A is any nxn matrix with all entries equal to 1 then 0 is an eigenvalue of A anda)multiplicity of 0 is n - 1b)multiplicity of 0 is 1c)multiplicity of 0 is nd)multiplicity of 0 is 0Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Explanation:

To prove that 0 is an eigenvalue of matrix A, we need to show that there exists a nonzero vector x such that Ax = 0.

Let's consider the matrix A with all entries equal to 1.

Matrix A:
[1 1 1 ... 1]
[1 1 1 ... 1]
[1 1 1 ... 1]
...
[1 1 1 ... 1]

Step 1:
To find the eigenvalues of matrix A, we need to solve the equation (A - λI)x = 0, where λ is the eigenvalue and I is the identity matrix.

Step 2:
Substituting A and I into the equation, we get:
[1-λ 1 1 ... 1]
[1 1-λ 1 ... 1]
[1 1 1-λ ... 1]
...
[1 1 1 ... 1-λ] * [x1, x2, x3, ..., xn] = 0

Step 3:
Expanding this equation, we get n equations:
(1-λ)x1 + x2 + x3 + ... + xn = 0
x1 + (1-λ)x2 + x3 + ... + xn = 0
x1 + x2 + (1-λ)x3 + ... + xn = 0
...
x1 + x2 + x3 + ... + (1-λ)xn = 0

Step 4:
Simplifying these equations, we get:
(1-λ)(x1 + x2 + x3 + ... + xn) = 0

Step 5:
Since all entries of matrix A are equal to 1, the sum of all elements in each row is n.

Step 6:
Therefore, the equation becomes:
(1-λ)(n) = 0

Step 7:
Solving this equation, we get:
1-λ = 0
λ = 1

Conclusion:
From the above steps, we can see that the only eigenvalue of matrix A is 1. Therefore, 0 is an eigenvalue of A with multiplicity n-1.

Hence, the correct answer is option A) multiplicity of 0 is n - 1.

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