3
$\begingroup$

I am to prove via induction that for any $n \times n$ matrix $A$, the characteristic polynomial of $A$ has

  • degree $n$;

  • $(-1)^n$ as the coefficient of the $\lambda ^n$ terms;

  • $(-1)^{n-1}\cdot \text{Trace}(A)$ as the coefficient of $\lambda ^{n-1}$.

I have shown the first two, but can't seem to prove the third. Could somebody give me a hint?

$\endgroup$
0
$\begingroup$

Eigenvalue arguments tend to work well when the base field $\Bbb F$, with $A \in M_{n \times n}(\Bbb F)$, is algebraically closed, so that all roots of the characteristic polynomial are guaranteed to exist in $\Bbb F$. Here's an inductive demonstration not based on eigenvalues and eigenvectors which works over any field $\Bbb F$:

First, the result is clearly true if $n = 2$, for if

$A = \begin{bmatrix} a_{11} & a_{22} \\ a_{21} & a_{22} \end{bmatrix}, \tag{1}$

then

$A - \lambda I = \begin{bmatrix} a_{11} - \lambda & a_{22} \\ a_{21} & a_{22} - \lambda\end{bmatrix}, \tag{2}$

so the characteristic polynomial $p_A(\lambda)$ is

$p_A(\lambda) = \det(\begin{bmatrix} a_{11} - \lambda & a_{22} \\ a_{21} & a_{22} - \lambda\end{bmatrix})$ $= \lambda^2 - (a_{11} + a_{22}) \lambda + (a_{11}a_{22} - a_{12}a_{21}) = \lambda^2 - \text{Tr}(A)\lambda + \det(A); \tag{3}$

the coefficient of $\lambda^2$ is $1 = (-1)^2$ and that of $\lambda = \lambda^1$ is $(-1)^1\text{Tr}(A) = (-1)^{2 - 1}\text{Tr}(A)$, and $\deg p_A(\lambda) = 2$ in this the base case for the induction. Now assume that $\deg p_A(\lambda) = k$, the coefficient of $\lambda^k$ is $(-1)^k$, and the coefficient of $\lambda^{k - 1}$ in $p_A(\lambda)$ is $(-1)^{k - 1}\text{Tr}(A)$ for all $A \in M_{k \times k}(\Bbb F)$, and consider some $A \in M_{(k + 1) \times (k + 1)}(\Bbb F)$; its characteristic polynomial is

$p_A(\lambda) = \det(A - \lambda I)$ $= \det(\begin{bmatrix} a_{11} - \lambda & a_{12} & \ldots & a_{1 \; k} & a_{1 \; (k + 1)} \\ a_{21} & a_{22} - \lambda & \ldots & a_{2 \; k} & a_{2 \; (k + 1)} \\ a_{31} & a_{32} & a_{33} - \lambda & \ldots & a_{3 \; (k + 1)} \\ \ldots \\ a_{(k + 1) \; 1} & a_{(k + 1) \; 2} & \ldots & a_{(k + 1) \; k} & a_{(k + 1) \; (k + 1)} - \lambda \end{bmatrix}). \tag{4}$

When we evaluate the determinant in (4) by expanding in minors along the first row, we see that

$p_A(\lambda) = (a_{11} - \lambda) \det(\begin{bmatrix} a_{22} - \lambda & \ldots & a_{2 \; k} & a_{2 \; (k + 1)} \\ a_{32} & a_{33} - \lambda & \ldots & a_{3 \; (k + 1)} \\ \ldots \\ a_{(k + 1) \; 2} & \ldots & a_{(k + 1) \; k} & a_{(k + 1) \; (k + 1)} - \lambda \end{bmatrix}) + \Theta(\lambda), \tag{5}$

where $\Theta(\lambda)$ is a polynomial of degree at most $k - 1$ in $\lambda$, since deleting the row and column containing $a_{1 \; j}$, $2 \le j \le k + 1$ leaves a matrix with at most $k - 1$ entries of the form $a_{ii} - \lambda$. Furthermore, our inductive hypothesis implies that

$\det(\begin{bmatrix} a_{22} - \lambda & \ldots & a_{2 \; k} & a_{2 \; (k + 1)} \\ a_{32} & a_{33} - \lambda & \ldots & a_{3 \; (k + 1)} \\ \ldots \\ a_{(k + 1) \; 2} & \ldots & a_{(k + 1) \; k} & a_{(k + 1) \; (k + 1)} - \lambda \end{bmatrix})$ $= (-1)^k \lambda^k + (-1)^{k - 1} \text{Tr} (A_{11}) \lambda^{k - 1} +\theta(\lambda), \tag{6}$

where $A_{11}$ is the matrix obtained from $A$ by deleting the first row and column, and $\theta(\lambda)$ is a polynomial of degree at most $k - 2$ in $\lambda$. Thus the $\lambda^{k + 1}$ and $\lambda^k$ terms of $p_A(\lambda)$ are given by the product

$(a_{11} - \lambda)((-1)^k \lambda^k + (-1)^{k - 1} \text{Tr} (A_{11}) \lambda^{k - 1})$ $= (-1)^{k + 1}\lambda^{k + 1} + a_{11}(-1)^k \lambda^k + (-1)^k \text{Tr} (A_{11})\lambda^k + a_{11}(-1)^{k - 1} \text{Tr} (A_{11}) \lambda^{k - 1}$ $= (-1)^{k + 1}\lambda^{k + 1} + (-1)^k \text{Tr} (A)\lambda^k + a_{11}(-1)^{k - 1} \text{Tr} (A_{11}) \lambda^{k - 1} \tag{7};$

we see from (7) that not only is the coefficient of $\lambda^k$ $(-1)^k \text{Tr} (A)$ and , but that the coefficient of $\lambda^{k + 1}$ is $(-1)^{k + 1}$ and also that the degree of $p_A(\lambda)$ is $k + 1$, thus inductively establishing all three of the bullet points in the question. And all this over an arbitrary field $\Bbb F$! QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!!

$\endgroup$
0
$\begingroup$

Assuming that we have $n$ eigenvalues $c_1,c_2,...,c_n$, we can express the characteristic polynomial as

$\sum_{k=0}^na_k\lambda^k=\prod_{k=1}^n(\lambda-c_k)$ (Eq. $2$)

where $a_k$ is the coefficient of $\lambda^k$

The trace of a matrix $A$ is the sum of its eigenvalues, hence

$Trace(A)=\sum_{k=1}^nc_k$

If you were to expand the RHS of (Eq. $2$), what do you think the coefficient of $\lambda^{n-1}$ will be in terms of the eigenvalues, given that the coefficient of $\lambda^n$ is $(-1)^n$. Hint: Vieta's formula

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.