# Need Help: New kind of Matrix formula?

Question asked by ErAnushka in #Coffee Room on Sep 3, 2012

ErAnushka Β· Sep 3, 2012

Rank C3 - EXPERT

In my book, I've:

Okay, as far as I know, symmetric matrix remains the same even if we transpose it.

And the skew symmetric matrix is the same as it's -1 x the transpose of itself.

How do I suppose to solve this?

here [] =

| 2 -4 9 |

| 14 7 13 |

| 3 5 11 |

but in book, they used these formula,

B = 1/2 (A+A')

C = 1/2 (A-A')

**Problem: Express the matrix [] as a sum of symmetric and skew symmetric matrices.**Okay, as far as I know, symmetric matrix remains the same even if we transpose it.

And the skew symmetric matrix is the same as it's -1 x the transpose of itself.

How do I suppose to solve this?

here [] =

| 2 -4 9 |

| 14 7 13 |

| 3 5 11 |

*I think, first I should write the transpose of matrix which is the symmetric matrix and then write the transpose again multiplied by -1.**After that I can add both matrix and prove that it is equal to the [] matrix.*but in book, they used these formula,

B = 1/2 (A+A')

C = 1/2 (A-A')

**I'm not sure what is this... and where it came from**😨**Anyone can explain me? My method is right or the method used in the book?**☕ Posted in: #Coffee RoomErAnushka Β· Sep 3, 2012

Rank C3 - EXPERT

No problem I got π

A direct proof is as follows:

Let X be the given square matrix

Let X=A+B where A is symmetric and B is skew symmetric

That is A'=A and B' = - B .......(1)

Therefore,

X'=A'+B' ( using the property of transpose of a matrix)

or, X'=A - B from (1) above

Now, as X is square , therefore, X+X' and X-X' are defined.

Therefore X+X'=A+B +A-B= 2A

and X-X'=A+B- (A-B) = 2B

Therefrore A=(X+X')/2

and B= (X-X')/2

A direct proof is as follows:

Let X be the given square matrix

Let X=A+B where A is symmetric and B is skew symmetric

That is A'=A and B' = - B .......(1)

Therefore,

X'=A'+B' ( using the property of transpose of a matrix)

or, X'=A - B from (1) above

Now, as X is square , therefore, X+X' and X-X' are defined.

Therefore X+X'=A+B +A-B= 2A

and X-X'=A+B- (A-B) = 2B

Therefrore A=(X+X')/2

and B= (X-X')/2

**It is easy to verify that A'=A and B'= - B****That is A is symmetric and B is skew symmetric****Therefore , X can be uniquely expressed as sum of a symmetric matrix and a skew symmetric matrix, which is****X =(X+X')/2 + (X-X')/2****I understood the green part π , but the red part is little confusingπ .. A'=A and B' = -B but how? we got only A=(X+X')/2 and B= (X-X')/2.... ??? π¨**silverscorpion Β· Sep 4, 2012

Rank A3 - PRO

I'm not sure what you are not getting here.. A = A' and B = -B' are assumptions that you made in the beginning, right? (ie., you have assumed that A is symmetric and B is skew symmetric)

You can probably take the transpose of (X+X')/2 and (X-X')/2, and check that they are indeed the same as A and -B.. but, here it's actually assumed that it is true.

You can probably take the transpose of (X+X')/2 and (X-X')/2, and check that they are indeed the same as A and -B.. but, here it's actually assumed that it is true.

ErAnushka Β· Sep 4, 2012

Rank C3 - EXPERT

Omg wait... I got it πsilverscorpionI'm not sure what you are not getting here.. A = A' and B = -B' are assumptions that you made in the beginning, right? (ie., you have assumed that A is symmetric and B is skew symmetric)

You can probably take the transpose of (X+X')/2 and (X-X')/2, and check that they are indeed the same as A and -B.. but, here it's actually assumed that it is true.

We are actually proving that any square matrix is a sum of symmetric matrix and skew symmetric matrix, so here, A is symmetric matrix and B is skew symmetric matrix.

We got both value, so,

A+B = (X+X')/2 + (X-X')/2

A+B = (X+X' + X-X') /2

A+B = 2X/2

A+B = X

Here X is a square matrix! OMG! I'm so stupid π

By the by thanks for the reply, I got the idea at the moment of reading this again π