Matrices — common mistakes

An eigenvector is not unique; any non-zero scalar multiple is also a valid eigenvector. For example, for $\lambda_1=2$, $\begin{pmatrix} 2 \\ 2 \end{pmatrix}$ or $\begin{pmatrix} -5 \\ -5 \end{pmatrix}$ are also correct. Always choose the simplest integer values. A great check: Sum of eigenvalues = $2+3=5$. Trace of $\mathbf{A}$ = $4+1=5$. Product of eigenvalues = $2 \times 3 = 6$. Determinant of $\mathbf{A}$ = $(4)(1) - (-2)(1) = 6$. They match!

When finding the inverse using the Cayley-Hamilton theorem, a common mistake is to forget the identity matrix $\mathbf{I}$ on the constant term. The equation must be an equation of matrices, so a scalar constant $c_0$ becomes $c_0\mathbf{I}$. Also, check if the constant term is zero. If $c_0 = \det(\mathbf{A}) = 0$, the matrix is singular and has no inverse.

An eigenvector represents a direction that is unchanged by the linear transformation represented by the matrix. The vector itself is stretched or shrunk by a factor equal to its corresponding eigenvalue, but it still points along the same line through the origin. This line is called an invariant line.

Yes. An eigenvalue of zero means that there is a non-zero vector $\mathbf{x}$ such that $\mathbf{A}\mathbf{x} = 0\mathbf{x} = \mathbf{0}$. This means that the matrix maps a non-zero vector to the zero vector. This is only possible if the matrix is singular, i.e., its determinant is zero. The product of eigenvalues equals the determinant, so if one eigenvalue is zero, the determinant must be zero.

If an eigenvalue $\lambda$ is a repeated root of the characteristic equation (e.g., has algebraic multiplicity 2), you might find one or two linearly independent eigenvectors for it. If you can find two linearly independent eigenvectors, they will form an 'invariant plane'. If you can only find one, it corresponds to an invariant line. The number of linearly independent eigenvectors for a given eigenvalue is its 'geometric multiplicity'.