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Consider ann×n matrix such that the sum of the entries in each row is . Show that the vector

e→=11..1

InRn is an eigenvector of A. What is the corresponding eigenvalue?

Short Answer

Expert verified

The eigenvalue of the matrix with A the vectore→=11..1 isλ=1

Step by step solution

01

Finding the ith coordinate of Ae

For an n×nmatrix A and scalar λ,λ is an eigenvalue of A, if there exist a non-zero vectorv→ inRn such that,

Av→=λv→orAv→-0→orAv→-λlnv→=0→or(A-λ±ôn)v→=0→

For the given vectore→=11..1 in the matrix A,

Theith coordinate of Ae will be,

1×ai,1+1×ai,2,+…1×ai,n=1×∫j=1nai,i=1×1=1

02

Finding the eigenvalue λ

So the vector will be,

Ae=11..1=e

Therefore, e is an eigenvalue of Matrix A, with the corresponding eigenvalueλ=1

Thus, the eigenvalue of the matrix A with the vectorrole="math" localid="1659588113834" e→=11..1 isλ=1

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