Mixing
Determinant of a matrix of the form $M = I + xx^t$
$begingroup$
$$M=begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \ x_2x_1 & 1+x_2^2 &...&x_2x_n \...&...& &...& \x_nx_1 & x_nx_2& ...&1+x_n^2&end{pmatrix}.$$
So I noticed that $M$ is a symmetric matrix and $ M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$. That's all I can get.
linear-algebra determinant
edited Jan 22 at 23:02
Silverfish
1,0271223
asked Jan 22 at 20:21
Yibei HeYibei He
3139
$endgroup$
add a comment |
$begingroup$
$$M=begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \ x_2x_1 & 1+x_2^2 &...&x_2x_n \...&...& &...& \x_nx_1 & x_nx_2& ...&1+x_n^2&end{pmatrix}.$$
So I noticed that $M$ is a symmetric matrix and $ M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$. That's all I can get.
linear-algebra determinant
edited Jan 22 at 23:02
Silverfish
1,0271223
asked Jan 22 at 20:21
Yibei HeYibei He
3139
$endgroup$
1
$begingroup$
Can you determine the characteristic polynomial of $M-I$?
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:24
$begingroup$
Could you give me some more clue about the polynomial?
$endgroup$
– Yibei He
Jan 22 at 20:30
add a comment |
$begingroup$
$$M=begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \ x_2x_1 & 1+x_2^2 &...&x_2x_n \...&...& &...& \x_nx_1 & x_nx_2& ...&1+x_n^2&end{pmatrix}.$$
So I noticed that $M$ is a symmetric matrix and $ M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$. That's all I can get.
linear-algebra determinant
edited Jan 22 at 23:02
Silverfish
1,0271223
asked Jan 22 at 20:21
Yibei HeYibei He
3139
$endgroup$
$$M=begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \ x_2x_1 & 1+x_2^2 &...&x_2x_n \...&...& &...& \x_nx_1 & x_nx_2& ...&1+x_n^2&end{pmatrix}.$$
So I noticed that $M$ is a symmetric matrix and $ M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$. That's all I can get.
linear-algebra determinant
linear-algebra determinant
edited Jan 22 at 23:02
Silverfish
1,0271223
asked Jan 22 at 20:21
Yibei HeYibei He
3139
edited Jan 22 at 23:02
Silverfish
1,0271223
asked Jan 22 at 20:21
Yibei HeYibei He
3139
edited Jan 22 at 23:02
Silverfish
1,0271223
edited Jan 22 at 23:02
Silverfish
1,0271223
edited Jan 22 at 23:02
Silverfish
1,0271223
1,0271223
asked Jan 22 at 20:21
Yibei HeYibei He
3139
asked Jan 22 at 20:21
Yibei HeYibei He
3139
asked Jan 22 at 20:21
Yibei HeYibei He
3139
3139
1
$begingroup$
Can you determine the characteristic polynomial of $M-I$?
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:24
$begingroup$
Could you give me some more clue about the polynomial?
$endgroup$
– Yibei He
Jan 22 at 20:30
add a comment |
1
$begingroup$
Can you determine the characteristic polynomial of $M-I$?
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:24
$begingroup$
Could you give me some more clue about the polynomial?
$endgroup$
– Yibei He
Jan 22 at 20:30
1
1
$begingroup$
Can you determine the characteristic polynomial of $M-I$?
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:24
$begingroup$
Can you determine the characteristic polynomial of $M-I$?
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:24
$begingroup$
Could you give me some more clue about the polynomial?
$endgroup$
– Yibei He
Jan 22 at 20:30
$begingroup$
Could you give me some more clue about the polynomial?
$endgroup$
– Yibei He
Jan 22 at 20:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).
For a column vector $x = (x_1,...,x_n)$, and an $ntimes n $ identity matrix $I$, we have
$$
M = I + x x^T
$$
hence, in view of the aforementioned lemma,
$$
det M = (1 + x^T I x)det I = 1 + |x|^2 = 1 + x_1^2 + cdots +x_n^2.
$$
edited Jan 23 at 11:37
Rodrigo de Azevedo
13k41960
answered Jan 22 at 20:47
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
Consider the matrix
$$N=M-I.$$
It has rank $le1$, so its characteristic polynomial $P(lambda)=det(lambda I-N)$
has the form $lambda^n-alambda^{n-1}$. But $a$ is the trace of $N$, so
$$P(lambda)=lambda^n-(x_1^2+cdots+x_n^2)lambda^{n-1}.$$
Now
$$det M=det(I+N)=(-1)^nP(-1)=1+x_1^2+cdots+x_n^2.$$
answered Jan 22 at 20:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
Denote by $M_n$ the matrix we want to compute the determinant. Letting $mathbf{x}=(x_1,dots,x_{n-1})$, we express $M_n$ as a block matrix:
$$
M_n=pmatrix{M_{n-1}& x_nmathbf x^t\
x_nmathbf{x}&1+x_n^2}.
$$
Since $pmatrix{ x_nmathbf x^t\1+x_n^2}=pmatrix{ 0\1}+pmatrix{ mathbf x_nx^t\x_n^2}$, it follows by multi-linearity of the determinant that
$$
detleft(M_nright)=detpmatrix{M_{n-1}& 0\
x_nmathbf{x}&1}+detpmatrix{M_{n-1}& x_nmathbf x^t\
x_nmathbf{x}&x_n^2}.
$$
The first term of the right hand side is $detleft(M_{n-1}right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that
$$
detleft(M_{n}right)=detleft(M_{n-1}right)+x_n^2detpmatrix{M_{n-1}& mathbf x^t\
mathbf{x}&1}.
$$
The last determinant is one: one can make the linear combinations $C_ileftarrow x_iC_n$ to see this.
answered Jan 22 at 21:46
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
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
$begingroup$
You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).
For a column vector $x = (x_1,...,x_n)$, and an $ntimes n $ identity matrix $I$, we have
$$
M = I + x x^T
$$
hence, in view of the aforementioned lemma,
$$
det M = (1 + x^T I x)det I = 1 + |x|^2 = 1 + x_1^2 + cdots +x_n^2.
$$
edited Jan 23 at 11:37
Rodrigo de Azevedo
13k41960
answered Jan 22 at 20:47
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).
For a column vector $x = (x_1,...,x_n)$, and an $ntimes n $ identity matrix $I$, we have
$$
M = I + x x^T
$$
hence, in view of the aforementioned lemma,
$$
det M = (1 + x^T I x)det I = 1 + |x|^2 = 1 + x_1^2 + cdots +x_n^2.
$$
edited Jan 23 at 11:37
Rodrigo de Azevedo
13k41960
answered Jan 22 at 20:47
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).
For a column vector $x = (x_1,...,x_n)$, and an $ntimes n $ identity matrix $I$, we have
$$
M = I + x x^T
$$
hence, in view of the aforementioned lemma,
$$
det M = (1 + x^T I x)det I = 1 + |x|^2 = 1 + x_1^2 + cdots +x_n^2.
$$
edited Jan 23 at 11:37
Rodrigo de Azevedo
13k41960
answered Jan 22 at 20:47
HaykHayk
2,6271214
$endgroup$
You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).
For a column vector $x = (x_1,...,x_n)$, and an $ntimes n $ identity matrix $I$, we have
$$
M = I + x x^T
$$
hence, in view of the aforementioned lemma,
$$
det M = (1 + x^T I x)det I = 1 + |x|^2 = 1 + x_1^2 + cdots +x_n^2.
$$
edited Jan 23 at 11:37
Rodrigo de Azevedo
13k41960
answered Jan 22 at 20:47
HaykHayk
2,6271214
edited Jan 23 at 11:37
Rodrigo de Azevedo
13k41960
edited Jan 23 at 11:37
Rodrigo de Azevedo
13k41960
edited Jan 23 at 11:37
Rodrigo de Azevedo
13k41960
13k41960
answered Jan 22 at 20:47
HaykHayk
2,6271214
answered Jan 22 at 20:47
HaykHayk
2,6271214
answered Jan 22 at 20:47
HaykHayk
2,6271214
2,6271214
add a comment |
add a comment |
$begingroup$
Consider the matrix
$$N=M-I.$$
It has rank $le1$, so its characteristic polynomial $P(lambda)=det(lambda I-N)$
has the form $lambda^n-alambda^{n-1}$. But $a$ is the trace of $N$, so
$$P(lambda)=lambda^n-(x_1^2+cdots+x_n^2)lambda^{n-1}.$$
Now
$$det M=det(I+N)=(-1)^nP(-1)=1+x_1^2+cdots+x_n^2.$$
answered Jan 22 at 20:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
Consider the matrix
$$N=M-I.$$
It has rank $le1$, so its characteristic polynomial $P(lambda)=det(lambda I-N)$
has the form $lambda^n-alambda^{n-1}$. But $a$ is the trace of $N$, so
$$P(lambda)=lambda^n-(x_1^2+cdots+x_n^2)lambda^{n-1}.$$
Now
$$det M=det(I+N)=(-1)^nP(-1)=1+x_1^2+cdots+x_n^2.$$
answered Jan 22 at 20:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
Consider the matrix
$$N=M-I.$$
It has rank $le1$, so its characteristic polynomial $P(lambda)=det(lambda I-N)$
has the form $lambda^n-alambda^{n-1}$. But $a$ is the trace of $N$, so
$$P(lambda)=lambda^n-(x_1^2+cdots+x_n^2)lambda^{n-1}.$$
Now
$$det M=det(I+N)=(-1)^nP(-1)=1+x_1^2+cdots+x_n^2.$$
answered Jan 22 at 20:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
Consider the matrix
$$N=M-I.$$
It has rank $le1$, so its characteristic polynomial $P(lambda)=det(lambda I-N)$
has the form $lambda^n-alambda^{n-1}$. But $a$ is the trace of $N$, so
$$P(lambda)=lambda^n-(x_1^2+cdots+x_n^2)lambda^{n-1}.$$
Now
$$det M=det(I+N)=(-1)^nP(-1)=1+x_1^2+cdots+x_n^2.$$
answered Jan 22 at 20:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 22 at 20:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 22 at 20:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 22 at 20:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
add a comment |
add a comment |
$begingroup$
Denote by $M_n$ the matrix we want to compute the determinant. Letting $mathbf{x}=(x_1,dots,x_{n-1})$, we express $M_n$ as a block matrix:
$$
M_n=pmatrix{M_{n-1}& x_nmathbf x^t\
x_nmathbf{x}&1+x_n^2}.
$$
Since $pmatrix{ x_nmathbf x^t\1+x_n^2}=pmatrix{ 0\1}+pmatrix{ mathbf x_nx^t\x_n^2}$, it follows by multi-linearity of the determinant that
$$
detleft(M_nright)=detpmatrix{M_{n-1}& 0\
x_nmathbf{x}&1}+detpmatrix{M_{n-1}& x_nmathbf x^t\
x_nmathbf{x}&x_n^2}.
$$
The first term of the right hand side is $detleft(M_{n-1}right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that
$$
detleft(M_{n}right)=detleft(M_{n-1}right)+x_n^2detpmatrix{M_{n-1}& mathbf x^t\
mathbf{x}&1}.
$$
The last determinant is one: one can make the linear combinations $C_ileftarrow x_iC_n$ to see this.
answered Jan 22 at 21:46
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
add a comment |
$begingroup$
Denote by $M_n$ the matrix we want to compute the determinant. Letting $mathbf{x}=(x_1,dots,x_{n-1})$, we express $M_n$ as a block matrix:
$$
M_n=pmatrix{M_{n-1}& x_nmathbf x^t\
x_nmathbf{x}&1+x_n^2}.
$$
Since $pmatrix{ x_nmathbf x^t\1+x_n^2}=pmatrix{ 0\1}+pmatrix{ mathbf x_nx^t\x_n^2}$, it follows by multi-linearity of the determinant that
$$
detleft(M_nright)=detpmatrix{M_{n-1}& 0\
x_nmathbf{x}&1}+detpmatrix{M_{n-1}& x_nmathbf x^t\
x_nmathbf{x}&x_n^2}.
$$
The first term of the right hand side is $detleft(M_{n-1}right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that
$$
detleft(M_{n}right)=detleft(M_{n-1}right)+x_n^2detpmatrix{M_{n-1}& mathbf x^t\
mathbf{x}&1}.
$$
The last determinant is one: one can make the linear combinations $C_ileftarrow x_iC_n$ to see this.
answered Jan 22 at 21:46
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
add a comment |
$begingroup$
Denote by $M_n$ the matrix we want to compute the determinant. Letting $mathbf{x}=(x_1,dots,x_{n-1})$, we express $M_n$ as a block matrix:
$$
M_n=pmatrix{M_{n-1}& x_nmathbf x^t\
x_nmathbf{x}&1+x_n^2}.
$$
Since $pmatrix{ x_nmathbf x^t\1+x_n^2}=pmatrix{ 0\1}+pmatrix{ mathbf x_nx^t\x_n^2}$, it follows by multi-linearity of the determinant that
$$
detleft(M_nright)=detpmatrix{M_{n-1}& 0\
x_nmathbf{x}&1}+detpmatrix{M_{n-1}& x_nmathbf x^t\
x_nmathbf{x}&x_n^2}.
$$
The first term of the right hand side is $detleft(M_{n-1}right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that
$$
detleft(M_{n}right)=detleft(M_{n-1}right)+x_n^2detpmatrix{M_{n-1}& mathbf x^t\
mathbf{x}&1}.
$$
The last determinant is one: one can make the linear combinations $C_ileftarrow x_iC_n$ to see this.
answered Jan 22 at 21:46
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
Denote by $M_n$ the matrix we want to compute the determinant. Letting $mathbf{x}=(x_1,dots,x_{n-1})$, we express $M_n$ as a block matrix:
$$
M_n=pmatrix{M_{n-1}& x_nmathbf x^t\
x_nmathbf{x}&1+x_n^2}.
$$
Since $pmatrix{ x_nmathbf x^t\1+x_n^2}=pmatrix{ 0\1}+pmatrix{ mathbf x_nx^t\x_n^2}$, it follows by multi-linearity of the determinant that
$$
detleft(M_nright)=detpmatrix{M_{n-1}& 0\
x_nmathbf{x}&1}+detpmatrix{M_{n-1}& x_nmathbf x^t\
x_nmathbf{x}&x_n^2}.
$$
The first term of the right hand side is $detleft(M_{n-1}right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that
$$
detleft(M_{n}right)=detleft(M_{n-1}right)+x_n^2detpmatrix{M_{n-1}& mathbf x^t\
mathbf{x}&1}.
$$
The last determinant is one: one can make the linear combinations $C_ileftarrow x_iC_n$ to see this.
answered Jan 22 at 21:46
Davide GiraudoDavide Giraudo
127k17154268
answered Jan 22 at 21:46
Davide GiraudoDavide Giraudo
127k17154268
answered Jan 22 at 21:46
Davide GiraudoDavide Giraudo
127k17154268
answered Jan 22 at 21:46
Davide GiraudoDavide Giraudo
127k17154268
127k17154268
add a comment |
add a comment |
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1
$begingroup$
Can you determine the characteristic polynomial of $M-I$?
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:24
$begingroup$
Could you give me some more clue about the polynomial?
$endgroup$
– Yibei He
Jan 22 at 20:30