Diagonalization

Computing $A^{100}$ by repeated multiplication is ninety-nine matrix products. If $A$ has a basis of eigenvectors there is a shortcut: rewrite it as $PD{P}^{-1}$ with $D$ diagonal, and

where $D^{100}$ is $n$ numbers raised to the hundredth power. The inner factors cancel in pairs, so nothing else survives.

That is the payoff, and it extends: any polynomial in $A$, the matrix exponential, the long-run behaviour of a repeated process. What makes it work is that in eigenvector coordinates the map never mixes one coordinate into another.

The question is when such a basis exists. The answer is already available from the previous unit, compare the two multiplicities for each eigenvalue, and the two ways it can fail are not alike.

Why $AP=PD$ is the natural form. Read both sides column by column. The $j$th column of $AP$ is $A{v}_j$. The $j$th column of $PD$ is ${\lambda }_j{v}_j$, because multiplying on the right by a diagonal matrix scales each column. So $AP=PD$ is precisely the list of statements $A{v}_j={\lambda }_j{v}_j$. The factorisation is the eigenvector equations written at once.

This is also why $P$ must be invertible: rearranging to $A=PD{P}^{-1}$ requires it, and $P$ is invertible exactly when its columns are independent, which is exactly the condition that the eigenvectors form a basis.

Left and right multiplication differ. $PD$ scales the columns of $P$; $DP$ scales its rows. Only the first matches the eigenvector equations, so writing $A=P^{-1}DP$ instead of $PD{P}^{-1}$ is a genuine error rather than a notational preference. It diagonalises with respect to the wrong basis.

The pairing is positional. Column $j$ of $P$ and entry $(j,j)$ of $D$ must belong to the same eigenvalue. Reordering the eigenvectors is allowed provided the diagonal is reordered identically; changing one without the other produces a $P$ and $D$ that fail on multiplication, and the failure is silent until the product is checked.

$P$ is never unique. Scaling any eigenvector by a nonzero constant gives another valid $P$, since the eigenvector equation is unaffected. An eigenspace of dimension 2 or more admits infinitely many choices of basis. So "the diagonalization" is a mild abuse: $D$ is determined up to the order of its entries, and $P$ is not determined at all.

Similar matrices. $A$ and $B$ are similar when $B=P^{-1}AP$ for some invertible $P$. Similar matrices are the same map written in two bases, which is why they share eigenvalues, characteristic polynomial, determinant, trace and rank. Diagonalization asks whether the simplest possible representative, a diagonal matrix, is available in the similarity class.

Theorem. $A$ is diagonalizable over $F$ if and only if the eigenspace dimensions sum to $n$; equivalently, the characteristic polynomial splits over $F$ and geometric multiplicity equals algebraic multiplicity for every eigenvalue.

Proof. Suppose $A=PD{P}^{-1}$. The columns of $P$ are independent and each satisfies $A{v}_j={\lambda }_j{v}_j$, so they are $n$ independent eigenvectors and the eigenspaces jointly contain a basis; their dimensions therefore sum to at least $n$, and to exactly $n$ since they are independent subspaces of an $n$-dimensional space.

Conversely, if the dimensions sum to $n$, collect a basis from each eigenspace. Eigenvectors for distinct eigenvalues are independent, so the collection is independent, and having $n$ elements it is a basis. Taking those as the columns of $P$ gives $AP=PD$ with $P$ invertible. $◼$

Lemma (independence across eigenvalues). Eigenvectors belonging to distinct eigenvalues are linearly independent.

Proof sketch. Suppose not, and take a shortest vanishing combination $\sum _i{c}_i{v}_i=0$ with all $c_i\ne 0$ and the ${\lambda }_i$ distinct. Apply $A$ to get $\sum _i{c}_i{\lambda }_i{v}_i=0$, and subtract ${\lambda }_1$ times the original to get $\sum _{i\ge 2}{c}_i({\lambda }_i-{\lambda }_1)v_i=0$. Every coefficient is nonzero because the eigenvalues are distinct, and the combination is shorter, contradicting minimality. $◼$

Corollary. $n$ distinct eigenvalues force diagonalizability: each has algebraic multiplicity 1, so its geometric multiplicity is 1 as well by the bound from the previous unit, and the total is $n$.

The converse fails. $I$ has one eigenvalue of algebraic multiplicity $n$, geometric multiplicity $n$, and is already diagonal. Repeated eigenvalues are compatible with diagonalizability whenever the eigenspace is large enough, so a repeated root is a reason to compute the eigenspace, never a verdict on its own.

What fails when the criterion fails. Two separate things, which the non-example block separates. The polynomial may not split over $F$, in which case a larger field repairs it. Or some eigenvalue may have geometric multiplicity below its algebraic multiplicity, in which case nothing repairs it: the rank of $A-\lambda I$ does not change under field extension, so a defective matrix is defective over every field containing its eigenvalues. The best available representative is then the Jordan form, which is block-diagonal rather than diagonal.

1. Find the eigenvalues with their algebraic multiplicities, over the stated field.
2. If the polynomial does not split, stop: not diagonalizable over this field. Say whether it would be over a larger one.
3. For each eigenvalue, compute a basis of $E_{\lambda }$. The number of basis vectors is the geometric multiplicity.
4. Sum the geometric multiplicities. Equal to $n$: diagonalizable. Less than $n$: defective, and no field repairs it.
5. Build $P$ with those basis vectors as columns, in any order you choose.
6. Build $D$ with the matching eigenvalues on the diagonal, in the same order. Each eigenvalue repeated as often as its eigenvectors appear.
7. Check. Verify $AP=PD$ column by column. This is cheaper than forming $P^{-1}$ and catches a pairing error immediately.

Shortcuts for step 1–4.

When you need $P^{-1}$. Only for $A=PD{P}^{-1}$ itself or for computing powers. For a $2\times 2$,

Computing a power. $A^k=P{D}^k{P}^{-1}$, with $D^k$ obtained by raising each diagonal entry to the $k$th power. Multiply left to right: $P$ times $D^k$ scales the columns of $P$, then multiply by $P^{-1}$. The result must have integer entries when $A$ does, which is a useful check on the arithmetic.

The check worth running at the end. The trace of $D$ must equal the trace of $A$, and the product of the diagonal of $D$ must equal $detA$. Both are free and catch a mis-transcribed eigenvalue.

Steps 1–4: the verdict. From the eigenvalue unit, $p(\lambda )={\lambda }^2-7\lambda +10$ with roots $\lambda =5$ and $\lambda =2$, each of algebraic multiplicity 1, and

Two distinct real eigenvalues, so the geometric multiplicities are each 1 and sum to $2=n$. Diagonalizable over $\mathbb{R}$, and the corollary would have told us so from distinctness alone.

Steps 5–6: build $P$ and $D$. Taking the eigenvectors in the order $(1,1)$ then $(1,-2)$:

Column 1 of $P$ is the eigenvector for 5, and $D_{11}=5$. The pairing is what makes the next step work.

Step 7: check $AP=PD$.

The columns say exactly $A(1,1{)}^{\mathsf{T}}=5(1,1{)}^{\mathsf{T}}$ and $A(1,-2{)}^{\mathsf{T}}=2(1,-2{)}^{\mathsf{T}}$.

The inverse. $detP=(1)(-2)-(1)(1)=-3$, so

Confirming the factorisation: , and equivalently .

The fifth power. $D^5=\mathrm{diag}(5^5,2^5)=\mathrm{diag}(3125,32)$, so

Computing $A^5$ by four successive multiplications gives the same matrix. The entries are integers, as they must be since $A$ has integer entries. A useful check that the fractions in $P^{-1}$ cancelled correctly.

Why the cancellation works.

and the same collapse repeats, leaving $A^k=P{D}^k{P}^{-1}$ for every $k$. Nothing about the argument depends on $k$ being small.

Both eigenvalues exceed 1, so every nonzero vector grows under repetition; since , the $E_5$ component grows faster and dominates. After enough applications almost every vector points nearly along $(1,1)$, which is visible in $A^5$, whose columns are close to proportional to $(1,1)$ relative to their size.

Distinct eigenvalues: decided without computing an eigenspace. is lower triangular, so its eigenvalues are $2$, $3$ and $-1$, three distinct values in a $3\times 3$. Each $det(B-\lambda I)=0$. By the corollary it is diagonalizable, and no eigenspace need be computed to know that. Computing them is still required to build $P$.

Already diagonal. $\mathrm{diag}(7,-2)$ is diagonalizable with $P=I$ and $D$ equal to itself. This is the case a procedure should recognise rather than grind through.

Repeated eigenvalue, still diagonalizable. has the single eigenvalue 3 with algebraic multiplicity 2. Here $S-3I$ is the zero matrix, so its kernel is all of ${\mathbb{R}}^2$ and the geometric multiplicity is 2 as well. Every nonzero vector is an eigenvector; any basis of ${\mathbb{R}}^2$ serves as the columns of $P$. The sum of geometric multiplicities is 2, so the criterion is met.

This is the case that refutes "repeated root means defective". The repetition is compatible with diagonalizability; the size of the eigenspace decides it.

Repeated eigenvalue, defective. has the same characteristic polynomial $(\lambda -3{)}^2$ as $S$, same eigenvalue, same algebraic multiplicity. But has rank 1, so its kernel is one-dimensional: $E_3=\mathrm{span}\{(1,0)\}$.

The geometric multiplicities sum to 1, short of 2, so $Q$ is not diagonalizable. Any attempted $P$ would need two columns drawn from a one-dimensional space, making them dependent and $P$ singular, for instance has determinant 0.

The decisive pair. $S$ and $Q$ have identical characteristic polynomials and opposite verdicts. Nothing visible in the polynomial distinguishes them; the rank of $A-\lambda I$ does. That is why step 3 of the procedure cannot be skipped whenever an eigenvalue repeats.

Obstruction 1: the polynomial does not split. has $p(\lambda )={\lambda }^2+1$, with no real roots. Over $\mathbb{R}$ there are no eigenvalues at all, so certainly no basis of eigenvectors, and $R$ is not diagonalizable there.

Over $\mathbb{C}$ the roots are $\pm i$, distinct, so $R$ is diagonalizable with $D=\mathrm{diag}(i,-i)$ and columns $(1,-i)$ and $(1,i)$. The obstruction was the field, and enlarging it removed the obstruction entirely.

Obstruction 2: not enough eigenvectors. has $p(\lambda )=(\lambda -3{)}^2$, which splits over $\mathbb{R}$ already. There is no larger field to move to. The failure is that $\mathrm{rank}(Q-3I)=1$, so $E_3$ is one-dimensional and only one independent eigenvector exists.

Rank is unchanged by field extension, so $Q$ is not diagonalizable over $\mathbb{C}$ either, nor over any field containing 3. This obstruction is permanent, and the Jordan form , already in that form, is the best available.

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Mistake: concluding from the repeated root. " $(\lambda -3{)}^2$, so not diagonalizable." $S=3I$ has the same polynomial and is diagonal. The root's multiplicity permits a shortfall without implying one; only $\dim E_{\lambda }$ decides.

Mistake: mismatching $P$ and $D$. Listing eigenvectors as columns in one order and eigenvalues on the diagonal in another. Each column then satisfies the wrong equation, $AP\ne PD$, and the error is silent until the product is checked, which is why the check is step 7 rather than optional.

Mistake: writing $A=P^{-1}DP$. The correct form is $A=PD{P}^{-1}$ with the eigenvectors as the columns of $P$. The reversed version diagonalises with respect to the wrong basis and gives wrong answers for powers, while looking symmetric enough to pass unexamined.

The distinction to carry. Missing roots are a property of the field and are removable. Missing eigenvectors are a property of the matrix and are not. Reporting "not diagonalizable" without saying which one applies leaves out the part that decides whether anything can be done about it.