Now we calculate the two determinants in the second equality separately.

The easiest way is to note that the determinant of a triangular matrix is the product of its diagonal entries.
Thus the first determinant is $(-\lambda)^{n-1}$ and the second determinant is $1$.

(If you don’t know this fact, then use the first row cofactor expansion inductively to compute the first determinant. For the second one, use the first column cofactor expansion inductively.)

Thus we obtain $\det(A-\lambda I)=(-1)^n\lambda^n+(-1)^{n+1}=(-1)^n(\lambda^n-1)$.
Therefore eigenvalues are $n$-th roots of unity $e^{2\pi i/n}$ for $i=0,1,\dots, n-1$.

Comment.

The original determinant is not in a good shape for induction but once we apply the 1st row cofactor expansion the smaller determinants obtained are better suited for induction.
When I say in the proof “inductively”, I meant that you need to use mathematical induction to prove the claim (more) rigorously.

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Let $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are linearly independent $3$-dimensional vectors. Suppose that we have
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