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Horn and Johnson in "Matrix Analysis" leave as a exercise this proof:

Let A be a real matrix with characteristic polynomial $p(\lambda)$ where $$p(\lambda) = \lambda^n + c_{1}\lambda^{n-1} + c_2\lambda^{n-2}+\cdots+c_n.$$ Let $E_k = \sum C_{kk}$ where $C_{kk}$ means a $k$-by-$k$ principal minor of $A$, and the summation is over all $k$-by-$k$ principal minors. Then, $$p(\lambda) = \lambda^n + E_1\lambda^{n-1}+E_2\lambda^{n-2}+\cdots+E_n.$$

The authors say that this can be proved by mathematical induction, using the Laplace expansion.

I have written out the base case and the induction hypothesis. My assumption is that the induction is on the dimension of the matrix, although now I am not sure at this point as I don't know what to do from here.

Could anyone give me a hint or help as to what I should do next?

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    what is the correct statement for $n=2?$ – 2017-01-26
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    oh, here the word should be principal – 2017-01-26
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    Consider for example a $5 \times 5$ matrix: let us consideer for example the terms $a\lambda^3$ in the expansion of $\det(A-\lambda I)$for example come from selecting $3$ times a term in the diagonal $a_{ii}-\lambda$ and $2$ times elements that are not on the diagonal. The number of ways one can do that will lead to the expansion of different principal minors. – 2017-01-26
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    @JeanMarie I see, the separate 2013 film is discussed https://en.wikipedia.org/wiki/Home_from_Home_%282013_film%29 as well as https://en.wikipedia.org/wiki/Heimat_%28film_series%29#The_Other_Heimat while the long sequence before that https://en.wikipedia.org/wiki/Heimat_%28film_series%29 Also http://www.die-andere-heimat.de/ and http://www.imdb.com/title/tt1998204/ – 2017-01-26
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    @JeanMarie while I can see how this would work, it doesn't use Laplace expansion, which is ok, but I just curious how it is suppose to factor in. Also I guess I don't understand how to make then inductive step, not so much that yes you can figure this out combinatorially. – 2017-01-27

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Your question amounts to prove that the determinant of the matrix $A-\lambda E$ has the form:

$$ p(\lambda) = (-1)^n\lambda^n+E_1\lambda^{n-1}+E_2\lambda^{n-2}+ \dots +E_{n-1}\lambda + E_n$$

Please notice the $(-1)^n$ sign in front of the first $\lambda$ that is originally missing in your question but which is crucial for the theorem to be proved.

As correctly suggested by Horn and Johnson, the above can be proved by using mathematical induction (on the size of sqare matrix A) applying the Laplace exapansion in the inductive step as follows:

Base case $size(A)=2$ (we omit $n=1$ being trivial)

Consider the determinant of an $2x2$ matrix $det(A-\lambda E)$ $$ \begin{vmatrix} a-\lambda & b \\ d & e-\lambda \\ \end{vmatrix} $$ If we consider $det(A-\lambda E)$ as follows $$ \begin{vmatrix} a-\lambda & b+0\\ d+0 & e-\lambda \\ \end{vmatrix} $$ we can see each column of $det(A-\lambda E)$ as a linear combination of two columns of numbers and, therefore, we can apply the linear property of determinants by "separating out the powers of $\lambda$" (as illustrated in https://math.stackexchange.com/a/144130/557814) by obtaining: $$ \begin{vmatrix} a-\lambda & b \\ c & d-\lambda \end{vmatrix} = \begin{vmatrix} a & b \\ c & d-\lambda \end{vmatrix} + \begin{vmatrix} -\lambda & b \\ 0 & d-\lambda \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} + %% \begin{vmatrix} a & 0 \\ c & -\lambda \end{vmatrix} + %% \begin{vmatrix} -\lambda & b \\ 0 & d \end{vmatrix} + \begin{vmatrix} -\lambda & 0 \\ 0 & -\lambda \end{vmatrix} = \\ (-1)^2\lambda^2 +E_1\lambda{2-1}+E_2 $$

Inductive step (size(A) $n$ assuming the for matrixes with size $n-1$ the hypothesis is true)

Consider the following determinant: $$ \begin{vmatrix} a_{1,1}-\lambda & a_{1,2} & \dots & a_{1,n-1} & a_{1,n} \\ a_{2,1} & a_{2,2}-\lambda & \dots & a_{2,n-1} & a_{2,n} \\ \vdots & \vdots & \dots & \vdots & \vdots \\ a_{n-1,1} & a_{n-1,2} & \dots & a_{n-1,n-1}-\lambda & a_{n-1,n} \\ a_{n,1} & a_{n,2} & \dots & a_{n,n-1} & a_{n,n} - \lambda \\ \end{vmatrix} $$ Now, by applying the linear property of determinants (as done the base case) we have: $$ \begin{vmatrix} a_{1,1}-\lambda & a_{1,2} & \dots & a_{1,n-1} & a_{1,n} \\ a_{2,1} & a_{2,2}-\lambda & \dots & a_{2,n-1} & a_{2,n} \\ \vdots & \vdots & \dots & \vdots & \vdots \\ a_{n-1,1} & a_{n-1,2} & \dots & a_{n-1,n-1}-\lambda & a_{n-1,n} \\ a_{n,1} & a_{n,2} & \dots & a_{n,n-1} & a_{n,n} - \lambda \\ \end{vmatrix} = \\ \begin{vmatrix} a_{1,1} & a_{1,2} & \dots & a_{1,n-1} & a_{1,n} \\ a_{2,1} & a_{2,2} & \dots & a_{2,n-1} & a_{2,n} \\ \vdots & \vdots & \dots & \vdots & \vdots \\ a_{n-1,1} & a_{n-1,2} & \dots & a_{n-1,n-1} & a_{n-1,n} \\ a_{n,1} & a_{n,2} & \dots & a_{n,n-1} & a_{n,n}\\ \end{vmatrix} + \begin{vmatrix} a_{1,1}-\lambda & a_{1,2} & \dots & a_{1,n-1} & 0 \\ a_{2,1} & a_{2,2}-\lambda & \dots & a_{2,n-1} & 0 \\ \vdots & \vdots & \dots & \vdots & \vdots \\ a_{n-1,1} & a_{n-1,2} & \dots & a_{n-1,n-1}-\lambda & 0 \\ a_{n,1} & a_{n,2} & \dots & a_{n,n-1} & - \lambda \\ \end{vmatrix} $$ Now, by applying the laplace expansion on the last column of the second determinant in the above formula we obtain: $$ det(A) -\lambda det(M_{n,n}-\lambda E) $$ where $M_{n,n}$ is the matrix that results by deleting the $n$th row and column fomr A. But the above, by the inductive hypothesis equals to: $$ det(A) -\lambda[(-1)^{n-1}\lambda^{n-1}+E_1\lambda^{(n-1)-1}+E_2\lambda^{(n-1)-2}+\dots+E_{n-1})] = \\ (-1)^{n}\lambda^n +E_1\lambda^{n-1}+E_2\lambda^{n-2}+\dots+E_{n-1}\lambda+E_n = p(\lambda) $$

QED