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