Inverse of a matrix is same as conjugate transpose of matrix [closed]

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

I have a matrix of the form
$$
X=beginpmatrixA&B\C&Dendpmatrix
$$
What are the conditions on $A,B,C$ and $D$ such that $X^-1$ = $X^*$ i.e. inverse of a matrix is equal to conjugate transpose? $A,B,C$ and $D$ are square matrices of same size.

linear-algebra matrices matrix-equations matrix-calculus

share|cite|improve this question

asked Jul 16 at 15:32

Dushyant Sahoo

558

closed as off-topic by Alex Francisco, Strants, José Carlos Santos, Xander Henderson, Adrian Keister Jul 20 at 1:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Strants, José Carlos Santos, Xander Henderson, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
1
down vote

favorite

I have a matrix of the form
$$
X=beginpmatrixA&B\C&Dendpmatrix
$$
What are the conditions on $A,B,C$ and $D$ such that $X^-1$ = $X^*$ i.e. inverse of a matrix is equal to conjugate transpose? $A,B,C$ and $D$ are square matrices of same size.

linear-algebra matrices matrix-equations matrix-calculus

share|cite|improve this question

asked Jul 16 at 15:32

Dushyant Sahoo

558

closed as off-topic by Alex Francisco, Strants, José Carlos Santos, Xander Henderson, Adrian Keister Jul 20 at 1:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Strants, José Carlos Santos, Xander Henderson, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I have a matrix of the form
$$
X=beginpmatrixA&B\C&Dendpmatrix
$$
What are the conditions on $A,B,C$ and $D$ such that $X^-1$ = $X^*$ i.e. inverse of a matrix is equal to conjugate transpose? $A,B,C$ and $D$ are square matrices of same size.

linear-algebra matrices matrix-equations matrix-calculus

share|cite|improve this question

asked Jul 16 at 15:32

Dushyant Sahoo

558

I have a matrix of the form
$$
X=beginpmatrixA&B\C&Dendpmatrix
$$
What are the conditions on $A,B,C$ and $D$ such that $X^-1$ = $X^*$ i.e. inverse of a matrix is equal to conjugate transpose? $A,B,C$ and $D$ are square matrices of same size.

linear-algebra matrices matrix-equations matrix-calculus

share|cite|improve this question

asked Jul 16 at 15:32

Dushyant Sahoo

558

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 15:32

Dushyant Sahoo

558

asked Jul 16 at 15:32

Dushyant Sahoo

558

asked Jul 16 at 15:32

Dushyant Sahoo

558

558

closed as off-topic by Alex Francisco, Strants, José Carlos Santos, Xander Henderson, Adrian Keister Jul 20 at 1:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Strants, José Carlos Santos, Xander Henderson, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Alex Francisco, Strants, José Carlos Santos, Xander Henderson, Adrian Keister Jul 20 at 1:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Strants, José Carlos Santos, Xander Henderson, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

The best way I know to describe the required conditions on the four block matrices (that is, in their most general form), is to consider the four SVDs
$$beginarrayl
underline overline bfA = underline overline bfU _Aunderline overline bfLambda _Aunderline overline bfV _A^ + quad - underline overline bfB = underline overline bfU _Bunderline overline bfLambda _Bunderline overline bfV _B^ + \
underline overline bfC = underline overline bfU _Cunderline overline bfLambda _Cunderline overline bfV _C^ + quad underline overline bfD = underline overline bfU _Dunderline overline bfLambda _Dunderline overline bfV _D^ +
endarray$$
Then for $underline overline bfX $ to be unitary, it is required that
$$beginarrayl
underline overline bfLambda _D = underline overline bfLambda _Aquad quad underline overline bfLambda _B = underline overline bfLambda _C = left( underline overline bfI - underline overline bfLambda _A^2 right)^frac12\
underline overline bfU _B = underline overline bfU _Aquad quad underline overline bfU _C = underline overline bfU _D\
underline overline bfV _C = underline overline bfV _Aquad quad underline overline bfV _B = underline overline bfV _D
endarray$$
See here for formal proof of this result, which is not trivial. Note that I have taken a few liberties here with the specific meaning of “SVD” – for example, this requires the two off-diagonal singular value matrices to be in increasing order rather than standard decreasing order, and there are several normalizations required to insure all the singular vector matrices are actually precisely identical (rather than different columns being multiplied by various arbitrary choices of phase) – so you won’t necessarily get this immediately if you just take an example of a unitary matrix and use a standard SVD routine on the four block matrices. Note also that the requirements on the singular values imply that they must all be between zero and one.

share|cite|improve this answer

edited Jul 16 at 17:34

answered Jul 16 at 16:50

John Polcari

382111

add a comment | 

up vote
1
down vote

If $X^-1 = X^ast$, then $XX^ast = I$

$$
XX^ast = beginbmatrixA & B \ C & DendbmatrixbeginbmatrixA^ast & C^ast\ B^ast & D^astendbmatrix = beginbmatrix AA^ast + BB^ast & AC^ast + BD^ast \ CA^ast + DB^ast & CC^ast + DD^ast endbmatrix = I
$$

If this is so, we require the diagonal blocks to be $I$ and off-diagonals to be $0$. One way to achieve this is to let $B$ and $C$ be zero blocks, with $A$ and $D$ unitary.

share|cite|improve this answer

answered Jul 16 at 15:52

n casale

134

• 
  
  
  

  
  

  
  
  
  
  
  

  
  However, we also require X*X=I, which leads to another set of conditions.
  – John Polcari
  Jul 16 at 15:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JohnPolcari If $X$ is square then $X X^* = I$ implies that $X^* X = I$.
  – littleO
  Jul 16 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  after multiplying $X^ast X$, the above conditions will also satisfy $X^ast X = XX^ast = I$
  – n casale
  Jul 16 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lttleO agreed, but it will turn out that the conditions on the the block matrices will be different. For example, it is required that (A*)A+(C*)C=I.
  – John Polcari
  Jul 16 at 16:06
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The best way I know to describe the required conditions on the four block matrices (that is, in their most general form), is to consider the four SVDs
$$beginarrayl
underline overline bfA = underline overline bfU _Aunderline overline bfLambda _Aunderline overline bfV _A^ + quad - underline overline bfB = underline overline bfU _Bunderline overline bfLambda _Bunderline overline bfV _B^ + \
underline overline bfC = underline overline bfU _Cunderline overline bfLambda _Cunderline overline bfV _C^ + quad underline overline bfD = underline overline bfU _Dunderline overline bfLambda _Dunderline overline bfV _D^ +
endarray$$
Then for $underline overline bfX $ to be unitary, it is required that
$$beginarrayl
underline overline bfLambda _D = underline overline bfLambda _Aquad quad underline overline bfLambda _B = underline overline bfLambda _C = left( underline overline bfI - underline overline bfLambda _A^2 right)^frac12\
underline overline bfU _B = underline overline bfU _Aquad quad underline overline bfU _C = underline overline bfU _D\
underline overline bfV _C = underline overline bfV _Aquad quad underline overline bfV _B = underline overline bfV _D
endarray$$
See here for formal proof of this result, which is not trivial. Note that I have taken a few liberties here with the specific meaning of “SVD” – for example, this requires the two off-diagonal singular value matrices to be in increasing order rather than standard decreasing order, and there are several normalizations required to insure all the singular vector matrices are actually precisely identical (rather than different columns being multiplied by various arbitrary choices of phase) – so you won’t necessarily get this immediately if you just take an example of a unitary matrix and use a standard SVD routine on the four block matrices. Note also that the requirements on the singular values imply that they must all be between zero and one.

share|cite|improve this answer

edited Jul 16 at 17:34

answered Jul 16 at 16:50

John Polcari

382111

add a comment | 

up vote
1
down vote



accepted

The best way I know to describe the required conditions on the four block matrices (that is, in their most general form), is to consider the four SVDs
$$beginarrayl
underline overline bfA = underline overline bfU _Aunderline overline bfLambda _Aunderline overline bfV _A^ + quad - underline overline bfB = underline overline bfU _Bunderline overline bfLambda _Bunderline overline bfV _B^ + \
underline overline bfC = underline overline bfU _Cunderline overline bfLambda _Cunderline overline bfV _C^ + quad underline overline bfD = underline overline bfU _Dunderline overline bfLambda _Dunderline overline bfV _D^ +
endarray$$
Then for $underline overline bfX $ to be unitary, it is required that
$$beginarrayl
underline overline bfLambda _D = underline overline bfLambda _Aquad quad underline overline bfLambda _B = underline overline bfLambda _C = left( underline overline bfI - underline overline bfLambda _A^2 right)^frac12\
underline overline bfU _B = underline overline bfU _Aquad quad underline overline bfU _C = underline overline bfU _D\
underline overline bfV _C = underline overline bfV _Aquad quad underline overline bfV _B = underline overline bfV _D
endarray$$
See here for formal proof of this result, which is not trivial. Note that I have taken a few liberties here with the specific meaning of “SVD” – for example, this requires the two off-diagonal singular value matrices to be in increasing order rather than standard decreasing order, and there are several normalizations required to insure all the singular vector matrices are actually precisely identical (rather than different columns being multiplied by various arbitrary choices of phase) – so you won’t necessarily get this immediately if you just take an example of a unitary matrix and use a standard SVD routine on the four block matrices. Note also that the requirements on the singular values imply that they must all be between zero and one.

share|cite|improve this answer

edited Jul 16 at 17:34

answered Jul 16 at 16:50

John Polcari

382111

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The best way I know to describe the required conditions on the four block matrices (that is, in their most general form), is to consider the four SVDs
$$beginarrayl
underline overline bfA = underline overline bfU _Aunderline overline bfLambda _Aunderline overline bfV _A^ + quad - underline overline bfB = underline overline bfU _Bunderline overline bfLambda _Bunderline overline bfV _B^ + \
underline overline bfC = underline overline bfU _Cunderline overline bfLambda _Cunderline overline bfV _C^ + quad underline overline bfD = underline overline bfU _Dunderline overline bfLambda _Dunderline overline bfV _D^ +
endarray$$
Then for $underline overline bfX $ to be unitary, it is required that
$$beginarrayl
underline overline bfLambda _D = underline overline bfLambda _Aquad quad underline overline bfLambda _B = underline overline bfLambda _C = left( underline overline bfI - underline overline bfLambda _A^2 right)^frac12\
underline overline bfU _B = underline overline bfU _Aquad quad underline overline bfU _C = underline overline bfU _D\
underline overline bfV _C = underline overline bfV _Aquad quad underline overline bfV _B = underline overline bfV _D
endarray$$
See here for formal proof of this result, which is not trivial. Note that I have taken a few liberties here with the specific meaning of “SVD” – for example, this requires the two off-diagonal singular value matrices to be in increasing order rather than standard decreasing order, and there are several normalizations required to insure all the singular vector matrices are actually precisely identical (rather than different columns being multiplied by various arbitrary choices of phase) – so you won’t necessarily get this immediately if you just take an example of a unitary matrix and use a standard SVD routine on the four block matrices. Note also that the requirements on the singular values imply that they must all be between zero and one.

share|cite|improve this answer

edited Jul 16 at 17:34

answered Jul 16 at 16:50

John Polcari

382111

The best way I know to describe the required conditions on the four block matrices (that is, in their most general form), is to consider the four SVDs
$$beginarrayl
underline overline bfA = underline overline bfU _Aunderline overline bfLambda _Aunderline overline bfV _A^ + quad - underline overline bfB = underline overline bfU _Bunderline overline bfLambda _Bunderline overline bfV _B^ + \
underline overline bfC = underline overline bfU _Cunderline overline bfLambda _Cunderline overline bfV _C^ + quad underline overline bfD = underline overline bfU _Dunderline overline bfLambda _Dunderline overline bfV _D^ +
endarray$$
Then for $underline overline bfX $ to be unitary, it is required that
$$beginarrayl
underline overline bfLambda _D = underline overline bfLambda _Aquad quad underline overline bfLambda _B = underline overline bfLambda _C = left( underline overline bfI - underline overline bfLambda _A^2 right)^frac12\
underline overline bfU _B = underline overline bfU _Aquad quad underline overline bfU _C = underline overline bfU _D\
underline overline bfV _C = underline overline bfV _Aquad quad underline overline bfV _B = underline overline bfV _D
endarray$$
See here for formal proof of this result, which is not trivial. Note that I have taken a few liberties here with the specific meaning of “SVD” – for example, this requires the two off-diagonal singular value matrices to be in increasing order rather than standard decreasing order, and there are several normalizations required to insure all the singular vector matrices are actually precisely identical (rather than different columns being multiplied by various arbitrary choices of phase) – so you won’t necessarily get this immediately if you just take an example of a unitary matrix and use a standard SVD routine on the four block matrices. Note also that the requirements on the singular values imply that they must all be between zero and one.

share|cite|improve this answer

edited Jul 16 at 17:34

answered Jul 16 at 16:50

John Polcari

382111

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 17:34

edited Jul 16 at 17:34

edited Jul 16 at 17:34

answered Jul 16 at 16:50

John Polcari

382111

answered Jul 16 at 16:50

John Polcari

382111

answered Jul 16 at 16:50

John Polcari

382111

382111

add a comment | 

add a comment | 

up vote
1
down vote

If $X^-1 = X^ast$, then $XX^ast = I$

$$
XX^ast = beginbmatrixA & B \ C & DendbmatrixbeginbmatrixA^ast & C^ast\ B^ast & D^astendbmatrix = beginbmatrix AA^ast + BB^ast & AC^ast + BD^ast \ CA^ast + DB^ast & CC^ast + DD^ast endbmatrix = I
$$

If this is so, we require the diagonal blocks to be $I$ and off-diagonals to be $0$. One way to achieve this is to let $B$ and $C$ be zero blocks, with $A$ and $D$ unitary.

share|cite|improve this answer

answered Jul 16 at 15:52

n casale

134

• 
  
  
  

  
  

  
  
  
  
  
  

  
  However, we also require X*X=I, which leads to another set of conditions.
  – John Polcari
  Jul 16 at 15:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JohnPolcari If $X$ is square then $X X^* = I$ implies that $X^* X = I$.
  – littleO
  Jul 16 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  after multiplying $X^ast X$, the above conditions will also satisfy $X^ast X = XX^ast = I$
  – n casale
  Jul 16 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lttleO agreed, but it will turn out that the conditions on the the block matrices will be different. For example, it is required that (A*)A+(C*)C=I.
  – John Polcari
  Jul 16 at 16:06
  

  
  
  
  

add a comment | 

up vote
1
down vote

If $X^-1 = X^ast$, then $XX^ast = I$

$$
XX^ast = beginbmatrixA & B \ C & DendbmatrixbeginbmatrixA^ast & C^ast\ B^ast & D^astendbmatrix = beginbmatrix AA^ast + BB^ast & AC^ast + BD^ast \ CA^ast + DB^ast & CC^ast + DD^ast endbmatrix = I
$$

If this is so, we require the diagonal blocks to be $I$ and off-diagonals to be $0$. One way to achieve this is to let $B$ and $C$ be zero blocks, with $A$ and $D$ unitary.

share|cite|improve this answer

answered Jul 16 at 15:52

n casale

134

• 
  
  
  

  
  

  
  
  
  
  
  

  
  However, we also require X*X=I, which leads to another set of conditions.
  – John Polcari
  Jul 16 at 15:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JohnPolcari If $X$ is square then $X X^* = I$ implies that $X^* X = I$.
  – littleO
  Jul 16 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  after multiplying $X^ast X$, the above conditions will also satisfy $X^ast X = XX^ast = I$
  – n casale
  Jul 16 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lttleO agreed, but it will turn out that the conditions on the the block matrices will be different. For example, it is required that (A*)A+(C*)C=I.
  – John Polcari
  Jul 16 at 16:06
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

If $X^-1 = X^ast$, then $XX^ast = I$

$$
XX^ast = beginbmatrixA & B \ C & DendbmatrixbeginbmatrixA^ast & C^ast\ B^ast & D^astendbmatrix = beginbmatrix AA^ast + BB^ast & AC^ast + BD^ast \ CA^ast + DB^ast & CC^ast + DD^ast endbmatrix = I
$$

If this is so, we require the diagonal blocks to be $I$ and off-diagonals to be $0$. One way to achieve this is to let $B$ and $C$ be zero blocks, with $A$ and $D$ unitary.

share|cite|improve this answer

answered Jul 16 at 15:52

n casale

134

If $X^-1 = X^ast$, then $XX^ast = I$

$$
XX^ast = beginbmatrixA & B \ C & DendbmatrixbeginbmatrixA^ast & C^ast\ B^ast & D^astendbmatrix = beginbmatrix AA^ast + BB^ast & AC^ast + BD^ast \ CA^ast + DB^ast & CC^ast + DD^ast endbmatrix = I
$$

If this is so, we require the diagonal blocks to be $I$ and off-diagonals to be $0$. One way to achieve this is to let $B$ and $C$ be zero blocks, with $A$ and $D$ unitary.

share|cite|improve this answer

answered Jul 16 at 15:52

n casale

134

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 15:52

n casale

134

answered Jul 16 at 15:52

n casale

134

answered Jul 16 at 15:52

n casale

134

134

• 
  
  
  

  
  

  
  
  
  
  
  

  
  However, we also require X*X=I, which leads to another set of conditions.
  – John Polcari
  Jul 16 at 15:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JohnPolcari If $X$ is square then $X X^* = I$ implies that $X^* X = I$.
  – littleO
  Jul 16 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  after multiplying $X^ast X$, the above conditions will also satisfy $X^ast X = XX^ast = I$
  – n casale
  Jul 16 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lttleO agreed, but it will turn out that the conditions on the the block matrices will be different. For example, it is required that (A*)A+(C*)C=I.
  – John Polcari
  Jul 16 at 16:06
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  However, we also require X*X=I, which leads to another set of conditions.
  – John Polcari
  Jul 16 at 15:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JohnPolcari If $X$ is square then $X X^* = I$ implies that $X^* X = I$.
  – littleO
  Jul 16 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  after multiplying $X^ast X$, the above conditions will also satisfy $X^ast X = XX^ast = I$
  – n casale
  Jul 16 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lttleO agreed, but it will turn out that the conditions on the the block matrices will be different. For example, it is required that (A*)A+(C*)C=I.
  – John Polcari
  Jul 16 at 16:06
  

  
  
  
  

However, we also require X*X=I, which leads to another set of conditions.
– John Polcari
Jul 16 at 15:56

However, we also require X*X=I, which leads to another set of conditions.
– John Polcari
Jul 16 at 15:56

@JohnPolcari If $X$ is square then $X X^* = I$ implies that $X^* X = I$.
– littleO
Jul 16 at 16:00

@JohnPolcari If $X$ is square then $X X^* = I$ implies that $X^* X = I$.
– littleO
Jul 16 at 16:00

after multiplying $X^ast X$, the above conditions will also satisfy $X^ast X = XX^ast = I$
– n casale
Jul 16 at 16:01

after multiplying $X^ast X$, the above conditions will also satisfy $X^ast X = XX^ast = I$
– n casale
Jul 16 at 16:01

@lttleO agreed, but it will turn out that the conditions on the the block matrices will be different. For example, it is required that (A*)A+(C*)C=I.
– John Polcari
Jul 16 at 16:06

@lttleO agreed, but it will turn out that the conditions on the the block matrices will be different. For example, it is required that (A*)A+(C*)C=I.
– John Polcari
Jul 16 at 16:06

add a comment |