' multiplicities, how I get the determinant from there, or the scaling to compute $det a$? If it's the first part, the trick is that $Ev=0$ only places one constraint on an eigenvector, so the eigenspace has dimension $n-1$, while any $lambda=n$ eigenvector is parallel to the all-$1$s vector.
$endgroup$
– J.G.
Dec 23 '18 at 17:45
$begingroup$
@Omer Which part is unclear: how I get the eigenvalues' multiplicities, how I get the determinant from there, or the scaling to compute $det a$? If it's the first part, the trick is that $Ev=0$ only places one constraint on an eigenvector, so the eigenspace has dimension $n-1$, while any $lambda=n$ eigenvector is parallel to the all-$1$s vector.
$endgroup$
– J.G.
Dec 23 '18 at 17:45
add a comment |
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1
$begingroup$
You may start from factoring out $2j$ from each column in the determinant.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:05
$begingroup$
@A.Γ. sorry, I did not write the question well. I needed to write a(j,k) insted of a(k,j) but this is still the same, so now i need to factor 2j from each row. I have done that, and i get a messy matrix with much 1's, i understood that on the diagnole there are 0's and the other are 1's, but how can i calculate the determinant of that matrix?
$endgroup$
– Omer
Dec 23 '18 at 17:21
$begingroup$
If you add the identity matrix to the result, you will get a very nice matrix.
$endgroup$
– A.Γ.
Dec 23 '18 at 17:22
$begingroup$
@A.Γ. Yes, I see, I will get a matrix of 1's, and its determinant of this is 0. but, there is no formula for det(A+B) as an expression of detA and detB, nothing that I was tought at least... so how does that help me?
$endgroup$
– Omer
Dec 23 '18 at 17:29