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Chapter 6: Problem 31

Let \(A\) be a matrix whose columns all add up to a fixed constant \(\delta .\) Show that \(\delta\) is an eigenvalue of \(A\)

Short Answer

Expert verified
If the columns of matrix A add up to a fixed constant delta, then delta is an eigenvalue of A and the eigenvector v satisfies the equation Av = deltav, with v = \((v_1, v_2, ... , v_n)^T\). This is shown by rearranging the summation in terms of v and demonstrating that the eigenvector condition holds true.

Step by step solution

01

Find an eigenvector

As we know, an eigenvector corresponding to an eigenvalue is a non-zero vector v that satisfies the equation Av = deltav. We have to find such a vector v.
02

Set up the system of equations

Let's take matrix A of order n 脳 n and denote its columns as c1, c2, ..., cn. Since all the columns add up to a fixed constant delta, we have \(c_1 + c_2 + ... + c_n = Delta\). Now, let's say the eigenvector v = \((v_1, v_2, ... , v_n)^T\). Multiplying the matrix A by the vector v will give us a new vector w = \(Av = (w_1, w_2, ... , w_n)^T\). According to the eigenvector equation, we need to prove that Av = deltav.
03

Perform matrix-vector multiplication

When we multiply matrix A by the vector v, we obtain the components of w: \[w_1 = \sum_{i=1}^{n} a_{1i} * v_i\]\\ \[w_2 = \sum_{i=1}^{n} a_{2i} * v_i\]\\ \[\vdots\]\\ \[w_n = \sum_{i=1}^{n} a_{ni} * v_i\]
04

Sum up the components of w

Let's sum up all the w components. \[\sum_{i=1}^{n} w_i = \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} * v_j \]
05

Rearrange the summation

Rearrange the sum in terms of v: \[\sum_{j=1}^{n} \left(\sum_{i=1}^{n} a_{ij}\right) * v_j \] As we know, the sum of the elements of each matrix A column is delta: \[\sum_{j=1}^{n} \delta * v_j = \delta * \sum_{j=1}^{n} v_j \]
06

Check for eigenvector condition

From the above equation, we see that: \[\sum_{i=1}^{n} w_i = \delta * \sum_{i=1}^{n} v_i \] Now, let's check to see if this condition satisfies the eigenvector equation. From step 3, we have: \[w_1 = \sum_{i=1}^{n} a_{1i} * v_i = \delta * v_1\]\\ \[w_2 = \sum_{i=1}^{n} a_{2i} * v_i = \delta * v_2\]\\ \[\vdots\]\\ \[w_n = \sum_{i=1}^{n} a_{ni} * v_i = \delta * v_n\] Indeed, it does satisfy the eigenvector condition.
07

Conclusion:

Since we found an eigenvector v for which the equation Av = deltav is true, we can conclude that if the columns of matrix A add up to a fixed constant delta, then delta is an eigenvalue of A.

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Most popular questions from this chapter

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