2

Their multipliciti For the matrix, list the real Eigen values, repeated according4 -2 0The real eigenvalues are_...

Question

Their multipliciti For the matrix, list the real Eigen values, repeated according4 -2 0The real eigenvalues are_

their multipliciti For the matrix, list the real Eigen values, repeated according 4 -2 0 The real eigenvalues are_



Answers

Deal with the eigenvalue/eigenvector problem for $n \times n$ real skew-symmetric matrices. Determine all eigenvalues and corresponding eigenvectors of the matrix $$ A=\left[\begin{array}{rrr} 0 & 4 & -4 \\ -4 & 0 & -2 \\ 4 & 2 & 0 \end{array}\right] $$

This problem gives us a matrix and asks us to sell for its guiding values and Eigen vectors. We do this first by finding the characters to polynomial, which is found by taking the determinant of the matrix a minus lander times the identity matrix. This is going to give us the determinant of the matrix with negative land all along of diagonal and then to everywhere else. Something for this will give us opponent. You don't Negative land. A cute plus 16 plus 12 Landau, which is equivalent to negative Lambda months to Lambda minus four. Lead two plus two. We're gonna set that equal to zero. That gives us our Eigen values to equal for and negative to where Negative two has an algebraic multiplicity of to So in order, solve for the Eigen vectors, we have to take a minus land of times. I times I director ex national equal the zero factor. So first, with the four eyed in space, we plug this in, and that will give us the Matrix Negative. 4 to 2 to negative for two to to negative four times x one. And then I'm gonna divide um, every road by two expected performed Yashin elimination and just make this solving easiest. We get negative. 211 one night of 2111 negative too. Times X one equals the director. That means that negative to a one plus B one plus C one equals zero. I want minus to B one plus C 10 and a one plus B one minus two. C one. It was zero when we sold for the system of equations. We will find our wagon vector X one, 111 Then we do the same for Lambda equal to negative two. Doing this will give us the Matrix, uh, with twos in every row. So there we go times the director X two, and then we can perform gashing elimination. Um, by subtracting the second subtracting the first row from the second and third rose and also divided by to give us 111 and then zeros everywhere else. And that should equal the zero factor. In doing so, we can get to major cities that span this, um, argon space and those air going to equal negative 101 and night of 110 as they're too linearly. Independent solutions. And that's our final answer


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