Further matrix algebra (Further Pure 2)

Explanation

• A non-zero vector $\mathbf{v}$ is an eigenvector of the square matrix $\mathbf{M}$ with eigenvalue $\lambda$ when $\mathbf{M}\mathbf{v}=\lambda\mathbf{v}$.
• Rearranging gives $\left(\mathbf{M}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0}$, which has a non-zero solution only when $\det\left(\mathbf{M}-\lambda\mathbf{I}\right)=0$; that equation is the characteristic equation and its roots are the eigenvalues.
• For a $2\times2$ matrix it reduces to $\lambda^{2}-\left(\operatorname{tr}\mathbf{M}\right)\lambda+\det\mathbf{M}=0$, which is a quick way to write it down and a quick way to check your expansion.
• Having found each $\lambda$, substitute it back and solve the resulting homogeneous system; the solution set is a line for a $2\times2$ matrix with distinct eigenvalues, and may be a line or a plane in the $3\times3$ case.
• Eigenvectors are only determined up to a non-zero scalar multiple, so any convenient scaling is acceptable unless the question asks for a unit vector.

Explanation

• If an $n\times n$ matrix $\mathbf{M}$ has $n$ linearly independent eigenvectors, put them in the columns of a matrix $\mathbf{P}$ and put the corresponding eigenvalues, in the same order, on the leading diagonal of $\mathbf{D}$. Then $\mathbf{M}\mathbf{P}=\mathbf{P}\mathbf{D}$, so $\mathbf{P}^{-1}\mathbf{M}\mathbf{P}=\mathbf{D}$ and $\mathbf{M}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}$.
• The order of the columns of $\mathbf{P}$ and of the entries of $\mathbf{D}$ must match. The main use is computing powers: $\mathbf{M}^{n}=\mathbf{P}\mathbf{D}^{n}\mathbf{P}^{-1}$, and $\mathbf{D}^{n}$ is found by raising each diagonal entry to the power $n$.
• A matrix with $n$ distinct eigenvalues is always diagonalisable, because eigenvectors belonging to distinct eigenvalues are linearly independent.
• A matrix with a repeated eigenvalue is diagonalisable only if that eigenvalue's eigenspace has dimension equal to its multiplicity.
• When $\mathbf{M}$ is real and symmetric the eigenvectors can be chosen orthonormal, and then $\mathbf{P}^{-1}=\mathbf{P}^{\mathrm{T}}$.

Explanation

• The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation: if $\det\left(\mathbf{M}-\lambda\mathbf{I}\right)$ expands to a polynomial in $\lambda$, replacing $\lambda$ by $\mathbf{M}$ and the constant term by that constant times $\mathbf{I}$ gives the zero matrix. For a $2\times2$ matrix this reads $\mathbf{M}^{2}-\left(\operatorname{tr}\mathbf{M}\right)\mathbf{M}+\left(\det\mathbf{M}\right)\mathbf{I}=\mathbf{0}$.
• Two applications dominate the examination. First, high powers: rearranging gives $\mathbf{M}^{2}$ as a linear combination of $\mathbf{M}$ and $\mathbf{I}$, and repeated substitution expresses $\mathbf{M}^{3}$, $\mathbf{M}^{4}$ and so on in the same form without ever multiplying matrices.
• Second, inverses: multiplying the identity through by $\mathbf{M}^{-1}$ gives $\mathbf{M}^{-1}$ as a polynomial in $\mathbf{M}$, which works whenever the constant term, that is $\det\mathbf{M}$, is non-zero.
• Take care with the sign convention.
• Expanding $\det\left(\mathbf{M}-\lambda\mathbf{I}\right)$ and setting $\lambda=0$ always leaves $\det\mathbf{M}$, whatever the order; for odd order that expansion opens with $-\lambda^{n}$, so multiply through by $-1$ to reach the monic form $\lambda^{n}-\left(\operatorname{tr}\mathbf{M}\right)\lambda^{n-1}+\cdots$, whose constant term is $\left(-1\right)^{n}\det\mathbf{M}$.