Let A be a 4¡Á4 matrix with real entries that has all 1's on the main diagonal (i.e. a11 = a22 = a33 = a44 = 1). If A is singular and ¦Ë1 = 3 + 2i is an eigenvalue of A, then what, if anything, is it possible to conclude about the values of the remaining eigenvalues ¦Ë2, ¦Ë3, and ¦Ë4 ?

Answer: ¦Ë1 = 3 + 2i, ¦Ë2 = 3 ¨C 2i , ¦Ë3 = 0, ¦Ë4 = -2.

Could someone explain why that is the answer? I understand that given ¦Ë1, ¦Ë2 must be an eigenvalue, but how should I have known that 0 and -2 are eigenvalues?

Thanks.

To find the eigenvalues of a matrix, we need to solve the characteristic equation. The characteristic equation is obtained by setting the determinant of the matrix minus ¦ËI (where ¦Ë is an eigenvalue and I is the identity matrix) equal to zero.

In this case, let's consider the matrix A with eigenvalue ¦Ë1 = 3 + 2i. The matrix A is a 4x4 matrix with 1's on the main diagonal. So, the matrix A - ¦ËI can be written as:

A - ¦ËI = $\begin{bmatrix} 1-3-2i & 0 & 0 & 0 \\ 0 & 1-3-2i & 0 & 0 \\ 0 & 0 & 1-3-2i & 0 \\ 0 & 0 & 0 & 1-3-2i \end{bmatrix} = \begin{bmatrix} -2-2i & 0 & 0 & 0 \\ 0 & -2-2i & 0 & 0 \\ 0 & 0 & -2-2i & 0 \\ 0 & 0 & 0 & -2-2i \end{bmatrix}$

Now, we need to find the determinant of A - ¦ËI and set it equal to zero:

det(A - ¦ËI) = (-2-2i)(-2-2i)(-2-2i)(-2-2i) = 0

Expanding this equation gives us:

(-2-2i)^4 = 0

Simplifying further:

(-2-2i)^2 = 0

(4+8i-4) = 0

8i = 0

Since 8i = 0 implies that i = 0, we can conclude that 0 is an eigenvalue.

Similarly, for the eigenvalue ¦Ë1 = 3 + 2i, we have an eigenvalue ¦Ë2 = 3 - 2i. This is because complex eigenvalues always come in conjugate pairs.

Finally, since the matrix A is singular, the sum of its eigenvalues must be zero. We already have the eigenvalues ¦Ë1 = 3 + 2i, ¦Ë2 = 3 - 2i, and ¦Ë3 = 0. Therefore, the remaining eigenvalue ¦Ë4 must be -2, as the sum of these values is 5 + 0 - 2 = 3 = 1 (which is invalid for a singular matrix).

Hence, the eigenvalues of matrix A are ¦Ë1 = 3 + 2i, ¦Ë2 = 3 - 2i, ¦Ë3 = 0, and ¦Ë4 = -2.

If the matrix is singular, then the determinant must be zero.

One way to get singularity is to have two identical rows, such as:

1 1 0 0
1 1 0 0
0 0 1 0
0 0 0 1

Now we find the eigenvalues by equating to zero the determinant of the following matrix:

1-λ 1 0 0
1 1-λ 0 0
0 0 1-λ 0
0 0 0 1-λ

which is
((1-λ)^2-1)(1-λ)(1-λ)=0

Since two solutions of λ are known, we can substitute into the above equation as follows:

((1-λ)^2-1)(1-3+2i)(1-3-2i)=0 ...(1)

Solving equation (1) for remaining λ gives:
λ=0, -2