Tessellate S.T.E.M.S (2019) - Mathematics - Category C - Set 3 - Objective Problem 1

Level 2

Determine the number of possible values for the determinant of $A$ , given that $A$ is a $n\times n$ matrix with real entries such that $A^3 - 8A - 3I = 0$ , where $I$ is the identity and $0$ is the all-zero matrix.

\dfrac{n^2 - n + 6}{2}

\dfrac{n^2 + n + 4}{2}

(n+1)(n + 2)/2

n(n+1)/2

1 solution

Patrick Corn
Jan 31, 2019

If $\lambda$ is an eigenvalue of $A,$ then $\lambda^3-8\lambda + 3 = 0,$ so $(\lambda-3)(\lambda^2+3\lambda+1) = 0.$ This equation has three real roots, $3, \beta_1, \beta_2,$ where $\beta_1\beta_2 = 1.$

Now any $A$ is similar over the complex numbers to an upper triangular matrix with diagonal entries equal to its eigenvalues. (See e.g. the wiki on Jordan canonical form for a more precise statement.) So the determinant of $A,$ which is an invariant under similarity, must equal $3^a \beta_1^b \beta_2^c,$ where the exponents are nonnegative integers satisfying $a+b+c=n.$ Substituting $\beta_2 = 1/\beta_1$ gives $\det(A) = 3^a \beta_1^{a+2b-n}.$

So the number of distinct possible determinants is equal to the number of ordered pairs $(a,b)$ with $a,b \ge 0$ such that $a+b \le n$ (it is not hard to check that any two such ordered pairs produce different exponents, and different exponents yield different determinants, as the powers of $\beta_1$ are all irrational).

This number is of course $(n+1)(n+2)/2.$

Tessellate S.T.E.M.S (2019) - Mathematics - Category C - Set 3 - Objective Problem 1

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