# Multiplying eigenvalues with characteristic polynomial

The matrix A has eigenvalues $\lambda = \lambda_1, \lambda_2, \cdots , \lambda_n$ with the characteristic polynomial $$\lambda^n + a_{n-1}\lambda^{n-1}+\cdots+a_0=0$$ the eigenvalues of $At$ are $\lambda t$ and are solutions of $$(\lambda t)^n + (a_{n-1}t)(\lambda t)^{n-1}+\cdots+a_0 t^n=0$$

I do not understand why there is a $t$ after $a_{n-1}$, so why is it not $$(\lambda t)^n + a_{n-1}(\lambda t)^{n-1}+\cdots+a_0 t^n=0$$?

• What is $At$? Do you mean $tA$ (scalar t multiplied by matrix A) instead? Commented Dec 11, 2016 at 11:22
• @ald.li There is no particular reason to write scalars to the left or to the right of matrices when they are multiplied (scalar multiplication is a linear map, and it commutes with matrix multiplication); apparently the scalar is written to the right here. Commented Dec 11, 2016 at 17:22
• @MarcvanLeeuwen Sure. It's just that when $t$ is not defined $tA$ would hint that $t$ is a scalar. But $At$ does not really give such a hint, here the first intention is to treat $t$ as a vector to which $A$ is applied, but if this assumption fails, it can also mean other things, e.g., a matrix-valued variable. Commented Dec 11, 2016 at 20:05

Your formulation is not very clear. The characteristic polynomial is a polynomial, so it involves an indeterminate that is usually written $x$ or $X$, not $\lambda$ which in your setup is a scalar (one particular eigenvalue).

What it seems your question is about is how the characteristic polynomial changes if we scale a matrix $A$ by a scalar $t$. One way to see this is to check the definition of the characteristic polynomial. Being the determinant of $XI_n-A$, every contribution is obtained by multiplying $n$ factors, each of which either comes from $XI_n$ or from $-A$. To get a term involving a power $X^k$, one needs $k$ factors from $XI_n$ and the remaining $n-k$ from $-A$; then scaling $A$ by $t$ results in scaling the coefficient of $X^k$ in the characteristic polynomial by $t^{n-k}$.

A similar argument can be made using the known effect of scaling of eigenvalues. If the characteristic polynomial of $A$ splits as $(X-\lambda_1)\ldots(X-\lambda_n)$ then the characteristic polynomial of $At$ splits as $(X-t\lambda_1)\ldots(X-t\lambda_n)$, and again we see that the coefficient of $X^k$ gets multiplied by $t^{n-k}$.

If $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$, then they are the roots of the characteristic polynomial $p_A(\lambda)$. So we can write

$$\begin{array}{rcl}p_A(\lambda) & = & (\lambda - \lambda_1)(\lambda - \lambda_2)\cdots(\lambda - \lambda_n) \\ & = & \lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_1\lambda + a_0\end{array}$$

If $v_i$ is the eigenvector of $A$ corresponding to $\lambda_i$, that is, $Av_i = \lambda_i v_i$ for any $i$, then $v_i$ is an eigenvector for $tA$ with eigenvalue $t\lambda_i$ since $tAv_i = t\lambda_i v_i$. So the eigenvalues of $tA$ are $t\lambda_1,\ldots,t\lambda_n$.

The eigenvalues of $tA$ are precisely the roots of the characteristic polynomial of $tA$, so we get

$$\begin{array}{rcl}p_{tA}(\lambda) & = & (\lambda - t\lambda_1)(\lambda - t\lambda_2)\cdots(\lambda - t\lambda_n) \\ & = & \lambda^n + (ta_{n-1})\lambda^{n-1} + \cdots + (t^{n-1}a_1)\lambda + t^na_0\end{array}$$

The two polynomials you've written don't have the right eigenvalues as roots. $$(t\lambda)^n + (ta_{n-1})(t\lambda)^{n-1} + \cdots + (t^{n-1}a_1)t\lambda + t^na_0 = t^np_A(\lambda)$$ and so has roots $\lambda_1,\ldots,\lambda_n$, while the second has roots $t^{-1}\lambda_1,\ldots,t^{-1}\lambda_n$ as roots.

• When you said "minimal polynomial" did you mean "characteristic polynomial"? It is not that the eigenvalues are not also roots of the minimal polynomial, just that this is not what you appear to be using here. Commented Dec 11, 2016 at 17:28
• Yes, apologies.
– ODF
Commented Dec 11, 2016 at 17:28