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How to prove $text{det}(I+xy^{top}+wz^{top})=(1+y^{top}x)(1+z^{top}w)-(x^{top}z)(y^{top}w)$?

2

$begingroup$

Suppose $x,y,z,w$ are vectors in $mathbb{R}^n$ and $I$ is the identity matrix.

Show that $text{det}(I+xy^{top}+wz^{top})=(1+y^{top}x)(1+z^{top}w)-(x^{top}z)(y^{top}w)$.

linear-algebra matrices determinant

share|cite|improve this question

asked Mar 23 at 22:50

SaeedSaeed

1,149310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any attempts so far? Do you know any potentially useful theorems? For example, things like en.wikipedia.org/wiki/Matrix_determinant_lemma and en.wikipedia.org/wiki/Sherman–Morrison_formula.
  $endgroup$
  – Minus One-Twelfth
  Mar 23 at 22:54
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Suppose $x,y,z,w$ are vectors in $mathbb{R}^n$ and $I$ is the identity matrix.

Show that $text{det}(I+xy^{top}+wz^{top})=(1+y^{top}x)(1+z^{top}w)-(x^{top}z)(y^{top}w)$.

linear-algebra matrices determinant

share|cite|improve this question

asked Mar 23 at 22:50

SaeedSaeed

1,149310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any attempts so far? Do you know any potentially useful theorems? For example, things like en.wikipedia.org/wiki/Matrix_determinant_lemma and en.wikipedia.org/wiki/Sherman–Morrison_formula.
  $endgroup$
  – Minus One-Twelfth
  Mar 23 at 22:54
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $x,y,z,w$ are vectors in $mathbb{R}^n$ and $I$ is the identity matrix.

Show that $text{det}(I+xy^{top}+wz^{top})=(1+y^{top}x)(1+z^{top}w)-(x^{top}z)(y^{top}w)$.

linear-algebra matrices determinant

share|cite|improve this question

asked Mar 23 at 22:50

SaeedSaeed

1,149310

$endgroup$

share|cite|improve this question

asked Mar 23 at 22:50

SaeedSaeed

1,149310

share|cite|improve this question

asked Mar 23 at 22:50

SaeedSaeed

1,149310

asked Mar 23 at 22:50

SaeedSaeed

1,149310

asked Mar 23 at 22:50

SaeedSaeed

1,149310

1 Answer
1

active

oldest

votes

1

$begingroup$

Consider the matrix
$$tag1
begin{bmatrix} 1+y^Tx & y^Tw \ z^Tx & 1+z^Tw end{bmatrix}
=I + begin{bmatrix} y^T\ z^T end{bmatrix} begin{bmatrix} x & wend{bmatrix}
$$
For any $A,B$, we have the equality $det(I+AB)=det(I+BA)$. So the determinant in $(1)$ is equal to the determinant of
$$
I + begin{bmatrix} x & wend{bmatrix}begin{bmatrix} y^T\ z^T end{bmatrix}
=I+xy^T+wz^T.
$$

---

Proof of the equality $det(I+AB)=det(I+BA)$.

The matrices $AB$ and $BA$ have the same eigenvalues (padding with zeroes when the sizes are not equal). So $I+AB$ and $I+BA$ have the same eigenvalues, and if the two lists have different length then the remaining eigenvalues are $1$. As the determinant is the product of the eigenvalues, we get $det(I+AB)=det(I+BA)$.

share|cite|improve this answer

edited Mar 25 at 1:44

answered Mar 24 at 0:57

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

add a comment | 

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1

$begingroup$

Consider the matrix
$$tag1
begin{bmatrix} 1+y^Tx & y^Tw \ z^Tx & 1+z^Tw end{bmatrix}
=I + begin{bmatrix} y^T\ z^T end{bmatrix} begin{bmatrix} x & wend{bmatrix}
$$
For any $A,B$, we have the equality $det(I+AB)=det(I+BA)$. So the determinant in $(1)$ is equal to the determinant of
$$
I + begin{bmatrix} x & wend{bmatrix}begin{bmatrix} y^T\ z^T end{bmatrix}
=I+xy^T+wz^T.
$$

---

Proof of the equality $det(I+AB)=det(I+BA)$.

The matrices $AB$ and $BA$ have the same eigenvalues (padding with zeroes when the sizes are not equal). So $I+AB$ and $I+BA$ have the same eigenvalues, and if the two lists have different length then the remaining eigenvalues are $1$. As the determinant is the product of the eigenvalues, we get $det(I+AB)=det(I+BA)$.

share|cite|improve this answer

edited Mar 25 at 1:44

answered Mar 24 at 0:57

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

add a comment | 

1

$begingroup$

Consider the matrix
$$tag1
begin{bmatrix} 1+y^Tx & y^Tw \ z^Tx & 1+z^Tw end{bmatrix}
=I + begin{bmatrix} y^T\ z^T end{bmatrix} begin{bmatrix} x & wend{bmatrix}
$$
For any $A,B$, we have the equality $det(I+AB)=det(I+BA)$. So the determinant in $(1)$ is equal to the determinant of
$$
I + begin{bmatrix} x & wend{bmatrix}begin{bmatrix} y^T\ z^T end{bmatrix}
=I+xy^T+wz^T.
$$

---

Proof of the equality $det(I+AB)=det(I+BA)$.

The matrices $AB$ and $BA$ have the same eigenvalues (padding with zeroes when the sizes are not equal). So $I+AB$ and $I+BA$ have the same eigenvalues, and if the two lists have different length then the remaining eigenvalues are $1$. As the determinant is the product of the eigenvalues, we get $det(I+AB)=det(I+BA)$.

share|cite|improve this answer

edited Mar 25 at 1:44

answered Mar 24 at 0:57

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

add a comment | 

$begingroup$

Consider the matrix
$$tag1
begin{bmatrix} 1+y^Tx & y^Tw \ z^Tx & 1+z^Tw end{bmatrix}
=I + begin{bmatrix} y^T\ z^T end{bmatrix} begin{bmatrix} x & wend{bmatrix}
$$
For any $A,B$, we have the equality $det(I+AB)=det(I+BA)$. So the determinant in $(1)$ is equal to the determinant of
$$
I + begin{bmatrix} x & wend{bmatrix}begin{bmatrix} y^T\ z^T end{bmatrix}
=I+xy^T+wz^T.
$$

---

Proof of the equality $det(I+AB)=det(I+BA)$.

The matrices $AB$ and $BA$ have the same eigenvalues (padding with zeroes when the sizes are not equal). So $I+AB$ and $I+BA$ have the same eigenvalues, and if the two lists have different length then the remaining eigenvalues are $1$. As the determinant is the product of the eigenvalues, we get $det(I+AB)=det(I+BA)$.

share|cite|improve this answer

edited Mar 25 at 1:44

answered Mar 24 at 0:57

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

share|cite|improve this answer

edited Mar 25 at 1:44

answered Mar 24 at 0:57

Martin ArgeramiMartin Argerami

130k1184185

answered Mar 24 at 0:57

Martin ArgeramiMartin Argerami

130k1184185

answered Mar 24 at 0:57

Martin ArgeramiMartin Argerami

130k1184185