In linear algebra, eigensystems denote a set of problems that include finding eigenvectors and eigenvalues. The word eigen comes from German and means ‘own’, which will make sense when we formulate the problem more concretely. Informally, the main idea behind eigensystems is finding vectors that transform in a special way when we apply a certain transformation $\mathbf{A}$ on them. Specifically, we wish to find which vectors are affected the least by the transformation $\mathbf{A}$, and by least we mean that they are not rotated, but are only scaled by a factor $\lambda$. Formally, given a vector $\mathbf{v}$ and a transformation $\mathbf{A}$, this requirement can be written as:

$$\mathbf{A}\mathbf{v}=\lambda\mathbf{v}$$ Since on the right-hand side we multiply a vector by a scalar, we can equivalently add the identity matrix as $\lambda\rightarrow \lambda\mathbf{I}$. Rearranging terms gives us the following equation: $$(\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}$$ Assuming that $\mathbf{A}\in\mathbf{R}^{n\times n}$ and $\mathbf{v}\in\mathbb{R}^n$, we can rewrite the former equation in an expanded form: $$\matrix{(A_{11}-\lambda) & A_{12} & \cdots & A_{1n}\\ A_{21} & (A_{22}-\lambda) & \cdots & A_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ A_{n1} & A_{n2} & \cdots & (A_{nn}-\lambda)} \cv{v_1\\ \vdots \\ \vdots \\ v_n}= \cv{0\\ \vdots \\ \vdots \\ 0}$$ The equation above represents a system of linear equations, and the goal is to find vectors $\mathbf{v}=\bigl(v_1\;\;\cdots\;\; v_n\bigl)^{\top}$ and $\lambda$ that satisfy it. For example, the $m$-th equation is given by: $$\text{A}_{m1} v_1 + \cdots + \left(\text{A}_{mm}-\lambda\right) v_m + \cdots + \text{A}_{mn}v_n = 0.$$ What we can see is that in every equation, we have all the unknowns (elements of the vector Therefore, if the equations are not linearly independent, the only solution is the trivial one, i.e. $v_1=v_2=\ldots=v_n=0$. (Note: the equations of a linear system are independent if none of the equations can be derived algebraically from the others. In other words, we cannot write one equation as a linear combination of other equations.) Although this is a plausible solution, it is not very informative, as it just tells us that if we take a null vector and apply an arbitrary transformation, it will remain a null vector, which is a trivial statement.

The second option is that the system of linear equations is not indeed linearly independent. This implies that the columns of the matrix $\mathbf{A}-\lambda\mathbf{I}$ are linearly dependent, which in turn implies that the mapping is not bijective (does not have an inverse). As we have seen in the previous section, matrices that are not invertible have the determinant zero, and using this property, we can search for non-trivial solutions. Thus, we can start solving the problem by looking for values of $\lambda$ that satisfy the following equation:

$$\left| \mathbf{A} -\lambda \mathbf{I}\right| = 0.$$