Eigenvalue applications and synthesis

Diagonalization exists when eigenvectors are independent. If $n\times n$ matrix $A$ has $n$ linearly independent eigenvectors $\mathbf v_1,\dots,\mathbf v_n$ — equivalently, geometric multiplicity equals algebraic multiplicity for every eigenvalue — then stacking the eigenvectors as columns of $P=[\mathbf v_1\ \cdots\ \mathbf v_n]$ and the matching eigenvalues on the diagonal of $D$ gives

where: $P$ is the eigenvector matrix (column $j$ is the eigenvector for eigenvalue $j$), $D$ is diagonal with $D_{jj}=\lambda_j$ in the same order as $P$'s columns, and $P^{-1}$ undoes the change of basis. Every real symmetric (and every Hermitian) matrix is automatically diagonalizable; a distinct-eigenvalues matrix is too, since eigenvectors from distinct eigenvalues are always independent — that shortcut saves you from checking eigenspace dimensions by hand.

Why this makes powers cheap. A diagonal matrix's powers are trivial entrywise, $D^k=\operatorname{diag}(\lambda_1^k,\dots,\lambda_n^k)$, and substituting into $A=PDP^{-1}$ gives $A^k=PD^kP^{-1}$ for any $k$, using only one inversion of $P$ regardless of how large $k$ is. The same trick with $k=-1$ gives a second route to the inverse, $A^{-1}=PD^{-1}P^{-1}$, valid whenever every $\lambda_i\neq0$.

Cayley–Hamilton reframes the characteristic polynomial as an identity in $A$ itself. If $p(\lambda)=\det(A-\lambda I)$, then substituting $A$ for $\lambda$ (and $I$ for the constant term) gives the zero matrix, $p(A)=O$. For $2\times2$ matrices this is

where: $\operatorname{tr}A$ is the sum of the diagonal entries and $\det A$ is the determinant — both readable off $A$ directly, no eigenvalues required. Two standard payoffs: rearrange to isolate $I$ and divide by $\det A$ to get $A^{-1}$ (when $\det A\neq0$); or repeatedly substitute $A^2=(\operatorname{tr}A)A-(\det A)I$ to write any $A^k$ as $\alpha A+\beta I$, avoiding $k-1$ direct multiplications.

Two independent toolkits, not competitors. Diagonalization needs eigenvectors and only exists when $P$ is invertible, but once available it handles any power in one shot. Cayley–Hamilton needs nothing beyond $\operatorname{tr}A$ and $\det A$ for a $2\times2$ matrix, holds for every square matrix without exception, and reduces powers one step at a time rather than jumping straight to $A^k$. Recognizing which one a problem is steering you toward — an explicit "diagonalize and find $A^k$" instruction versus a bare "find $A^{-1}$" or "simplify $A^k$" — is itself half the battle in this topic.

Problem. Diagonalize $A=\begin{pmatrix}4&1\\2&3\end{pmatrix}$, use it to compute $A^3$, and then verify $A^{-1}$ two independent ways.

Step 1. From the eigenvalue computation for this matrix (previous topic), $\lambda_1=5$ with eigenvector $\mathbf v_1=(1,1)$, and $\lambda_2=2$ with eigenvector $\mathbf v_2=(1,-2)$. Build $P=\begin{pmatrix}1&1\\1&-2\end{pmatrix}$, $D=\begin{pmatrix}5&0\\0&2\end{pmatrix}$.

Step 2. Invert $P$: $\det P=1(-2)-1(1)=-3$, so $P^{-1}=\dfrac{1}{-3}\begin{pmatrix}-2&-1\\-1&1\end{pmatrix}=\begin{pmatrix}2/3&1/3\\1/3&-1/3\end{pmatrix}$.

Step 3. Apply $A^3=PD^3P^{-1}$ with $D^3=\begin{pmatrix}125&0\\0&8\end{pmatrix}$. First $PD^3=\begin{pmatrix}125&8\\125&-16\end{pmatrix}$, then multiplying by $P^{-1}$ and simplifying each entry gives

This matches direct multiplication ($A^2=\begin{pmatrix}18&7\\14&11\end{pmatrix}$, $A^3=A^2A$ gives the same matrix), confirming the diagonalization route.

Answer: $A^3=\begin{pmatrix}86&39\\78&47\end{pmatrix}$.

Competitive stretch

Find $A^{-1}$ two ways for the same $A$, to see diagonalization and Cayley–Hamilton agree. Via diagonalization, $A^{-1}=PD^{-1}P^{-1}$ with $D^{-1}=\operatorname{diag}(1/5,1/2)$, which works out to $\dfrac{1}{10}\begin{pmatrix}3&-1\\-2&4\end{pmatrix}$. Via Cayley–Hamilton, $\operatorname{tr}A=7$, $\det A=10$, so $A^2-7A+10I=O$, rearranged to $A^{-1}=\dfrac{7I-A}{10}=\dfrac{1}{10}\begin{pmatrix}3&-1\\-2&4\end{pmatrix}$ — the identical matrix, reached without ever computing an eigenvector. Notice the trade-off in effort: the diagonalization route needed the eigenvalues and eigenvectors from the previous topic plus two matrix multiplications, while the Cayley–Hamilton route needed only a trace and a determinant read straight off $A$. When only $A^{-1}$ is wanted — not a general power — Cayley–Hamilton is usually the faster path for a $2\times2$ matrix, since it skips the eigenvector computation entirely.

• Writing $A=PDP^{-1}$ when $A$ does not actually have $n$ independent eigenvectors — check geometric multiplicity equals algebraic multiplicity for every eigenvalue first, or $P$ is singular and the formula is meaningless.
• Mismatching column order: column $j$ of $P$ must be the eigenvector for the eigenvalue sitting in position $j$ of $D$ — swapping the order silently gives the wrong matrix.
• In Cayley–Hamilton, forgetting the constant term multiplies the identity matrix $I$, not the scalar $1$ — the equation is $p(A)=O$, a matrix equation, never a scalar one.
• Assuming the $A^k=PD^kP^{-1}$ shortcut works for a non-diagonalizable matrix — Cayley–Hamilton itself still holds (it needs no diagonalizability), but the clean power formula needs $P$ invertible.

Ask the tutor: want to work through diagonalizing a $3\times3$ matrix or using Cayley–Hamilton to simplify a higher power step by step?