What does a 0 eigenvalue mean?

A zero eigenvalue means the matrix in question is singular. The eigenvectors corresponding to the zero eigenvalues form the basis for the null space of the matrix.

How do you find the eigenvectors of a 4×4 matrix?

From the last equation (−2×4=0), it is clear that x4=0. Substitute into the third equation to get x3=0 and in the second to get x2=0. From the first equation, x1 remains as a free variable so vectors of the form (x1,0,0,0) are eigenvectors associated with the eigenvalue 5; pick e.g. (1,0,0,0).

How do you find the minimal polynomial of a 4×4 matrix?

If d = 1, then mipoly = x + b, therefore, M + b*Id = 0 but this is not true. If d = 2, then mipoly = xˆ2 + ax + b, therefore, Mˆ2 + aM + bId = 0 after doing some calculations, I came to the conclusion that the system has no non-zero solutions since the rank of the matrix is 3.

What if all eigenvalues are zero?

As we know the determinant of a matrix is equal to the products of all eigenvalues. So, if one or more eigenvalues are zero then the determinant is zero and that is a singular matrix. If all eigenvalues are zero then that is a Nilpotent Matrix. And for any such matrix A: A^k = 0 for some specific k.

How do you know if an eigenvalue is 0?

Vectors with eigenvalue 0 make up the nullspace of A; if A is singular, then A = 0 is an eigenvalue of A. Suppose P is the matrix of a projection onto a plane. For any x in the plane Px = x, so x is an eigenvector with eigenvalue 1.

How do you find the minimal polynomial?

The minimal polynomial is always well-defined and we have deg µA(X) ≤ n2. If we now replace A in this equation by the undeterminate X, we obtain a monic polynomial p(X) satisfying p(A) = 0 and the degree d of p is minimal by construction, hence p(X) = µA(X) by definition.

How do you find the minimal polynomial of a square matrix?

Let T be a square matrix. If the linear equation aT +bI = 0 has a solution a, b with a nonzero, then x + b/a is the minimal polynomial. Any solution with a = 0 must necessarily also have b=0 as well.

How do you know if a 4×4 matrix is diagonalizable?

A matrix is diagonalizable if and only if for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue. Meaning, if you find matrices with distinct eigenvalues (multiplicity = 1) you should quickly identify those as diagonizable.