# Quick ways to _verify_ determinant, minimal polynomial, characteristic polynomial, eigenvalues, eigenvectors ...

What are easy and quick ways to verify determinant, minimal polynomial, characteristic polynomial, eigenvalues, eigenvectors after calculating them?

So if I calculated determinant, minimal polynomial, characteristic polynomial, eigenvalues, eigenvectors, what are ways to be sure that I didn't do a major mistake? I don't want to verify my solutions all the way through, I just want a quick way which gives me that it is highly likely that the calculated determinant is right etc.

Let $A$ be a matrix $A \in \operatorname{Mat}(n, \mathbb{C})$,

let $\det(A)$ be the determinant of matrix $A$,

let $v_1, v_2, ..., v_k$ be eigenvectors of matrix $A$,

let $\lambda_1, \lambda_2, ..., \lambda_n$ be eigenvalues of matrix $A$,

let $\chi_A(t) = t^n + a_{n-1}t^{n-1}+\cdots + a_0 = (t-\lambda_1)\cdots(t-\lambda_n)$ be the characteristic polynomial of matrix $A$,

let $\mu_A(t)$ be the minimal polynomial of matrix $A$.

Verifications suggested so far:

eigenvectors / eigenvalues

• $\det(A) = \lambda_1^{m_1} \lambda_2^{m_2} \cdots \lambda_n^{m_l}$ where $m_i$ is the multiplicity of the corresponding eigenvalue
• $a_0 = (-1)^n\lambda_1\cdots\lambda_n$
• eigenvectors can be verified by multiplying with the matrix; the eigenvalues can be verified at the same time; i.e. $A v_i = \lambda_i v_i$

determinant

• $\det(A) = \lambda_1^{m_1} \lambda_2^{m_2} \cdots \lambda_l^{m_l}$ where $m_i$ is the multiplicity of the corresponding eigenvalue

characteristic / minimal polynomial

• $a_0 = (-1)^n\lambda_1\cdots\lambda_n$
• $\mu_A(A) = 0$ and $\chi_A(A) = 0$
• $\mu_A(t) \mid \chi_A(t)$
• The product of the eigenvalues (counting multiplicity) equals the determinant. The minimal polynomial should divide the characteristic polynomial. The constant term of the characteristic polynomial is $(-1)^n$ times the determinant; the coefficient of the term of degree $n-1$ is minus the sum of the eigenvalues (which is also the trace). That sort of thing? Feb 3, 2012 at 19:40
• Interesting question! Here's a paper which investigates complexity of computing and verifying the characteristic and minimal polynomials for an integer matrix: sciencedirect.com/science/article/pii/S0304397502004048
– dls
Feb 3, 2012 at 19:44
• ... and the matrix satisfies its minimal polynomial (and characteristic polynomial), i.e. if $P(\lambda) = a_0 + a_1 \lambda + \ldots a_n \lambda^n$ is the minimal or characteristic polynomial of matrix $A$, then $P(A) = a_0 I + a_1 A + \ldots a_n A^n = 0$. Feb 3, 2012 at 19:44
• Exactly that sort of thing! So multiplying all the eigenvalues gives the determinant? That's a good trick! Since I calculate the minimal polynomial the way that it should divide the characteristic polynomial that doesn't really help to detect mistakes in this calculation. What do you mean by the »constant term of the characteristic polynomial«; so if the polynomial is not written as linea factors? Didn't understand the last one yet.. Feb 3, 2012 at 20:00
• @meinzlein: The characteristic polynomial is a polynomial. If you write it out as $x^n + a_{n-1}x^{n-1}+\cdots + a_0$, then the "constant term" is $a_0$. Since the characteristic polynomial equals $(x-\lambda_1)\cdots(a-\lambda_n)$, where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues (in the algebraic closure of the underlying field), then we know that $a_0 = (-1)^n\lambda_1\cdots\lambda_n$ and $a_{n-1} = -(\lambda_1+\cdots+\lambda_n)$. The fact that the determinant equals the product of eigenvalues and the trace equals the sum follows by thinking about, say, the Jordan canonical form. Feb 3, 2012 at 20:13

You can use the Caley-Hamilton theorem. Put your matrix in the characteristic polynomial and you should find $0$. Explicitly, $\chi_A(A) = A^n + a_{n-1}A^{n-1}+\cdots + a_0I=0$. This checks the
• Note that the wrong characteristic polynomial can pass this test; e.g. if the minimal polynomial of a $4 \times 4$ matrix is $x(x-1)$, then plugging $A$ into any of $x^3 (x-1)$, $x^2 (x-1)^2$, and $x(x-1)^3$, or even $x(x-1)(x-2)(x-3)$ will give zero.