Sorry, our question is not clear enough to me. An example will surely be helpful.

If you want to develop different matrices with same eigenvalues, eigenvectors can help.

Ok, let's consider we have matrix A with dimension 4X4 wich has 4 eigenvalues, those 4 eigenvalues are located on right side of y-axis. What I want to do, is mirroring those eigenvalues to left side of y-axis by multiplying A matrix. How I can do it? 

Please see "Gershgorins circle theorem."

Dear Viera Ceranova you added a good example yesterday, I just checked it quickly. It was very good, could you send it again please?

Dear Ali Rabiei,

Thank you for your interesting question. I am now correcting my correction :)

Yesterday I deleted my solution, because I didn't want to mislead people. The example was good. However, with another real matrix I obtained inconvenient eigenvalues. (Apparently, there were a mistake.)

Today I did a general algebraic computation using wxMaxima. The characteristic polynomials are

• charpoly(A) = x4 - ax3 + bx2 - cx + d
• charpoly(B) = x4 + ax3 + bx2 + cx + d

Thus it seems that the principle I proposed holds for any real matrix A4x4. 

Proposition:

Consider an nxn matrix A=(aij) with real entries. Construct the matrix B=(bij) such that 

• bij = -aij if j=i+2k for an integer k
• bij = aij otherwise

Then (at least for n=2,3,4) it holds:

If  x+iy is an eigenvalue of A, then -x+iy is an eigenvalue of B.

The proof is easy, though tedious for n>3. 

I guess the question is :

How to mirroring the eigenvalues of matrix (y-axis)? 

Consider the matrix B= - A-bar, i.e., b-sub-ij= - complex conjugate of a-sub-ij for each i,j. Then the set of eigenvalues of B is the reflection in y-axis of the set of eigenvalues of A. In particular, if A has purely imaginary entries, then the set S of eigenvalues of A and  reflection of S  in y-axis are identical.

Prof. Singh is right.

Please see the following fact:

Proposition. If λ is an eigenvalue of A, then αλ is an eigenvalue of αA where α is any arbitrary scalar.

Proof. If λ is an eigenvalue of A, then there must exist an eigenvector x such that Ax =λx. Multiplying both sides of this equation by α we obtain  (αA)x=(αλ)x.                  ■

For α = -1, the eigenvalue of -A  is -λ.

@professor Wanicharpichat:

Multiplying by -1 is not always appropriate. 

The problem is, that if some entries of A are complex, then it can occur:

x+iy is a root of the characteristic polynomial, while x-iy is not.

Then the opposite -x-iy of the root does not have symmetrical position (to a root) with respect to y-axis. 

@Prof. Cernanova:

Yes, your comment is correct.

If x is a eigenvector corresponding to a complex eigenvalue a + bi, b ≠ 0 of real matrix A. Then conjugate(x) is an eigenvector corresponding to the eigenvalue a - bi.

Ok thankyou all a lot of your answering.

What if we have one eigenvalue on right side and another on left side, my goal is to keep all eigenvalues on left side,can I handle A vector so I can mirror just the eigenvalue on right side to left side?

Each eigenvalue of a positive definite matrix is a positive real number. 

If A is a positive definite matrix then (from our previous Proposition) all eigenvalues of -A are negative real numbers.

In view of my accepted simple answer to your original question, for your new variant of your question, I think  that  literature on Stable polynomials may be helpful.