Given Trace of powers of A, how to find its determinant

Question

Hello everyone, I am given trace of $A$, $A^2$, and $A^3$, how do I find determinant of $A$ from this. $A$ is $3\times 3$ matrix. Can this be solved using norm ?

Thanks in advance.

Answer 1

Let $\lambda_i$ be eigenvalues of $A$. Use relations between trace and eigenvalues like $$ Tr(A)=\lambda_1+\lambda_2+\lambda_3 $$ $$ Tr(A^2)=\lambda_1^2+\lambda_2^2+\lambda_3^2 $$ $$ Tr(A^3)=\lambda_1^3+\lambda_2^3+\lambda_3^3 $$ Solve above to get $\lambda_i$ and $$ \det(A)=\lambda_1\lambda_2\lambda_3 $$ You can see this method can easily be extended to $n\times n$ matrix.

Answer 2

Hints:

Let $A$ be a $n \times n$ matrix with eigenvalues $\lambda_1, \dots, \lambda_n$ and $k\in \mathbb{N}$:

$$Tr(A^k) = \sum_{i=1}^n \lambda_i^k$$ $$\det(A^k) = \prod_{i=1}^n \lambda_i^k$$

Answer 3

You can use Newton's identities. These let you write the characteristic polynomial of a matrix from trace powers. A recursive equation and more info can be found here https://en.wikipedia.org/wiki/Newton%27s_identities

For a 3x3 matrix, defining $p_k = Tr(A^k)$, the characteristic polynomial is: $$\lambda^3-p_1\lambda^2+\frac{1}{2}(p_1^2-p_2)\lambda-\frac{1}{6}(p_1^3-3p_1p_2+2p_3)$$

To find the determinant, we simply look at the zero'th degree term: $$\det(A) = \frac{1}{6}(p_1^3-3p_1p_2+2p_3)$$.

In general, for $n \times n$ matrices, the determinant is always a rational polynomial in the $p_k$'s.

In fact, the intermediate terms in the recurive equation on Wikipedia are both the other coefficients in the characteristic polynomial (with the sign alternating), and the equations for finding the determinant in smaller dimensions. (i.e the degree one term in the above characteristic polynomial is same as the formula for finding the determinant of a 2x2 matrix given its trace power)

You might have noticed that each term in the polynomial of $p_k$ corresponds to an unordered partition of n. For example, the three terms in $$\frac{1}{6}(p_1^3-3p_1p_2+2p_3)$$ correspond to the three ways you can write three as a sum of integers, $3=3$, $3= 2+1$, and $3=1+1+1$. Upon some reflection, you may realize that these are the only terms allowed by unit analysis. To do this, imagine that each entry of $A$ is a physical quantity measured in a unit, say metres, and consider the units of the $p_k$ and the determinant of $A$.

If you know a bit of group theory, there is a direct formula for the coefficients in front of a partition.

To find the coefficient of a term, consider its corresponding partition. Each partition $\mu$ of $n$ corresponds to a conjugacy class $C_\mu$ of the symmetric group $S_n$. The coefficient is: $$\frac{1}{n!}\times parity(C_\mu)\times size(C_\mu)$$ Here the parity of $C_\mu$ is the shared parity of all of its elements.