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Show that rank of skew-symetric is even number

Rank of skew-symmetric matrixThe rank of skew-symmetric matrix is evenProof if $AB+BA=0$ Then atleast one of the matrices are singular.Proving properties about matrix $A$ s.t. $A^2 = -I$Perturbation theory for a symetric rank-one updateShow that $r$ is the rank of the $n$x$n$ matrix $Aiff A$ has a nonsingular $r$x$r$ submatrixEvery skew-symmetric matrix has even rankShow that conic C has rank 2If $A$ skew hermitian matrix of order $n$ and $n$ be an even then prove that $det A$ is a real numberIf $A in M_n(mathbb C)$ is skew symmetric, $adj(A)$ is symmetric or skew symmetric depending on $n$ being even or oddWhy must n be even? Skew Symmetric matricesProving numerically that the rank of a matrix has a certain value

2

$begingroup$

$$A = -A^T$$
I assume that $A$ is not singular.

So $$det{A} neq 0$$ Then $$ det(A) = det(-A^T) = det(-I_{n} A^T) = (-1)^ndet(A^T) = (-1)^ndet(A)$$

So I get that $n$ must be even.

But what about odd $n$? I know it has to be singular matrix. Hints?

matrices matrix-rank

share|cite|improve this question

edited yesterday

Mårten W

2,56241837

asked Dec 1 '15 at 16:45

tomtomtomtom

22318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you assume that $A$ is nonsingular? Many skew-symmetric matrices (including, e.g., the zero matrix) are singular.
  $endgroup$
  – Travis
  Dec 1 '15 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming that I have half problem solved :)
  $endgroup$
  – tomtom
  Dec 1 '15 at 17:49
  

  
  
  
  

add a comment | 

2

$begingroup$

$$A = -A^T$$
I assume that $A$ is not singular.

So $$det{A} neq 0$$ Then $$ det(A) = det(-A^T) = det(-I_{n} A^T) = (-1)^ndet(A^T) = (-1)^ndet(A)$$

So I get that $n$ must be even.

But what about odd $n$? I know it has to be singular matrix. Hints?

matrices matrix-rank

share|cite|improve this question

edited yesterday

Mårten W

2,56241837

asked Dec 1 '15 at 16:45

tomtomtomtom

22318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you assume that $A$ is nonsingular? Many skew-symmetric matrices (including, e.g., the zero matrix) are singular.
  $endgroup$
  – Travis
  Dec 1 '15 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming that I have half problem solved :)
  $endgroup$
  – tomtom
  Dec 1 '15 at 17:49
  

  
  
  
  

add a comment | 

$begingroup$

$$A = -A^T$$
I assume that $A$ is not singular.

So $$det{A} neq 0$$ Then $$ det(A) = det(-A^T) = det(-I_{n} A^T) = (-1)^ndet(A^T) = (-1)^ndet(A)$$

So I get that $n$ must be even.

But what about odd $n$? I know it has to be singular matrix. Hints?

matrices matrix-rank

share|cite|improve this question

edited yesterday

Mårten W

2,56241837

asked Dec 1 '15 at 16:45

tomtomtomtom

22318

$endgroup$

share|cite|improve this question

edited yesterday

Mårten W

2,56241837

asked Dec 1 '15 at 16:45

tomtomtomtom

22318

share|cite|improve this question

edited yesterday

Mårten W

2,56241837

asked Dec 1 '15 at 16:45

tomtomtomtom

22318

edited yesterday

Mårten W

2,56241837

edited yesterday

Mårten W

2,56241837

asked Dec 1 '15 at 16:45

tomtomtomtom

22318

asked Dec 1 '15 at 16:45

tomtomtomtom

22318

2 Answers
2

active

oldest

votes

2

$begingroup$

Note: This answer is essentially based on this one by Jason DeVito. I have merely added some details.

I assume $A$ is real matrix. Note that it's rank as a real matrix equals its rank when considered as a complex matrix.

So from now on we consider $A$ as a complex matrix.

It is proved here that all the eigenvalues of $A$ are purely imaginary. Also, we know that for a real matrix, complex eigenvalues come in conjugate pairs. (Since the coefficients of the characteristic polynomial are real).

Since skew-symmetric matrices are digonalizable over $mathbb{C}$, we get there is an even number of non-zero eigenvalues $pm y_1 i,pm y_2 i,...,pm y_k i$ different from zero. Since the rank of a matrix is invariant under similarity, we get that $rank(A)$ equals the rank of it's diagonal form, which is trivially $2k$.

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

answered Dec 1 '15 at 18:12

Asaf ShacharAsaf Shachar

5,75331141

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have not covered eigenvalues and eigenvectors. All I can use are determinants and simple matrix properties.
  $endgroup$
  – tomtom
  Dec 1 '15 at 22:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok... I will try to think of some alternative explanation. Which characterizations of ranks do you know?
  $endgroup$
  – Asaf Shachar
  Dec 1 '15 at 22:49
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have covered kernels and images. Additions of subspaces. Determinants. Laplace extension.
  $endgroup$
  – tomtom
  Dec 1 '15 at 23:05
  

  
  
  
  

add a comment | 

1

$begingroup$

Here's another proof, avoiding use of eigenvalues/eigenvectors:

Let $A$ be a skew-symmetric matrix, and consider the alternating bilinear form $$B(v,w) := langle Av, w rangle = -langle v, Aw rangle$$ where $langle cdot, cdot rangle$ is the standard inner product on $mathbb{R}^n$. Let $W := text{Im}(A)$ and let $P: mathbb{R}^n to W$ denote the orthogonal projection.

For any nonzero $w = Au in W$, note that $$B(Pu, w) = B(u,w) = langle w, w rangle > 0$$ where we have used that $B(u-Pu, w) = 0$ since $u-Pu in W^{perp}$. Thus, the restriction of $B$ to $W$ is nondegenerate. It follows that $text{rank}(A) = dim(W)$ is even, since the matrix of $Bvert_{W}$ is skew-symmetric with nonzero determinant (and any odd-dimensional skew-symmetric matrix has determinant zero).

share|cite|improve this answer

answered Sep 8 '18 at 16:37

Sameer KailasaSameer Kailasa

5,57321843

$endgroup$

add a comment | 

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2

$begingroup$

Note: This answer is essentially based on this one by Jason DeVito. I have merely added some details.

I assume $A$ is real matrix. Note that it's rank as a real matrix equals its rank when considered as a complex matrix.

So from now on we consider $A$ as a complex matrix.

It is proved here that all the eigenvalues of $A$ are purely imaginary. Also, we know that for a real matrix, complex eigenvalues come in conjugate pairs. (Since the coefficients of the characteristic polynomial are real).

Since skew-symmetric matrices are digonalizable over $mathbb{C}$, we get there is an even number of non-zero eigenvalues $pm y_1 i,pm y_2 i,...,pm y_k i$ different from zero. Since the rank of a matrix is invariant under similarity, we get that $rank(A)$ equals the rank of it's diagonal form, which is trivially $2k$.

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

answered Dec 1 '15 at 18:12

Asaf ShacharAsaf Shachar

5,75331141

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have not covered eigenvalues and eigenvectors. All I can use are determinants and simple matrix properties.
  $endgroup$
  – tomtom
  Dec 1 '15 at 22:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok... I will try to think of some alternative explanation. Which characterizations of ranks do you know?
  $endgroup$
  – Asaf Shachar
  Dec 1 '15 at 22:49
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have covered kernels and images. Additions of subspaces. Determinants. Laplace extension.
  $endgroup$
  – tomtom
  Dec 1 '15 at 23:05
  

  
  
  
  

add a comment | 

2

$begingroup$

Note: This answer is essentially based on this one by Jason DeVito. I have merely added some details.

I assume $A$ is real matrix. Note that it's rank as a real matrix equals its rank when considered as a complex matrix.

So from now on we consider $A$ as a complex matrix.

It is proved here that all the eigenvalues of $A$ are purely imaginary. Also, we know that for a real matrix, complex eigenvalues come in conjugate pairs. (Since the coefficients of the characteristic polynomial are real).

Since skew-symmetric matrices are digonalizable over $mathbb{C}$, we get there is an even number of non-zero eigenvalues $pm y_1 i,pm y_2 i,...,pm y_k i$ different from zero. Since the rank of a matrix is invariant under similarity, we get that $rank(A)$ equals the rank of it's diagonal form, which is trivially $2k$.

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

answered Dec 1 '15 at 18:12

Asaf ShacharAsaf Shachar

5,75331141

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have not covered eigenvalues and eigenvectors. All I can use are determinants and simple matrix properties.
  $endgroup$
  – tomtom
  Dec 1 '15 at 22:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok... I will try to think of some alternative explanation. Which characterizations of ranks do you know?
  $endgroup$
  – Asaf Shachar
  Dec 1 '15 at 22:49
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have covered kernels and images. Additions of subspaces. Determinants. Laplace extension.
  $endgroup$
  – tomtom
  Dec 1 '15 at 23:05
  

  
  
  
  

add a comment | 

$begingroup$

Note: This answer is essentially based on this one by Jason DeVito. I have merely added some details.

I assume $A$ is real matrix. Note that it's rank as a real matrix equals its rank when considered as a complex matrix.

So from now on we consider $A$ as a complex matrix.

It is proved here that all the eigenvalues of $A$ are purely imaginary. Also, we know that for a real matrix, complex eigenvalues come in conjugate pairs. (Since the coefficients of the characteristic polynomial are real).

Since skew-symmetric matrices are digonalizable over $mathbb{C}$, we get there is an even number of non-zero eigenvalues $pm y_1 i,pm y_2 i,...,pm y_k i$ different from zero. Since the rank of a matrix is invariant under similarity, we get that $rank(A)$ equals the rank of it's diagonal form, which is trivially $2k$.

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

answered Dec 1 '15 at 18:12

Asaf ShacharAsaf Shachar

5,75331141

$endgroup$

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

answered Dec 1 '15 at 18:12

Asaf ShacharAsaf Shachar

5,75331141

edited Apr 13 '17 at 12:21

Community♦

1

edited Apr 13 '17 at 12:21

Community♦

1

answered Dec 1 '15 at 18:12

Asaf ShacharAsaf Shachar

5,75331141

answered Dec 1 '15 at 18:12

Asaf ShacharAsaf Shachar

5,75331141

1

$begingroup$

Here's another proof, avoiding use of eigenvalues/eigenvectors:

Let $A$ be a skew-symmetric matrix, and consider the alternating bilinear form $$B(v,w) := langle Av, w rangle = -langle v, Aw rangle$$ where $langle cdot, cdot rangle$ is the standard inner product on $mathbb{R}^n$. Let $W := text{Im}(A)$ and let $P: mathbb{R}^n to W$ denote the orthogonal projection.

For any nonzero $w = Au in W$, note that $$B(Pu, w) = B(u,w) = langle w, w rangle > 0$$ where we have used that $B(u-Pu, w) = 0$ since $u-Pu in W^{perp}$. Thus, the restriction of $B$ to $W$ is nondegenerate. It follows that $text{rank}(A) = dim(W)$ is even, since the matrix of $Bvert_{W}$ is skew-symmetric with nonzero determinant (and any odd-dimensional skew-symmetric matrix has determinant zero).

share|cite|improve this answer

answered Sep 8 '18 at 16:37

Sameer KailasaSameer Kailasa

5,57321843

$endgroup$

add a comment | 

1

$begingroup$

Here's another proof, avoiding use of eigenvalues/eigenvectors:

Let $A$ be a skew-symmetric matrix, and consider the alternating bilinear form $$B(v,w) := langle Av, w rangle = -langle v, Aw rangle$$ where $langle cdot, cdot rangle$ is the standard inner product on $mathbb{R}^n$. Let $W := text{Im}(A)$ and let $P: mathbb{R}^n to W$ denote the orthogonal projection.

For any nonzero $w = Au in W$, note that $$B(Pu, w) = B(u,w) = langle w, w rangle > 0$$ where we have used that $B(u-Pu, w) = 0$ since $u-Pu in W^{perp}$. Thus, the restriction of $B$ to $W$ is nondegenerate. It follows that $text{rank}(A) = dim(W)$ is even, since the matrix of $Bvert_{W}$ is skew-symmetric with nonzero determinant (and any odd-dimensional skew-symmetric matrix has determinant zero).

share|cite|improve this answer

answered Sep 8 '18 at 16:37

Sameer KailasaSameer Kailasa

5,57321843

$endgroup$

add a comment | 

$begingroup$

Here's another proof, avoiding use of eigenvalues/eigenvectors:

Let $A$ be a skew-symmetric matrix, and consider the alternating bilinear form $$B(v,w) := langle Av, w rangle = -langle v, Aw rangle$$ where $langle cdot, cdot rangle$ is the standard inner product on $mathbb{R}^n$. Let $W := text{Im}(A)$ and let $P: mathbb{R}^n to W$ denote the orthogonal projection.

For any nonzero $w = Au in W$, note that $$B(Pu, w) = B(u,w) = langle w, w rangle > 0$$ where we have used that $B(u-Pu, w) = 0$ since $u-Pu in W^{perp}$. Thus, the restriction of $B$ to $W$ is nondegenerate. It follows that $text{rank}(A) = dim(W)$ is even, since the matrix of $Bvert_{W}$ is skew-symmetric with nonzero determinant (and any odd-dimensional skew-symmetric matrix has determinant zero).

share|cite|improve this answer

answered Sep 8 '18 at 16:37

Sameer KailasaSameer Kailasa

5,57321843

$endgroup$

share|cite|improve this answer

answered Sep 8 '18 at 16:37

Sameer KailasaSameer Kailasa

5,57321843

answered Sep 8 '18 at 16:37

Sameer KailasaSameer Kailasa

5,57321843

answered Sep 8 '18 at 16:37

Sameer KailasaSameer Kailasa

5,57321843