Suppose that A_n is the n×n matrix which has 2’s on the diagonal, and 1’s everywhere else:

A_2 matrix : Vector (2,1) and (1,2) A_3 matrix : Vector (2,1,1), (1,2,1), (1,1,2) A_4 matrix : Vector (2,1,1,1), (1,2,1,1), (1,1,2,1), (1,1,1,2) A_n matrix : vector (2,1,1…), (1,2,1,1…), (…)

and suppose that B_n is the n × n matrix which is just filled with minus-ones:

B_2 matrix : Vector (-1,-1), (-1,-1) B_3 matrix : Vector (-1,-1,-1), (-1,-1,-1), (-1,-1,-1) B_n matrix : Vector (-1,-1,-1,…), (…)

Explain why det(An) = det(In − Bn), where In is the n × n identity matrix.

If Pn(λ) is the characteristic polynomial of Bn, explain why det(An) = Pn(1).

Since Bn has rank 1, explain why this means that λn−1 has to divide Pn(λ).

Either by looking at the trace of Bn, or by seeing what happens to the vector ⃗v = (1,1,...,1) of all 1’s when you put it through Bn, find the value of a.

What is det(An)?

What is the determinant of the n × n matrix Cn which has 5’s on the diagonal, and 1’s everywhere else?

  • $\begingroup$ This isn't really on topic, but what do all of you think of the alternative "diagonable matrix"? It's in Horn and Johnson and I had a friend who said it instead just because it sounds funny. $\endgroup$ – Graphth Mar 3 '13 at 21:44
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    $\begingroup$ @Anna-Banana: What is the connection to diagonalizability? $\endgroup$ – Dennis Gulko Mar 3 '13 at 21:44
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    $\begingroup$ I think that $\det(A_n) = \det(I_n -B_n)$ because $A_n = I_n -B_n$. $\endgroup$ – Ben Mar 3 '13 at 21:45
  • $\begingroup$ Oh, I please: write your mathematics with LaTeX...! You can find directyions in the FAQ section. $\endgroup$ – DonAntonio Mar 3 '13 at 22:03

There is no mystery to this. The square matrix consisting of all 1's has the eigenvalue $n$ with multiplicity one and $0$ with multiplicity $(n-1).$ A basis of orthogonal eigenvectors, but of varying length, is: $$ n: \; \; (1,1,1, \ldots,1),$$ $$ 0: \; \; (1,-1,0,0,\ldots,0),$$ $$ 0: \; \; (1,1,-2,0,\ldots,0),$$ $$ 0: \; \; (1,1,1,-3,0,\ldots,0),$$ $$ 0: \; \; (1,1,1,1,-4,\ldots,0),$$ $$ 0: \; \; (1,1,1,1,1,\ldots,1,1-n).$$

To get $w$ on the main diagonal you add $(w-1)I,$ which does not alter the eigenvectors but adds $(w-1)$ to each eigenvalue. The determinant is the product of the eigenvalues, giving $$ (n+w-1) (w-1)^{n-1}. $$

I see. I missed the evident homework questions interspersed. Sigh. Call the matrix of all ones $T,$ there is no difficulty dealing with $\alpha T + \beta I.$ Please do those yourself. An eigenvector of $T$ with eigenvalue $\lambda$ is also an eigenvector of $\alpha T + \beta I$ with eigenvalue $\alpha \lambda + \beta .$


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