Asked by Anonymous
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.
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.
Answers
Answered by
MathMate
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
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
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