# Expansion of a determinant in powers of $\lambda$ (from Courant-Hilbert Vol I)

On page 20-21 of volume I of Courant & Hilbert's "Methods of Mathematical Physics" they say:

If we expand the determinants $\Delta(u,y;\lambda)$ and $\Delta(\lambda)$ in powers of $\lambda$, we obtain the expressions: $$\Delta(u,y;\lambda)= \Delta_1(u,y)-\lambda\Delta_2(u,y)+\lambda^2\Delta_2(u,y)-\ldots +(-1)^n\lambda^{n-1}\Delta_n(u,y),$$ $$\Delta(\lambda)=1-\lambda\Delta_1+\lambda^2\Delta_2-\ldots + (-1)^n \lambda^n \Delta_n,$$ where \begin{align*} \Delta_h(u,y) &= \sum \begin{vmatrix} 0 & u_{p_1} & \cdots & u_{p_h} \\ y_{p_1} & t_{p_1p_1} & \cdots & t_{p_1p_h} \\ \vdots & \vdots & \ddots & \vdots \\ y_{p_h} & t_{p_hp_1} & \cdots & t_{p_hp_h} \\ \end{vmatrix}\\ \end{align*} and \begin{align*} \Delta_h &= \sum \begin{vmatrix} t_{p_1p_1} & t_{p_1p_2} & \cdots & t_{p_1p_h} \\ t_{p_2p_1} & t_{p_2p_2} & \cdots & t_{p_2p_h} \\ \vdots & \vdots & \ddots &\vdots \\ t_{p_hp_1} & t_{p_hp_2} & \cdots & t_{p_hp_h} \\ \end{vmatrix}\\ \end{align*}. The summations here are extended over all integers $p_1,p_2,\ldots , p_h$ from $1$ to $n$ with $p_1<p_2<\ldots < p_h$.

Where:

\begin{align*} \Delta(u,y;\lambda) &= \begin{vmatrix} 0 & u_1 & \cdots & u_n \\ y_1 & 1-\lambda t_{11} & \cdots & -\lambda t_{1n} \\ \vdots & \vdots & \ddots &\vdots \\ y_n & -\lambda t_{n1} & \cdots & 1-\lambda t_{nn} \\ \end{vmatrix}\\ \end{align*} and \begin{align*} \Delta(\lambda) &= \begin{vmatrix} 1-\lambda t_{11} & -\lambda t_{12} & \cdots & -\lambda t_{1n} \\ -\lambda t_{21} & 1-\lambda t_{22} & \cdots & -\lambda t_{2n} \\ \vdots & \vdots & \ddots &\vdots \\ -\lambda t_{n1} & -\lambda t_{n2} & \cdots & 1-\lambda t_{nn} \\ \end{vmatrix}\\ \end{align*}

How to prove these two identities? with and without induction?

• First, the book's name is "Courant&Hilbert", not the other way around. This is important in general, in particular in your case as: two, you didn't even bother to give the name of the book, which is "Methods of Mathematical Physics" . – DonAntonio Jun 17 '16 at 9:26
• Are there more books that they coauthored? – MathematicalPhysicist Jun 17 '16 at 9:45
• @Ma I've no idea, but if you mention a book then I think it is worthwhile to be a little more careful and write the name(s) of the authors as they appear (most probably in alphabetical order) and the name of the book. People don't need to guess whether there's one single book by this or that author and its name, in my opinion. – DonAntonio Jun 17 '16 at 9:47
• @Joanpemo I edited my original post and added the name of the title of the book I am adressing my question to. – MathematicalPhysicist Jun 17 '16 at 9:51

## 1 Answer

I will consider the simpler case of $\Delta(\lambda)$, but the approach should work for $\Delta(u,y;\lambda)$ as well. First, note that this determinant can be written in terms of column vectors as $$\Delta(\lambda)=\det(e_1-\lambda t_1,e_2-\lambda t_2,\cdots,e_n-\lambda t_n)$$ where $e_k$ is the $k$th basis column vector and $t_k=(t_{1k},t_{2k},\cdots,t_{nk})^T$. Since the determinant is a multilinear function of its column vectors, we have

\begin{align} \Delta(\lambda) =\det(e_1,e_2,\cdots,e_n) &-\lambda\left[\det(t_1,e_2,\cdots,e_n)+\cdots+\det(e_1,e_2,\cdots,t_n)\right]\\ &+\lambda^2\left[\det(t_1,t_2,\cdots,e_n)+\cdots+\det(e_1,e_2,\cdots,t_{n-1},t_n)\right]\\ &-\cdots+(-\lambda)^n \det(t_1,t_2,\cdots,t_n). \end{align} All that remains is to identify each coefficient in this expansion with $\Delta_n$; this can be shown by taking each determinant and successively expanding-by-cofactors along any column containing $e_k$.

• What is Laplace expanding? – MathematicalPhysicist Jun 17 '16 at 17:06
• Laplace expansion = expansion by cofactors. I think I'll use the latter phrase instead, though. (Reading that Wiki page, I suspect what I should probably referencing is the more general notion of 'expansion by complementary minors.') @MathematicalPhysicist – Semiclassical Jun 17 '16 at 17:10