Repeated Eigenvalues and Complete Diagonalization

Proof of the finite real matrix product laws

Let $m,n,p,q$ be positive integers, let $A\in\mathbb R^{m\times n}$, $B\in\mathbb R^{n\times p}$, and $C\in\mathbb R^{p\times q}$. All the following arguments use the row-by-column definition and finite real sums. For any output position $(i,\ell)$,

$$\begin{aligned} ((AB)C)_{i\ell} &=\sum_{k=1}^{p}\left(\sum_{j=1}^{n}a_{ij}b_{jk}\right)c_{k\ell}\\ &=\sum_{j=1}^{n}a_{ij}\left(\sum_{k=1}^{p}b_{jk}c_{k\ell}\right) =(A(BC))_{i\ell}. \end{aligned}$$

The middle equality only reorders two finite sums and uses real distributivity. Both products are $m\times q$, so equality at every position proves associativity. No factor has changed its order.

For $D\in\mathbb R^{n\times p}$, the same definition gives

$$[A(B+D)]_{ik}=\sum_{j=1}^{n}a_{ij}(b_{jk}+d_{jk}) =(AB)_{ik}+(AD)_{ik}.$$

For $E\in\mathbb R^{m\times n}$, it gives

$$[(A+E)B]_{ik}=\sum_{j=1}^{n}(a_{ij}+e_{ij})b_{jk} =(AB)_{ik}+(EB)_{ik}.$$

Thus both distributive laws hold with the displayed matching sizes. For every real scalar $c$,

$$[(cA)B]_{ik}=\sum_j(ca_{ij})b_{jk} =c\sum_j a_{ij}b_{jk} =\sum_j a_{ij}(cb_{jk})=[A(cB)]_{ik},$$

proving $(cA)B=c(AB)=A(cB)$ by entrywise equality. In particular, for $x,y\in\mathbb R^n$ and real $s,t$,

$$[A(sx+ty)]_i=\sum_j a_{ij}(sx_j+ty_j) =s(Ax)_i+t(Ay)_i.$$

This proves the matrix-vector linearity used when comparing solutions of the same linear system. All identities concern compatible finite real matrices; they do not assert commutativity or cancellation of arbitrary matrix factors.

Real inner products, their norms and the needed inequalities

An inner product on a real vector space is a map $\langle u,v\rangle\in\mathbb R$ that is symmetric, linear in each argument, and positive definite: $\langle v,v\rangle>0$ for $v\ne0$. Linearity gives $\langle0,v\rangle=0$. Define $\|v\|=\sqrt{\langle v,v\rangle}$. It is zero exactly at zero and satisfies $\|cv\|=|c|\|v\|$. The standard example on $\mathbb R^m$ is $\langle u,v\rangle=\sum_{i=1}^mu_iv_i$. Vectors are orthogonal when their inner product is zero, and a list is orthonormal when distinct members are orthogonal and each member has norm one.

If $v\ne0$, insert $t=\langle u,v\rangle/\|v\|^2$ into the nonnegative square $\|u-tv\|^2$. Expanding by bilinearity gives $\|u\|^2-\langle u,v\rangle^2/\|v\|^2\ge0$, hence $|\langle u,v\rangle|\le\|u\|\|v\|$. If $v=0$ the same inequality holds directly. Equality for $v\ne0$ means $\|u-tv\|=0$, equivalently $u=tv$; the zero-vector cases also give equality. Finally expand $\|u+v\|^2$ and apply the bound to its cross term to obtain $\|u+v\|^2\le(\|u\|+\|v\|)^2$. Both sides to be square-rooted are nonnegative, so the triangle inequality follows. Thus this is a norm, and normalization of any nonzero vector divides by a positive real number. No complex bilinear convention is being used.

Finite basis existence and extension, without a counting assumption

Let $V$ be a real vector space supplied with a finite spanning list $t_1,\ldots,t_N$. First prove that every independent list $u_1,\ldots,u_k$ has length at most $N$. After $j$ insertions, maintain a spanning list consisting of $u_1,\ldots,u_j$ and $N-j$ surviving old vectors. Expand $u_{j+1}$ in this list. Some coefficient on a surviving old vector is nonzero; otherwise $u_{j+1}$ would be in the span of the earlier $u_i$, contradicting independence. Solve for that old vector and replace it by $u_{j+1}$. The new list spans the same space and keeps every earlier insertion. If $k>N$, after $N$ steps the first $N$ independent vectors already span $V$, contradicting independence of $u_{N+1}$. This proves the bound, including $N=0$.

Every subspace $W$ has a finite basis. Start with the empty independent list. If its current span is not $W$, choose one vector of $W$ outside that span. A relation with nonzero coefficient on the new vector would express it in the old span, so the enlarged list remains independent. The proved bound permits at most $N$ additions; therefore the process stops with an independent spanning list for $W$. If $W=\{0\}$, it stops immediately with the empty basis. Starting instead with any given independent list in $W$ proves that list extends to a basis of $W$. Only a bounded finite selection is used. Finally, if two bases have sizes $p,q$, use each as the finite spanning list in the bound to obtain $p\le q$ and $q\le p$. Thus basis size, called dimension, is well-defined. This supplies the existence, extension and dimension facts used in rank–nullity.

Build the determinant from smaller matrices

Definition

Submatrix, minor, and cofactor

Let $A=[a_{ij}]$ be an $n\times n$ matrix with $n\ge 2$. For indices $1\le i,j\le n$, delete row $i$ and column $j$ simultaneously. The resulting $(n-1)\times(n-1)$ matrix is the submatrix $A(i\mid j)$.

The minor belonging to $a_{ij}$ is the scalar

$$M_{ij}=\det\bigl(A(i\mid j)\bigr).$$

The cofactor belonging to $a_{ij}$ is the signed scalar

$$C_{ij}=(-1)^{i+j}M_{ij}.$$

The signs form a checkerboard whose upper-left sign is positive:

$$\begin{bmatrix} +&-&+&\cdots\\ -&+&-&\cdots\\ +&-&+&\cdots\\ \vdots&\vdots&\vdots&\ddots \end{bmatrix}.$$

The notation $A(i\mid j)$ names a matrix, $M_{ij}$ is a determinant of that matrix, and $C_{ij}$ is that scalar after the sign has been attached. Keeping these three objects distinct prevents a common double-sign error.

Definition

Recursive definition of the determinant

For a $1\times1$ matrix $[a_{11}]$, define

$$\det([a_{11}])=a_{11}.$$

For an $n\times n$ matrix with $n>1$, define its determinant by expansion along the first row:

$$\det(A)=\sum_{j=1}^{n}a_{1j}C_{1j} =\sum_{j=1}^{n}(-1)^{1+j}a_{1j}M_{1j}.$$

This is a definition, not yet a theorem about other rows or columns. Each minor in the sum has size $n-1$, so repeated use of the definition eventually reaches the $1\times1$ base case. No $0\times0$ determinant convention is needed: the submatrix and minor definitions are used only for $n\ge2$.

Arbitrary cofactor expansion from the recursive determinant

Let $A$ be real $n\times n$, $n\ge1$, with determinant defined by the one-by-one base case and first-row recursion. Let $S_n$ be the permutations of $1,\ldots,n$. Define the inversion number $\operatorname{inv}(\sigma)$ as the number of pairs $r<s$ with $\sigma(r)>\sigma(s)$, and put $\operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}$.

Suppose $\sigma(i)=j$ and delete position $i$ and value $j$, keeping the relative orders of the remaining positions and values; the resulting permutation is $\tau\in S_{n-1}$. If $u$ entries before position $i$ are smaller than $j$, the inversions removed are $(i-1-u)+(j-1-u)=i+j-2-2u$. Consequently

$$\operatorname{sgn}(\sigma)=(-1)^{i+j}\operatorname{sgn}(\tau).$$

Define $P_n(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{r=1}^n a_{r,\sigma(r)}$. For $n=1$, it is the recursive determinant. For $n>1$, grouping by $\sigma(1)=j$ and using the sign identity gives $P_n(A)=\sum_j(-1)^{1+j}a_{1j}P_{n-1}(A(1\mid j))$. Induction therefore proves $P_n(A)=\det A$.

Now fix any row $i$ and group the same finite permutation sum by $\sigma(i)=j$. Deletion is a bijection to all permutations of the corresponding minor, so

$$\det A=\sum_{j=1}^n(-1)^{i+j}a_{ij}\det(A(i\mid j)).$$

For a fixed column $j$, each permutation has a unique position $i$ with $\sigma(i)=j$. Grouping by that position gives

$$\det A=\sum_{i=1}^n(-1)^{i+j}a_{ij}\det(A(i\mid j)).$$

Both expansion formulas concern $n\ge2$, so no empty determinant convention was introduced. This proves the general expansion theorem without assuming alternating row rules or the determinant product theorem.

Why the structural rules follow

Zero rows and zero columns

Theorem

A zero row or zero column gives determinant zero

If every entry in one row or one column of a square matrix is zero, then its determinant is zero.

Theorem

A row or column with one possible nonzero entry

For a square matrix of size $n\ge2$, suppose all entries in row $k$ are zero except possibly $c_{k\ell}$. Then

$$\det(C)=c_{k\ell}(-1)^{k+\ell}\det\bigl(C(k\mid\ell)\bigr).$$

The same formula holds when all entries in column $\ell$ are zero except possibly $c_{k\ell}$. The formula includes the case $c_{k\ell}=0$.

For $n=1$, a matrix with a zero row is $[0]$, whose determinant is $0$ by the base definition. For $n\ge2$, expand along that row using the arbitrary-row theorem. Every term has a zero entry as its factor, so every term is zero and hence $\det(A)=0$.

The zero-column statement can be proved directly by induction, which makes the inheritance of the zero column explicit. For $n=1$, the only $1\times1$ matrix with a zero column is $[0]$, and its determinant is $0$. Assume the statement for $n=k$. Let $C$ be a $(k+1)\times(k+1)$ matrix with zero column $\ell$. Expand along the first row. The term from column $\ell$ is zero. In every other term, delete row $1$ and a column $j\ne\ell$; the original zero column survives as a zero column of the resulting $k\times k$ submatrix (with its index shifted when $j\lt\ell$). By the induction hypothesis each such submatrix has determinant zero. Hence every term is zero, proving $\det(C)=0$.

For the one-support theorem, the row case follows by expanding along row $k$: all terms except the $\ell$-term vanish. For the column case, expand along row $k$. The $\ell$-term remains. For each $j\ne\ell$, the submatrix $C(k\mid j)$ inherits a zero column from column $\ell$ of $C$, so its determinant is zero by the zero-column result. Therefore only $c_{k\ell}(-1)^{k+\ell}\det(C(k\mid\ell))$ remains. This is the precise reason that a sparse row or column collapses the calculation to one minor.

Triangular matrices

Theorem

Triangular determinant theorem

If $T$ is upper triangular or lower triangular, then

$$\det(T)=t_{11}t_{22}\cdots t_{nn}.$$

We prove the upper-triangular statement by induction on $n$. For $n=1$, it is the determinant definition. Suppose it holds for upper-triangular $k\times k$ matrices, and let $T$ be upper triangular of size $k+1$. Its last row has zero entries except possibly $t_{k+1,k+1}$. Expanding along that last row gives

$$\det(T)=t_{k+1,k+1}\det\bigl(T(k+1\mid k+1)\bigr),$$

because the sign is $(-1)^{2k+2}=1$. The retained submatrix is still upper triangular, with diagonal $t_{11},\ldots,t_{kk}$. By induction its determinant is $t_{11}\cdots t_{kk}$, giving the required product through $t_{k+1,k+1}$. For a lower-triangular matrix, expand along the first row, whose only possible nonzero entry is $t_{11}$; the retained submatrix is lower triangular with diagonal $t_{22},\ldots,t_{nn}$, and the same induction proves the result.

Multilinearity in one fixed row

Theorem

Multilinearity in one fixed row

Fix every row except row $p$. If the new row is $\alpha u+\beta v$, then

$$\det(\text{same rows except }\alpha u+\beta v) =\alpha\det(\text{same rows except }u) +\beta\det(\text{same rows except }v).$$

For $n=1$, the base definition gives

$$\det([\alpha a+\beta b])=\alpha a+\beta b =\alpha\det([a])+\beta\det([b]).$$

For $n\ge2$, fix every row except row $p$. Let matrices $A,B,C$ agree in all those fixed rows, and suppose their $p$th rows satisfy $c_p=\alpha a_p+\beta b_p$. For each column $j$, deleting row $p$ gives the same submatrix:

$$A(p\mid j)=B(p\mid j)=C(p\mid j).$$

Expanding $\det(C)$ along row $p$ and using the entrywise relation $c_{pj}=\alpha a_{pj}+\beta b_{pj}$ gives

$$\begin{aligned} \det(C) &=\sum_{j=1}^{n}(\alpha a_{pj}+\beta b_{pj})(-1)^{p+j}\det(C(p\mid j))\\ &=\alpha\det(A)+\beta\det(B). \end{aligned}$$

The equality works precisely because every other row is held fixed; changing two rows at once is a different statement.

Adjacent swaps and repeated rows

Theorem

Repeated rows force determinant zero

If a square matrix has two identical rows, then its determinant is $0$.

Let $A$ have adjacent rows $r_q,r_{q+1}$, and let $B$ be obtained by swapping those two physical rows. Expand $\det(A)$ along row $q$. Expand $\det(B)$ along row $q+1$, where the row $r_q$ has moved. For each column $j$, the retained submatrix in the two expansions is the same matrix: deleting the position occupied by $r_q$ leaves the other row in the same order. However, the signs differ:

$$(-1)^{q+1+j}=-(-1)^{q+j}.$$

The entries and minors therefore match term by term with the opposite sign, so $\det(B)=-\det(A)$. Swapping rows in positions $p\lt q$ can be realised by $2(q-p)-1$ adjacent swaps, an odd number, and hence also changes the sign.

If rows $p$ and $q$ of $A$ are identical, the matrix obtained by swapping them is still $A$. The swap result then gives $\det(A)=-\det(A)$. Since the entries are real, this implies $2\det(A)=0$ and therefore $\det(A)=0$. This proof does not rely on a later row-replacement rule.

Indexing and strategy

For each term, keep three objects visible: the selected entry, its submatrix after deleting one row and one column, and the minor obtained by taking that submatrix determinant. Write the checkerboard sign from the original row and column positions before evaluating the minor. When a minor is expanded again, its local indices guide the inner sign; they do not replace the outer sign.

The arbitrary-row and arbitrary-column theorem makes the choice of expansion line strategic. Choose a line with many zeros or simple surviving minors, and state that choice before calculating. A zero entry removes a branch immediately, whereas a zero minor must be established by its own calculation. The recursive first-row formula defines the determinant; arbitrary-line expansion is the theorem that permits this computational choice.

The case in which the eigenspace is the whole space

Let $n\geq1$ be an integer, let $A\in\mathbb R^{n\times n}$, and let $\lambda$ be an eigenvalue. Write $g=\dim E_\lambda$. In the multiplicity proof, first separate $g=n$. A basis $v_1,\ldots,v_n$ of $E_\lambda$ is then a basis of $\mathbb R^n$. For $x=\sum_i c_iv_i$, linearity gives $Ax=\sum_i c_i\lambda v_i=\lambda x$, so $A=\lambda I_n$. Its monic characteristic polynomial is $\det(tI_n-A)=(t-\lambda)^n$, by the recursive determinant formula for a diagonal matrix. Consequently $a_\lambda=g=n$. For $g<n$, the block $B$ in the existing proof has positive size $n-g$, and that proof applies. No determinant of a $0\times0$ matrix is required.

Preparation and learning goals

You should be able to multiply matrices, calculate a determinant by elimination or cofactors, solve a homogeneous system, and find a basis of a null space. You should also know that a basis of $\mathbb R^n$ contains $n$ independent vectors. All matrices and vectors are real unless a different field is explicitly stated.

An eigenvalue calculation has two distinct stages. Factoring a characteristic polynomial finds possible scaling factors. Solving the associated homogeneous systems finds the directions on which those factors act. A repeated factor does not complete the second stage for us. This chapter develops a complete decision procedure, including repeated eigenvalues, parameter values where roots collide, singular matrices, and exact repeated-action formulas.

After studying it, you should be able to justify a diagonalization rather than merely display one, diagnose exactly why a proposed diagonalization fails, and interpret powers using the components of an initial vector in an eigenbasis. No Jordan canonical form is assumed.

1. Eigenvectors, eigenspaces, and two kinds of multiplicity

For a square $n\times n$ matrix $A$, an eigenvector for a scalar $\lambda$ is a nonzero vector $v$ such that $Av=\lambda v$. The associated eigenspace is

$$E_\lambda=\{v:(A-\lambda I)v=0\}=N(A-\lambda I).$$

The eigenspace is a subspace and includes zero; zero itself is not an eigenvector. The scalar $\lambda$ is an eigenvalue exactly when this null space is nonzero. By the determinant criterion for invertibility, that occurs exactly when

$$p_A(\lambda)=0,\qquad p_A(t)=\det(tI-A).$$

We use the monic characteristic polynomial $p_A(t)$ throughout. Using $\det(A-tI)$ instead changes the polynomial by $(-1)^n$, but not its roots or their multiplicities.

The algebraic multiplicity $a_\lambda$ is the multiplicity of $\lambda$ as a root of $p_A$. The geometric multiplicity $g_\lambda$ is $\dim E_\lambda$. The first comes from a polynomial; the second comes from a homogeneous system. For every eigenvalue,

$$1\le g_\lambda\le a_\lambda.$$

Here is a proof of the upper bound. Choose a basis $v_1,\ldots,v_g$ of $E_\lambda$ and extend it to a basis of the whole space. Let $P$ have that basis as columns. Because $Av_j=\lambda v_j$ for $j\le g$, the matrix in the new coordinates has block form

$$P^{-1}AP=\begin{pmatrix}\lambda I_g&C\\0&B\end{pmatrix}.$$

The characteristic determinant of this block triangular matrix is $(t-\lambda)^g\det(tI-B)$. Similarity preserves the characteristic polynomial, since

$$tI-P^{-1}AP=P^{-1}(tI-A)P$$

and determinant multiplicativity cancels $\det P^{-1}\det P$. Thus $(t-\lambda)^g$ divides $p_A(t)$, proving $g\le a_\lambda$. The lower bound follows from the definition of an eigenvalue.

This argument does not say a repeated root has too few eigenvectors. It says that a root of multiplicity three can contribute at most three independent eigenvectors. It may contribute one, two, or three; the null-space calculation decides.

2. Why eigenvectors for different eigenvalues fit together

There is an important independence statement for whole eigenspaces. Suppose $\lambda_1,\ldots,\lambda_s$ are distinct and $w_j\in E_{\lambda_j}$. If

$$w_1+\cdots+w_s=0,$$

then every $w_j$ is zero. We prove this by induction on $s$. The case $s=1$ is immediate. Apply $A-\lambda_sI$ to a relation with $s$ terms. This gives

$$\sum_{j=1}^{s-1}(\lambda_j-\lambda_s)w_j=0.$$

Each term still belongs to its corresponding eigenspace. By the induction hypothesis each is zero. Because the eigenvalues are distinct, $w_j=0$ for $j<s$, and the original relation then gives $w_s=0$.

Consequently, combining a basis from each distinct eigenspace produces an independent list. To see this explicitly, group any proposed linear dependence by eigenspace. The displayed result first makes each group vector zero; independence within each chosen basis then makes every coefficient zero. In particular,

$$\sum_{\lambda\text{ real eigenvalue}}g_\lambda\le n.$$

This is the reason we can count independent eigenvectors by adding eigenspace dimensions. We must not add the dimensions of arbitrary overlapping subspaces in the same way.

3. The complete diagonalization criterion

A real matrix is diagonalizable over $\mathbb R$ if $A=PDP^{-1}$ for an invertible real matrix $P$ and a real diagonal matrix $D$. This is equivalent to having a basis of real eigenvectors.

For the forward direction, multiply $A=PDP^{-1}$ by $P$. In $AP=PD$, column $j$ says that column $j$ of $P$ is an eigenvector with eigenvalue $D_{jj}$. Those columns form a basis because $P$ is invertible. Conversely, put any eigenbasis into the columns of $P$, with the corresponding eigenvalues in the same order down $D$. Columnwise multiplication gives $AP=PD$, and multiplying by $P^{-1}$ gives the diagonalization.

The criterion can therefore be stated in two equivalent ways:

1. The sum of the dimensions of the real eigenspaces is $n$.
2. The characteristic polynomial splits into real linear factors, and $g_\lambda=a_\lambda$ for every distinct eigenvalue.

To justify the equivalence, the independent-list result shows that the first condition supplies an eigenbasis. If the polynomial splits, its algebraic multiplicities sum to $n$. Since each $g_\lambda\le a_\lambda$, their sum equals $n$ exactly when every individual inequality is equality. If an eigenbasis exists, similarity to a real diagonal matrix also shows that the characteristic polynomial splits and has precisely those diagonal entries as roots.

Three useful consequences follow. First, $n$ distinct real eigenvalues are sufficient for diagonalization. Second, distinctness is not necessary: $3I_n$ is already diagonal, with one eigenvalue and an $n$-dimensional eigenspace. Third, invertibility is a different issue. A diagonal matrix may have zero entries; diagonalization does not require all eigenvalues to be nonzero.

A dependable calculation procedure

Compute and factor $p_A(t)$. For each distinct real root, solve $(A-\lambda I)v=0$ and record a basis and its dimension. If there are fewer than $n$ independent vectors in total, stop: there is no real diagonalization. Otherwise form $P$ and $D$, preserving the pairing between columns and eigenvalues. Verify $AP=PD$ and check that $P$ is invertible. Calculate $P^{-1}$ only after those checks. An incorrect inverse cannot repair an incorrect eigenbasis.

When parameters occur, isolate exceptional values before dividing by a parameter expression. A formula for a generic eigenspace can lose valid directions at a special value.

4. A repeated root that supplies a full eigenbasis

Worked example 1. Consider

$$A=\begin{pmatrix}2&0&1\\0&2&0\\0&0&3\end{pmatrix}.$$

It is triangular, so $p_A(t)=(t-2)^2(t-3)$. For $\lambda=2$ the system is

$$z=0,$$

with $x,y$ free. Thus $E_2=\operatorname{span}\{e_1,e_2\}$ and $g_2=2=a_2$. For $\lambda=3$, the equations are $-x+z=0$ and $-y=0$, giving

$$E_3=\operatorname{span}\{(1,0,1)^T\}.$$

There are three independent eigenvectors. A suitable choice is

$$P=\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix},\quad D=\operatorname{diag}(2,2,3),\quad P^{-1}=\begin{pmatrix}1&0&-1\\0&1&0\\0&0&1\end{pmatrix}.$$

The determinant of $P$ is one. Multiplying its three columns by $A$ verifies $AP=PD$, and multiplying $P$ by the displayed inverse gives $I$. The repeated eigenvalue is no obstacle because its eigenspace has the required dimension.

For every integer $k\ge0$, cancellation of consecutive $P^{-1}P$ factors gives $A^k=PD^kP^{-1}$. Therefore

$$A^k=\begin{pmatrix} 2^k&0&3^k-2^k\\0&2^k&0\\0&0&3^k \end{pmatrix}.$$

At $k=0$ this is $I$; at $k=1$ it is $A$. Those checks are necessary but do not replace the derivation for all $k$.

For an initial vector $x_0=(4,-1,2)^T$, its eigenbasis coordinates are $P^{-1}x_0=(2,-1,2)^T$. Thus

$$A^kx_0=(2\cdot2^k+2\cdot3^k,-2^k,2\cdot3^k)^T.$$

The coordinate coefficients are not generally the original coordinates. Here the third eigenvector has a nonzero first component, which is why the first component receives both growth factors.

5. The same characteristic polynomial without diagonalization

Worked example 2. Now let

$$B=\begin{pmatrix}2&1&0\\0&2&0\\0&0&3\end{pmatrix}.$$

Again $p_B(t)=(t-2)^2(t-3)$. But $(B-2I)v=0$ requires $y=z=0$, so $E_2=\operatorname{span}\{e_1\}$ has dimension one. Also $E_3=\operatorname{span}\{e_3\}$. The combined dimension is only two. No real diagonalization exists, even though all roots are real and $\det B=12\ne0$.

Failure of diagonalization does not mean that powers cannot be found. The upper-left block is $2I_2+N$, where

$$N=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad N^2=0.$$

Since $2I_2$ commutes with $N$, expanding a product of $k$ copies leaves only terms containing zero or one copy of $N$: a term with two copies vanishes. There are $k$ ways to choose the single copy. For $k\ge1$,

$$B^k=\begin{pmatrix}2^k&k2^{k-1}&0\\0&2^k&0\\0&0&3^k\end{pmatrix}.$$

Set $B^0=I$ separately. The additional factor $k$ is consistent with the missing eigenvector; it cannot be obtained by pretending there are two independent eigenvectors for the repeated root. This direct block calculation requires no general canonical-form theorem.

The matrices $A$ and $B$ have the same characteristic polynomial and determinant, but different eigenspace dimensions. Polynomial data alone do not determine diagonalizability.

6. Parameter values and colliding eigenvalues

Worked example 3. For real $s$, study

$$C_s=\begin{pmatrix}1&1&0\\0&s&0\\0&0&2\end{pmatrix}.$$

Its characteristic polynomial is $(t-1)(t-s)(t-2)$. If $s\ne1,2$, its three eigenvalues are distinct, so it is diagonalizable. We must still inspect both collision values.

At $s=2$, the eigenvalue two has multiplicity two. Solving $(C_2-2I)v=0$ gives $y=x$, with $z$ free. Thus $E_2=\operatorname{span}\{(1,1,0)^T,e_3\}$ has dimension two. Along with $E_1=\operatorname{span}\{e_1\}$ this gives an eigenbasis.

At $s=1$, solving $(C_1-I)v=0$ gives $y=z=0$. The repeated eigenvalue one has only one independent eigenvector. The remaining eigenvalue two supplies $e_3$, so diagonalization fails.

Hence $C_s$ is diagonalizable exactly when $s\ne1$. For all those values, one formula works, including $s=2$:

$$P_s=\begin{pmatrix}1&1&0\\0&s-1&0\\0&0&1\end{pmatrix},\qquad D_s=\operatorname{diag}(1,s,2).$$

The second column is an $s$-eigenvector because $C_s(1,s-1,0)^T=s(1,s-1,0)^T$. Moreover $\det P_s=s-1$, so the only excluded value is precisely the defective one. At $s=1$ this displayed second column becomes $e_1$, not a second independent vector. Counting columns without checking independence would conceal the failure.

For $s\ne1$ and integers $k\ge1$,

$$C_s^k=\begin{pmatrix} 1&\dfrac{s^k-1}{s-1}&0\\0&s^k&0\\0&0&2^k \end{pmatrix}.$$

The quotient is $1+s+\cdots+s^{k-1}$. Thus its value at $s=1$ extends to $k$, giving the correct direct formula for $C_1^k$. A diagonalizing matrix can become singular at a parameter value even though the power matrix itself has a perfectly finite limit. At $k=0$, every $C_s^0$ is $I$, including $s=0$; no convention about $0^0$ is needed.

7. Symmetry and the scalar field

If $S=S^T$, $Su=\lambda u$, and $Sv=\mu v$, then

$$\lambda u^Tv=(Su)^Tv=u^TSv=\mu u^Tv.$$

Distinct eigenvalues therefore give orthogonal eigenspaces. Within a repeated eigenspace, however, arbitrary independent eigenvectors need not be orthogonal. If $u,v\in E_\lambda$, then every linear combination of them remains in $E_\lambda$. Gram–Schmidt applied within that eigenspace therefore preserves the eigenvector property while making a chosen basis orthonormal.

Worked example 4. For

$$S=\begin{pmatrix}3&1&0\\1&3&0\\0&0&4\end{pmatrix},$$

the vectors $(1,1,0)^T$ and $e_3$ form a basis of $E_4$, while $(1,-1,0)^T$ spans $E_2$. Normalize them to obtain

$$q_1=\frac1{\sqrt2}(1,1,0)^T,\quad q_2=e_3,\quad q_3=\frac1{\sqrt2}(1,-1,0)^T.$$

With $Q=(q_1\ q_2\ q_3)$, direct dot products give $Q^TQ=I$, and $S=Q\operatorname{diag}(4,4,2)Q^T$. In contrast, $(1,1,0)^T$ and $(1,1,1)^T$ are two independent four-eigenvectors whose dot product is two, not zero. Repeated-eigenvalue directions are not automatically mutually perpendicular. We have constructed a complete basis in this example; a proof of the general real spectral theorem is not assumed here.

Finally, real and complex diagonalization are different assertions. The real rotation

$$R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$

has characteristic polynomial $t^2+1$, so it has no real eigenbasis. Over $\mathbb C$, the eigenvalues are $i,-i$, with eigenvectors $(1,-i)^T,(1,i)^T$ respectively, and distinctness gives a complex eigenbasis. This is a field distinction, not a failure of the criterion. Complex arithmetic is not needed for the real calculations above.

8. Exercises

1. A real $3\times3$ matrix has characteristic polynomial $(t-5)^2(t+1)$. Its five-eigenspace is a line. Can it be diagonalizable? State the missing dimension and explain whether zero may be listed as a second eigenvector.
2. Find every eigenspace of $T=\begin{pmatrix}1&2&0\\0&3&0\\0&0&3\end{pmatrix}$. Construct and verify a diagonalization.
3. Decide whether $U=\begin{pmatrix}4&0&0\\0&4&1\\0&0&4\end{pmatrix}$ is diagonalizable. Explain why the number of displayed diagonal entries is not the relevant eigenvector count.
4. For $V=\begin{pmatrix}0&1&0\\0&1&0\\0&0&2\end{pmatrix}$, find an eigenbasis, rank, determinant, and a diagonalization. Does singularity prevent your answer?
5. Let $P=\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}$ and $D=\operatorname{diag}(2,5,5)$. Compute $P^{-1}$ and $A=PDP^{-1}$. Find the whole five-eigenspace, not only one vector.
6. Reconsider $C_s$ from Worked example 3. Write explicit bases at $s=1$ and $s=2$, and explain why the two repeated-root cases have different outcomes.
7. Derive the power formula for $C_s$ for $s\ne1$ and $k\ge1$. Use the finite geometric sum to extend it to $s=1$. Explain why this does not give a diagonalization at $s=1$.
8. Suppose $B=H^{-1}AH$ with $H$ invertible. Prove that $v\mapsto H^{-1}v$ gives an isomorphism from $E_\lambda(A)$ to $E_\lambda(B)$. Deduce that diagonalizability is preserved by similarity, but need not be shared by matrices with the same characteristic polynomial.
9. For $S$ in Worked example 4, calculate $S^k$ explicitly for every integer $k\ge0$. Check its action on $e_3$ and at $k=0$.
10. Let $F=\begin{pmatrix}3/4&0&1/4\\0&1/2&0\\0&0&1/2\end{pmatrix}$. Decompose an arbitrary initial vector $(a,b,c)^T$ into eigenvectors, find $F^k(a,b,c)^T$ for $k\ge0$, and prove convergence to zero. State whether negative powers exist.

9. Complete solutions

1. The five-eigenspace has dimension one, whereas its algebraic multiplicity is two. The minus-one eigenspace has dimension at most one, so the total is at most two, short of the required three. The matrix is not diagonalizable. Zero belongs to each eigenspace but cannot be an eigenvector or increase the span of a list. A second independent five-eigenvector is what is missing.

2. The triangular characteristic polynomial is $(t-1)(t-3)^2$. For one, the second and third equations of $(T-I)v=0$ give $y=z=0$, hence $E_1=\operatorname{span}\{e_1\}$. For three, the only nonzero equation is $-2x+2y=0$, giving $E_3=\operatorname{span}\{(1,1,0)^T,e_3\}$. Set

$$P=\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix},\quad D=\operatorname{diag}(1,3,3),\quad P^{-1}=\begin{pmatrix}1&-1&0\\0&1&0\\0&0&1\end{pmatrix}.$$

The columns are scaled by $1,3,3$ under $T$, so $TP=PD$. Also $\det P=1$ and the displayed inverse multiplies $P$ to $I$, proving $T=PDP^{-1}$.

3. The only eigenvalue is four, of algebraic multiplicity three. The equation $(U-4I)v=0$ is $z=0$, so $E_4=\operatorname{span}\{e_1,e_2\}$ has dimension two. The matrix is not diagonalizable. Diagonal entries of a triangular matrix identify eigenvalues with multiplicity; they are not automatically three independent eigenvectors.

4. For zero, the equations give $y=z=0$, so use $e_1$. For one, they give $x=y,z=0$, so use $(1,1,0)^T$. For two, they give $x=y=0$, so use $e_3$. These three vectors form the columns of the same $P$ as in Solution 2, now with $D=\operatorname{diag}(0,1,2)$. Their direct images verify $VP=PD$, and $P$ is invertible. The original columns are zero, $(1,1,0)^T$, and $(0,0,2)^T$, so rank is two. The determinant is zero. Singularity prevents an inverse for $V$, not an eigenbasis.

5. Reversing the shear gives $P^{-1}=\begin{pmatrix}1&-1&0\\0&1&0\\0&0&1\end{pmatrix}$. Multiplication yields

$$A=\begin{pmatrix}2&3&0\\0&5&0\\0&0&5\end{pmatrix}.$$

The equation $(A-5I)v=0$ is $-3x+3y=0$, hence $E_5=\{(u,u,w)^T:u,w\in\mathbb R\}$. Its basis can be $(1,1,0)^T,e_3$. The one repeated root contributes two independent directions, exactly as required by $D$.

6. At $s=1$, $E_1=\operatorname{span}\{e_1\}$ and $E_2=\operatorname{span}\{e_3\}$, giving only two independent eigenvectors. At $s=2$, $E_1=\operatorname{span}\{e_1\}$ and $E_2=\operatorname{span}\{(1,1,0)^T,e_3\}$, giving three. The differing null-space dimensions, not the mere presence of a repeated factor, explain the differing outcomes.

7. For $s\ne1$,

$$P_s^{-1}=\begin{pmatrix}1&-1/(s-1)&0\\0&1/(s-1)&0\\0&0&1\end{pmatrix}.$$

Multiplying $P_s\operatorname{diag}(1,s^k,2^k)P_s^{-1}$ gives diagonal entries $1,s^k,2^k$ and upper-right entry of the two-by-two block $(s^k-1)/(s-1)$, with all other off-diagonal entries zero. The identity $s^k-1=(s-1)\sum_{j=0}^{k-1}s^j$ shows that this entry extends to $k$ at $s=1$. Direct multiplication, or the square-zero calculation from §5, gives exactly that power of $C_1$. The power formula does not create a missing eigenvector: $P_1$ remains singular and $E_1(C_1)$ remains one-dimensional.

8. If $Av=\lambda v$, then $B(H^{-1}v)=H^{-1}Av=\lambda H^{-1}v$. Thus the map sends the first eigenspace into the second. Its inverse is $w\mapsto Hw$, which gives the reverse implication. Both are linear and invertible, so the eigenspace dimensions agree for every eigenvalue. Transporting an eigenbasis through $H^{-1}$ preserves independence and spanning, proving similarity invariance of diagonalizability. Worked examples 1 and 2 have the same characteristic polynomial but different eigenspace dimensions and opposite outcomes, so equality of characteristic polynomials is not enough.

9. In the first two coordinates, $(1,1)^T$ and $(1,-1)^T$ have eigenvalues four and two. The third coordinate has eigenvalue four. Thus

$$S^k=\begin{pmatrix} (4^k+2^k)/2&(4^k-2^k)/2&0\\ (4^k-2^k)/2&(4^k+2^k)/2&0\\ 0&0&4^k \end{pmatrix}.$$

This follows by multiplying the orthogonal diagonalization, not by guessing entries from initial powers. It maps $e_3$ to $4^ke_3$, as required. At $k=0$ the two-by-two block is the identity and the final diagonal entry is one, so the full matrix is $I$.

10. The vector $e_1$ has eigenvalue $3/4$, while $e_2$ and $(-1,0,1)^T$ both have eigenvalue $1/2$. These form a basis: their column matrix is triangular with diagonal entries one. The decomposition is

$$(a,b,c)^T=(a+c)e_1+be_2+c(-1,0,1)^T.$$

Consequently

$$F^k(a,b,c)^T= \left((3/4)^k(a+c)-(1/2)^kc,\ (1/2)^kb,\ (1/2)^kc\right)^T.$$

At $k=0$ this returns the initial vector. Both geometric factors tend to zero, so each coordinate tends to zero for every fixed initial vector. All eigenvalues are nonzero, so $F$ is invertible and negative integer powers exist as well; they do not share the stated forward-time convergence.

Summary

Factorization locates eigenvalues; null spaces determine whether they supply a basis. Repeated roots require dimension checks, while different eigenspaces combine independently. A valid diagonalization pairs each column with the matching diagonal entry and permits exact power formulas. Singularity, repeated roots, and lack of real roots are separate issues. Parameter collisions must be examined without dividing away the exceptional case.

An explicit decay bound for the final repeated-action exercise

For the exercise matrix

$$F=\begin{pmatrix}3/4&0&1/4\\0&1/2&0\\0&0&1/2\end{pmatrix},$$

the already verified eigenbasis calculation gives, for every integer $k\geq0$,

$$F^k(a,b,c)^T=(3/4)^k(a+c)e_1+(1/2)^k b e_2+(1/2)^k c(-1,0,1)^T.$$

Here is a direct justification of the claimed limit. For $h\geq0$, induction gives $(1+h)^k\geq1+kh$: the step follows from $(1+kh)(1+h)=1+(k+1)h+kh^2$. Thus

$$0<(3/4)^k\leq\frac1{1+k/3},\qquad 0<(1/2)^k\leq\frac1{1+k}.$$

The triangle inequality and the norms of the three displayed directions give

$$\|F^k(a,b,c)^T\|\leq\frac{|a+c|}{1+k/3}+\frac{|b|+\sqrt2|c|}{1+k}.$$

For any fixed real $a,b,c$ and any $\varepsilon>0$, the Archimedean property permits $k$ large enough that each term with nonzero numerator is smaller than $\varepsilon/2$; a zero numerator already contributes zero. Hence the norm tends to zero. This proves the stated convergence for this matrix and all initial vectors without assuming an unproved general theorem about matrix powers.