Let A be a n × n matrix, Suppose that A has n distinct eigenvalues λ1, . . . , λn, and corresponding non-zero eigenvectorsx1, . . . , xn. Prove that {x1, . . . , xn} are linearly independent.Hint: You may use without proof the following property: If {y1, . . . , ym} are linearly dependentthen there exists some p such that 1 ≤ p < m, yp+1 ∈ span{y1, . . . , yp} and {y1, . . . , yp} arelinearly independent.

To prove that the set of eigenvectors {x1, . . . , xn} are linearly independent, we will use a proof by contradiction.

Assume that the set of eigenvectors {x1, . . . , xn} are linearly dependent. According to the hint, this means that there exists some p such that 1 ≤ p < n, xp+1 ∈ span{ x1, . . . , xp } and { x1, . . . , xp } are linearly independent.

This means that we can write xp+1 as a linear combination of the previous vectors:

xp+1 = a1x1 + a2x2 + ... + ap*xp

Now, let's multiply both sides of this equation by A (the matrix):

Axp+1 = a1Ax1 + a2Ax2 + ... + apA*xp

We know that Axi = λixi for each i (from the definition of eigenvalues and eigenvectors), so we can substitute these into the equation:

λp+1xp+1 = a1λ1x1 + a2λ2x2 + ... + apλp*xp

Subtract the original linear combination equation from this one:

λp+1xp+1 - (a1x1 + a2x2 + ... + apxp) = a1*(λ1 - λp+1)x1 + a2(λ2 - λp+1)x2 + ... + ap(λp - λp+1)*xp

The left side of this equation is zero, because xp+1 is a linear combination of the other vectors. So, we have:

0 = a1*(λ1 - λp+1)x1 + a2(λ2 - λp+1)x2 + ... + ap(λp - λp+1)*xp

This is a linear combination of {x1, . . . , xp}, which we assumed to be linearly independent. A linear combination of linearly independent vectors can only be zero if all the coefficients are zero.

But, since λi ≠ λp+1 for all i (because all eigenvalues are distinct), the coefficients a1*(λ1 - λp+1), a2*(λ2 - λp+1), ..., ap*(λp - λp+1) cannot all be zero unless a1, a2, ..., ap are all zero.

This contradicts our assumption that xp+1 can be written as a linear combination of the previous vectors. Therefore, our original assumption that the set of eigenvectors {x1, . . . , xn} are linearly dependent must be false.

So, we conclude that the set of eigenvectors {x1, . . . , xn} are linearly independent.