Section JCF Jordan Canonical Form

From A First Course in Linear Algebra
Version 1.01
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

This Section Under Construction
(Just Lacking JCF Examples)

We have seen in Section IS that generalized eigenspaces are invariant subspaces that in every instance have led to a direct sum decomposition of the domain of the associated linear transformation. This allows us to create a block diagonal matrix representation (Example ISMR4, Example ISMR6). We also know from Theorem RGEN that the restriction of a linear transformation to a generalized eigenspace is almost a nilpotent linear transformation. Of course, we understand nilpotent linear transformations very well from Section NLT and we have carefully determined a nice matrix representation for them.

So here is the game plan for the final push. Prove that the domain of a linear transformation always decomposes into a direct sum of generalized eigenspaces. We have unravelled Theorem RGEN at Theorem MRRGE so that we can formulate the matrix representations of the restrictions on the generalized eigenspaces using our storehouse of results about nilpotent linear transformations. Arrive at a matrix representation of any linear transformation that is block diagonal with each block being a Jordan block.

We will be strictly theoretical at first, proving two major theorems without any explanatory examples, so hang on. Then we can state our main result and move on to several interesting examples.

Subsection UTMR: Upper-Triangular Matrix Representation

Our theorems in this section will each assert that certain bases exist, but what we are really after is the matrix representation that arises from the basis (this is the style of Theorem CFNLT).

Theorem UTMR
Upper-Triangular Matrix Representation
Suppose that $T : V \mapsto V$ is a linear transformation. Then there is a basis, $B$ , for $V$ such that the matrix representation of $T$ relative to $B$ , $M_{B, B}^{T}$ , is an upper-triangular matrix. Each diagonal entry is an eigenvalue of $T$ , and if $λ$ is an eigenvalue of $T$ , then $λ$ occurs $α_{T} (λ)$ times on the diagonal. $□$

Proof We will establish this result using induction of the dimension of $V$ (Technique I). To start suppose that $dim (V) = 1$ . Choose any nonzero vector $v \in V$ , and then realize that $V = 〈\{v\}〉$ . Subsequently, we can describe $T$ completely by $T (v) = β v$ for some $β \in ℂ$ . Thus, we recognize $β$ as one eigenvalue of $T$ , and there are no others (Theorem ME). And $α_{β} (=) 1$ . Our description of $T$ also gives us a matrix representation relative to $\{v\}$ as the $1 \times 1$ matrix with lone entry equal to $β$ . This representation is upper-triangular and the diagonal entry is an eigenvalue of $T$ , occuring $α_{T} (β)$ times.

For the induction step let $dim (V) = m$ , and assume the theorem is true for every linear transformation defined on a vector space of dimension less than $m$ . By Theorem EMHE (suitably converted to the setting of a linear transformation), $T$ has at least one eigenvalue, denote this eigenvalue as $λ$ . (We will remark later about how critical this step is.) We now consider properties of the linear transformation $T - λ I_{V} : V \mapsto V$ .

Let $x$ be an eigenvector of $T$ for $λ$ . By definition $x \neq 0$ . Then

\begin{array}{l} T - λ I_{V} (x) & = T (x) - λ I_{V} (x) & Theorem VSLT \\ = T (x) - λ x & Definition IDLT \\ = λ x - λ x & Definition EELT \\ = 0 & Property AI \end{array}

So $T - λ I_{V}$ is not injective (Theorem KILT). With an argument on dimensions using Theorem RPNDD we can conlude that $T$ does not have full rank, $r (T) < m$ . Define $W$ to be the subspace of $V$ that is the range of $T - λ I_{V}$ , $W = ℛ (T - λ I_{V})$ . The range of a linear transformation is always invariant with respect to the linear transformation, but we want to establish that $W$ is also invariant with respect to just $T$ . To this end, suppose that $w \in W$ ,

\begin{array}{l} T (w) & = T (w) + 0 & Property Z \\ = T (w) + (λ I_{V} (w) - λ I_{V} (w)) & Property AI \\ = (T (w) - λ I_{V} (w)) + λ I_{V} (w) & Property AA \\ = (T (w) - λ I_{V} (w)) + λ w & Definition IDLT \\ = T - λ I_{V} (w) + λ w & Theorem VSLT \end{array}

Since $W$ is the range of $T - λ I_{V}$ , $T - λ I_{V} (w) \in W$ . And by Property SC, $λ w \in W$ . Finally, applying Property AC we see that $T (w) \in W$ and conclude that $W$ is invariant relative to $T$ (Definition IS).

Since $W$ is invariant relative to $T$ we can consider the restriction $T |_{W} : W \mapsto W$ (Definition LTR). $T |_{W}$ is a linear transformation defined on a vector space with dimension less than $m$ , so we can apply the induction hypothesis and conclude that $W$ has a basis, $C = \{w_{1}, w_{2}, w_{3}, \dots, w_{k}\}$ , such that the matrix representation of $T |_{W}$ relative to $C$ is an upper-triangular matrix.

By Theorem DSFOS there exists a second subspace of $V$ , which we will call $U$ , so that $V$ is a direct sum of $W$ and $U$ , $V = W \oplus U$ . Choose a basis $D = \{u_{1}, u_{2}, u_{3}, \dots, u_{ℓ}\}$ for $U$ . So $m = k + ℓ$ by Theorem DSD, and $B = C \cup D$ is basis for $V$ by Theorem DSLI and Theorem G. $B$ is the basis we desire. What does a matrix representation of $T$ look like, relative to $B$ ?

Since $W$ is invariant relative to $T$ , the first $k$ columns of $M_{B, B}^{T}$ will have the upper-triangular matrix representation of $T |_{W}$ using the basis $C$ , $m a t r i x r e p T |_{W} C C$ , in the first $k$ rows. The remaining $ℓ = m - k$ rows will be all zeros. The situation for $T$ on $D$ is not quite as pretty, but it is close.

For $1 \leq i \leq ℓ$ , consider

\begin{array}{l} ρ_{B} (T (u_{i})) & = ρ_{B} (T (u_{i}) + 0) & Property Z \\ = ρ_{B} (T (u_{i}) + (λ I_{V} (u_{i}) - λ I_{V} (u_{i}))) & Property AI \\ = ρ_{B} ((T (u_{i}) - λ I_{V} (u_{i})) + λ I_{V} (u_{i})) & Property AA \\ = ρ_{B} ((T (u_{i}) - λ I_{V} (u_{i})) + λ u_{i}) & Definition IDLT \\ = ρ_{B} (T - λ I_{V} (u_{i}) + λ u_{i}) & Theorem VSLT \\ = ρ_{B} (a_{1} w_{1} + a_{2} w_{2} + a_{3} w_{3} + \dots + a_{k} w_{k} + λ u_{i}) & Definition RLT \\ = [\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{k} \\ 0 \\ ⋮ \\ 0 \\ λ \\ 0 \\ ⋮ \\ 0 \end{matrix}] & Definition VR \end{array}

In the penultimate step of this proof, we have rewritten an element of the range of $T - λ I_{V}$ as a linear combination of the basis vectors for $W$ in $C$ , using the scalars $a_{1}, a_{2}, a_{3}, \dots, a_{ℓ}$ . If we incorporate these $ℓ$ column vectors into the matrix representation $M_{B, B}^{T}$ we find $ℓ$ occurences of $λ$ on the diagonal, and any nonzero entries lying in the first $k$ rows. Together with the $k \times k$ upper-triangular representation in the upper left-hand corner, the entire matrix representation is now clearly upper-triangular. This completes the induction step, so for any linear transformation there is a basis that creates a diagonal matrix representation.

We have one more statement in the conclusion of the theorem to verify. The eigenvalues, and their multiplicities, of $T$ can be computed with the techniques of Chapter E relative to any matrix representation (Theorem EER). We take this approach with our upper-triangular matrix representation $M_{B, B}^{T}$ . Let $d_{i}$ be the diagonal entry of $M_{B, B}^{T}$ in row and column $i$ . Then the characteristic polynomial, computed as a determinant (Definition CP), is

\begin{array}{l} p_{M_{B, B}^{T}} (x) & = (x - d_{1}) (x - d_{2}) (x - d_{3}) \dots (x - d_{m}) \end{array}

So each diagonal entry is an eigenvalue (Theorem EMRCP), and is repeated exactly $α_{T} (λ)$ times (Definition AME). $■$

A key step in this proof was the construction of the subspace $W$ with dimension strictly less than that of $V$ . This required an eigenvalue/eigenvector pair, which was guaranteed to us by Theorem EMHE. Digging deeper, the proof of Theorem EMHE requires that we can factor polynomials completely, into linear factors. This will not always happen if our set of scalars is the reals, $ℝ^{}$ . So this is our final explanation of our choice of the complex numbers, $ℂ$ , as our set of scalars. In $ℂ$ polynomials factor completely, so every matrix has at least one eigenvalue, and an inductive argument will get us to upper-triangular matrix representations.

The complex numbers are an example of an algebraically closed field which contains the roots of any polynomial created with coefficients from itself. If we had chosen to use the reals as our set of scalars, then we would arrive at matrix decompositions known as rational canonical form where the diagonal blocks are derived from certain relevant polynomials associated with the linear transformation. The theory is similar, but not identical to, what we have done here.

Subsection GESD: Generalized Eigenspace Decomposition

We now massage the basis from Theorem UTMR so that it yields an upper-triangular representation that is also block diagonal. The subspaces associated with each block will be generalized eigenspaces, so the most general result will be a decomposition of the domain of a linear transformation into a direct sum of generalized eigenspaces.

Theorem GESD
Generalized Eigenspace Decomposition
Suppose that $T (V) V$ is a linear transformation with distinct eigenvalues $λ_{1}, λ_{2}, λ_{3}, \dots, λ_{m}$ . Then

\begin{array}{l} V & = G_{T} (λ_{1}) \oplus G_{T} (λ_{2}) \oplus G_{T} (λ_{3}) \oplus \dots \oplus G_{T} (λ_{m}) \end{array}

□

Proof Suppose that $dim (V) = n$ and the $n$ (not necessarily distinct) eigenvalues of $T$ are $s c a l a r l i s t ρ n$ . We begin with a basis of $V$ that yields an upper-triangular matrix representation, as guaranteed by Theorem UTMR, $B = \{x_{1}, x_{2}, x_{3}, \dots, x_{n}\}$ . Since the matrix representation is upper-triangular, and the eigenvalues of the linear transformation are the diagonal elements we can choose this basis so that there are then scalars $a_{i j}$ , $1 \leq j \leq n$ , $1 \leq i \leq j - 1$ such that

\begin{array}{l} T (x_{j}) & = \sum_{i = 1}^{j - 1} a_{i j} x_{i} + ρ_{j} x_{j} \end{array}

We now define a new basis for $V$ which is just a slight variation in the basis $B$ . Choose any $k$ and $ℓ$ such that $1 \leq k < ℓ \leq n$ and $ρ_{k} \neq ρ_{ℓ}$ . Define the scalar $α = a_{k l} ∕ (ρ_{ℓ} - ρ_{k})$ . The new basis is $C = \{y_{1}, y_{2}, y_{3}, \dots, y_{n}\}$ where

\begin{array}{l} y_{j} & = x_{j}, j \neq ℓ, 1 \leq j \leq n & y_{ℓ} & = x_{ℓ} + α x_{k} \end{array}

We now compute the values of the linear transformation $T$ with inputs from $C$ , noting carefully the changed scalars in the linear combinations of $C$ describing the outputs. These changes will translate to minor changes in the matrix representation built using the basis $C$ . There are three cases to consider, depending on which column of the matrix representation we are examining. First, assume $j < ℓ$ . Then

\begin{array}{l} T (y_{j}) & = T (x_{j}) \\ = \sum_{i = 1}^{j - 1} a_{i j} x_{i} + ρ_{j} x_{j} \\ = \sum_{i = 1}^{j - 1} a_{i j} y_{i} + ρ_{j} y_{j} \end{array}

That seems a bit pointless. The first $ℓ - 1$ columns of the matrix representations of $T$ relative to $B$ and $C$ are identical. OK, if that was too easy, here’s the main act. Assume $j = ℓ$ . Then

\begin{array}{l} T (y_{ℓ}) & = T (x_{ℓ} + α x_{k}) \\ = T (x_{ℓ}) + α T (x_{k}) \\ = (\sum_{i = 1}^{ℓ - 1} a_{i ℓ} x_{i} + ρ_{ℓ} x_{ℓ}) + α (\sum_{i = 1}^{k - 1} a_{i k} x_{i} + ρ_{k} x_{k}) \\ = \sum_{i = 1}^{ℓ - 1} a_{i ℓ} x_{i} + ρ_{ℓ} x_{ℓ} + \sum_{i = 1}^{k - 1} α a_{i k} x_{i} + α ρ_{k} x_{k} \\ = \sum_{i = 1}^{ℓ - 1} a_{i ℓ} x_{i} + \sum_{i = 1}^{k - 1} α a_{i k} x_{i} + α ρ_{k} x_{k} + ρ_{ℓ} x_{ℓ} \\ = \sum_{\begin{array}{c} i = 1 \\ i \neq k \end{array}}^{ℓ - 1} a_{i ℓ} x_{i} + \sum_{i = 1}^{k - 1} α a_{i k} x_{i} + a_{k l} x_{k} + α ρ_{k} x_{k} + ρ_{ℓ} x_{ℓ} \\ = \sum_{\begin{array}{c} i = 1 \\ i \neq k \end{array}}^{ℓ - 1} a_{i ℓ} x_{i} + \sum_{i = 1}^{k - 1} α a_{i k} x_{i} + a_{k l} x_{k} + α ρ_{k} x_{k} - ρ_{ℓ} α x_{k} + ρ_{ℓ} α x_{k} ρ_{ℓ} x_{ℓ} \\ = \sum_{\begin{array}{c} i = 1 \\ i \neq k \end{array}}^{ℓ - 1} a_{i ℓ} x_{i} + \sum_{i = 1}^{k - 1} α a_{i k} x_{i} + (a_{k l} + α ρ_{k} - ρ_{ℓ} α) x_{k} + ρ_{ℓ} (α x_{k} + x_{ℓ}) \\ = \sum_{\begin{array}{c} i = 1 \\ i \neq k \end{array}}^{ℓ - 1} a_{i ℓ} x_{i} + \sum_{i = 1}^{k - 1} α a_{i k} x_{i} + (a_{k l} + α (ρ_{k} - ρ_{ℓ})) x_{k} + ρ_{ℓ} (x_{ℓ} + α x_{k}) \\ = \sum_{\begin{array}{c} i = 1 \\ i \neq k \end{array}}^{ℓ - 1} a_{i ℓ} y_{i} + \sum_{i = 1}^{k - 1} α a_{i k} y_{i} + (a_{k l} + α (ρ_{k} - ρ_{ℓ})) y_{k} + ρ_{ℓ} y_{ℓ} \end{array}

So how different are the matrix representations relative to $B$ and $C$ in column $ℓ$ ? For $i > k$ , the coefficient of $y_{i}$ is $a_{i j}$ , as in the representation relative to $B$ . It is a different story for $i \leq k$ , where the coefficients of $y_{i}$ may be very different. We are especially interested in the coefficient of $y_{k}$ . In fact, this whole first part of this proof is about this particular entry of the matrix representation. The coefficient of $y_{k}$ is

\begin{array}{l} a_{k l} + α (ρ_{k} - ρ_{ℓ}) & = a_{k l} + \frac{a_{k l}}{ρ_{ℓ} - ρ_{k}} (ρ_{k} - ρ_{ℓ}) \\ = a_{k l} + (- 1) a_{k l} \\ = 0 \end{array}

If the definition of $α$ was a mystery, then no more. In the matrix representation of $T$ relative to $C$ , the entry in column $ℓ$ , row $k$ is a zero. Nice. The only price we pay is that other entries in column $ℓ$ , specifically rows $1$ through $k - 1$ , may also change in a way we can’t control.

One more case to consider. Assume $j > ℓ$ . Then

\begin{array}{l} T (y_{j}) & = T (x_{j}) \\ = \sum_{i = 1}^{j - 1} a_{i j} x_{i} + ρ_{j} x_{j} \\ = \sum_{\begin{array}{c} i = 1 \\ i \neq ℓ, k \end{array}}^{j - 1} a_{i j} x_{i} + a_{ℓ j} x_{ℓ} + a_{k j} x_{k} + ρ_{j} x_{j} \\ = \sum_{\begin{array}{c} i = 1 \\ i \neq ℓ, k \end{array}}^{j - 1} a_{i j} x_{i} + a_{ℓ j} x_{ℓ} + α a_{ℓ j} x_{k} - α a_{ℓ j} x_{k} + a_{k j} x_{k} + ρ_{j} x_{j} \\ = \sum_{\begin{array}{c} i = 1 \\ i \neq ℓ, k \end{array}}^{j - 1} a_{i j} x_{i} + a_{ℓ j} (x_{ℓ} + α x_{k}) + (a_{k j} - α a_{ℓ j}) x_{k} + ρ_{j} x_{j} \\ = \sum_{\begin{array}{c} i = 1 \\ i \neq ℓ, k \end{array}}^{j - 1} a_{i j} y_{i} + a_{ℓ j} y_{ℓ} + (a_{k j} - α a_{ℓ j}) y_{k} + ρ_{j} y_{j} \end{array}

As before, we ask: how different are the matrix representations relative to $B$ and $C$ in column $j$ ? Only $y_{k}$ has a coefficent different from the corresponding coefficient when the basis is $B$ . So in the matrix representations, the only entries to change are in row $k$ , for columns $ℓ + 1$ through $n$ .

What have we accomplished? With a change of basis, we can place a zero in a desired entry (row $k$ , column $ℓ$ ) of the matrix representation, leaving most of the entries untouched. The only entries to possibly change are above the new zero entry, or to the right of the new zero entry. S Suppose we repeat this procedure, starting by “zeroing out” the entry above the diagonal in the second column and first wow. Then we move right to the third column, and zero out the element just above the diagonal in the second row. Next we zero out the element in the third column and first row. Then tackle the fourth column, work upwards from the diagonal, zeroing out elements as we go. Entries above, and to the right will repeatedly change, but newly created zeros will never get wrecked, since they are below, or just to the left of the entry we are working on. Similarly the values on the diagonal do not change either. This entire argument can be retooled in the language of change-of-basis matrices and similarity transformations, and this is the approach taken by Noble in his Applied Linear Algebra. It is interesting to concoct the change-of-basis matrix between the matrices $B$ and $C$ and compute the inverse.

Perhaps you have noticed that we have to be just a bit more careful than the previous paragraph suggests. The definition of $α$ has a denominator that cannot be zero, which restricts our maneuvers to zeroing out entries in row $k$ and column $ℓ$ only when $ρ_{k} \neq ρ_{ℓ}$ . So we do not necessarily arrive at a diagonal matrix. More carefully we can write

\begin{array}{l} T (y_{j}) & = \sum_{\begin{array}{c} i = 1 \\ i : ρ_{i} = ρ_{j} \end{array}}^{j - 1} b_{i j} y_{i} + ρ_{j} y_{j} \end{array}

where the $b_{i j}$ are our new coefficients after repeated changes, the $y_{j}$ are the new basis vectors, and the condition “ $i : ρ_{i} = ρ_{j}$ ” means that we only have terms in the sum involving vectors whose final coefficients are identical diagonal values (the eigenvalues). Now reorder the basis vectors carefully. Group together vectors that have equal diagonal entries in the matrix representation, but within each group preserve the order of the precursor basis. This grouping will create a block diagonal structure for the matrix representation, while otherwise preserving the order of the basis will retain the upper-triangular form of the representation. So we can arrive at a basis that yields a matrix representation that is upper-triangular and block diagonal, with the diagonal entries of each block all equal to a common eigenvalue of the linear transformation.

More carefully, employing the distinct eigenvalues of $T$ , $λ_{i}$ , $1 \leq i \leq m$ , we can assert there is a set of basis vectors for $V$ , $u_{i j}$ , $1 \leq i \leq m$ , $1 \leq j \leq α_{T} (λ_{i})$ , such that

\begin{array}{l} T (u_{i j}) & = \sum_{k = 1}^{j - 1} b_{i j k} u_{i k} + λ_{i} u_{i j} \end{array}

So the subspace $U_{i} = 〈\{u_{i j}| 1 \leq j \leq α_{T} (λ_{i})\}〉$ , $1 \leq i \leq m$ is an invariant subspace of $V$ relative to $T$ and the restriction $T |_{U_{i}}$ has an upper-triangular matrix representation relative to the basis $\{u_{i j}| 1 \leq j \leq α_{T} (λ_{i})\}$ where the diagonal entries are all equal to $λ_{i}$ . Notice too that with this definition,

\begin{array}{l} V & = U_{1} \oplus U_{2} \oplus U_{3} \oplus \dots \oplus U_{m} \end{array}

Whew. This is a good place to take a break, grab a cup of coffee, use the toilet, or go for a short stroll, before we show that $U_{i}$ is a subspace of the generalized eigenspace $G_{T} (λ_{i})$ . This will follow if we can prove that each of the basis vectors for $U_{i}$ is a generalized eigenvector of $T$ for $λ_{i}$ (Definition GEV). We need some power of $T - λ_{i} I_{V}$ that takes $u_{i j}$ to the zero vector. We prove by induction on $j$ (Technique I) the claim that ${(T - λ_{i} I_{V})}^{j} (u_{i j}) = 0$ . For $j = 1$ we have,

\begin{array}{l} (T - λ_{i} I_{V}) (u_{i 1}) & = T (u_{i 1}) - λ_{i} I_{V} (u_{i 1}) \\ = T (u_{i 1}) - λ_{i} u_{i 1} \\ = λ_{i} u_{i 1} - λ_{i} u_{i 1} \\ = 0 \end{array}

For the induction step, assume that if $k < j$ , then ${(T - λ_{i} I_{V})}^{k}$ takes $u_{i k}$ to the zero vector. Then

\begin{array}{l} {(T - λ_{i} I_{V})}^{j} (u_{i j}) & = {(T - λ_{i} I_{V})}^{j - 1} ((T - λ_{i} I_{V}) (u_{i j})) \\ = {(T - λ_{i} I_{V})}^{j - 1} (T (u_{i j}) - λ_{i} I_{V} (u_{i j})) \\ = {(T - λ_{i} I_{V})}^{j - 1} (T (u_{i j}) - λ_{i} u_{i j}) \\ = {(T - λ_{i} I_{V})}^{j - 1} (\sum_{k = 1}^{j - 1} b_{i j k} u_{i k} + λ_{i} u_{i j} - λ_{i} u_{i j}) \\ = {(T - λ_{i} I_{V})}^{j - 1} (\sum_{k = 1}^{j - 1} b_{i j k} u_{i k}) \\ = \sum_{k = 1}^{j - 1} b_{i j k} {(T - λ_{i} I_{V})}^{j - 1} (u_{i k}) \\ = \sum_{k = 1}^{j - 1} b_{i j k} {(T - λ_{i} I_{V})}^{j - 1 - k} ({(T - λ_{i} I_{V})}^{k} (u_{i k})) \\ = \sum_{k = 1}^{j - 1} b_{i j k} {(T - λ_{i} I_{V})}^{j - 1 - k} (0) \\ = \sum_{k = 1}^{j - 1} b_{i j k} 0 \\ = 0 \end{array}

This completes the induction step. Since every vector of the spaning set for $U_{i}$ is an element of the subspace $G_{T} (λ_{i})$ , Property AC and Property SC allow us to conclude that $U_{i} \subseteq G_{T} (λ_{i})$ . Then by Definition S, $U_{i}$ is a subspace of $G_{T} (λ_{i})$ . Notice that this inductive proof could be interpreted to say that every element of $U_{i}$ is a generalized eigenvector of $T$ for $λ_{i}$ , and the algebraic multiplicity of $λ_{i}$ is a sufficiently high power to demonstrate this via the definition for each vector.

We are now prepared for our final argument in this long proof. We wish to establish that the dimension of the subspace $G_{T} (λ_{i})$ is the algebraic multiplicity of $λ_{i}$ . This will be enough to show that $U_{i}$ and $G_{T} (λ_{i})$ are equal, and will finally provide the desired direct sum decomposition.

We will prove by induction (Technique I) the following claim. Suppose that $T : V \mapsto V$ is a linear transformation and $B$ is a basis for $V$ that provides an upper-triangular matrix representation of $T$ . The number of times any eigenvalue $λ$ occurs on the diagonal of the representation is greater than or equal to the dimension of the generalized eigenspace $G_{T} (λ)$ .

We will use the symbol $m$ for the dimension of $V$ so as to avoid confusion with our notation for the nullity. So $dim V = m$ and our proof will proced by induction on $m$ . Use the notation $#_{T} (λ)$ to count the number of times $λ$ occurs on the diagonal of a matrix representation of $T$ . We want to show that

\begin{array}{l} #_{T} (λ) & \geq dim (G_{T} (λ)) \\ = dim (K ({(T - λ)}^{m})) & Theorem GEK \\ = n ({(T - λ)}^{m}) & Definition NOLT \end{array}

For the base case, $dim V = 1$ . Every matrix representation of $T$ is an upper-triangular matrix with the lone eigenvalue of $T$ , $λ$ , as the diagonal entry. So $#_{T} (λ) = 1$ . The generalized eigenspace of $λ$ is not trivial (since by Theorem GEK it equals the regular eigenspace), and is a subspace of $V$ . With Theorem PSSD we see that $dim (G_{T} (λ)) = 1$ .

Now for the induction step, assume the claim is true for any linear transformation defined on a vector space with dimension $m - 1$ or less. Suppose that $B = \{v_{1}, v_{2}, v_{3}, \dots, v_{m}\}$ is a basis for $V$ that yields a diagonal matrix representation for $T$ with diagonal entries $λ_{1}, λ_{2}, λ_{3}, \dots, λ_{m}$ . Then $U = 〈\{v_{1}, v_{2}, v_{3}, \dots, v_{m - 1}\}〉$ is a subspace of $V$ that is invariant relative to $T$ . The restriction $T |_{U} : U \mapsto U$ is then a linear transformation defined on $U$ , a vector space of dimension $m - 1$ . A matrix representation of $T |_{U}$ relative to the basis $C = \{v_{1}, v_{2}, v_{3}, \dots, v_{m - 1}\}$ will be an upper-triangular matrix with diagonal entries $λ_{1}, λ_{2}, λ_{3}, \dots, λ_{m - 1}$ . We can therefore apply the induction hypothesis to $T |_{U}$ and its representation relative to $C$ .

Suppose that $λ$ is any eigenvalue of $T$ . Then suppose that $v \in K ({(T - λ_{V})}^{m})$ . As an element of $V$ , we can write $v$ as a linear combination of the basis elements of $B$ , or more compactly, there is a vector $u \in U$ and a scalar $α$ such that $v = u + α v_{m}$ . Then,

\begin{array}{l} α {(λ_{m} - λ)}^{m} v_{m} \\ = α {(T - λ I_{V})}^{m} (v_{m}) & Theorem EOMP \\ = 0 + α {(T - λ I_{V})}^{m} (v_{m}) & Property Z \\ = - {(T - λ I_{V})}^{m} (u) + {(T - λ I_{V})}^{m} (u) + α {(T - λ I_{V})}^{m} (v_{m}) & Property AI \\ = - {(T - λ I_{V})}^{m} (u) + {(T - λ I_{V})}^{m} (u + α v_{m}) & Theorem LTLC \\ = - {(T - λ I_{V})}^{m} (u) + {(T - λ I_{V})}^{m} (v) & Theorem LTLC \\ = - {(T - λ I_{V})}^{m} (u) + 0 & Definition KLT \\ = - {(T - λ I_{V})}^{m} (u) & Property Z \end{array}

The final expression in this string of equalities is an element of $U$ since $U$ is invariant relative to both $T$ and $I_{V}$ . The expression at the beginning is a scalar multiple of $v_{m}$ , and as such cannot be a nonzero element of $U$ without violating the linear independence of $B$ . So

\begin{array}{l} α {(λ_{m} - λ)}^{m} v_{m} & = 0 \end{array}

The vector $v_{m}$ is nonzero since $B$ is linearly independent, so Theorem SMEZV tells us that $α {(λ_{m} - λ)}^{m} = 0$ . From the properties of scalar multiplication, we are confronted with two possibilities.

Our first case is that $λ \neq λ_{m}$ . Notice then that $λ$ occurs the same number of times along the diagonal in the representations of $T |_{U}$ and $T$ . Now $α = 0$ and $v = u + 0 v_{m} = u$ . Since $v$ was chosen as an arbitrary element of $K ({(T - λ I_{V})}^{m})$ , Definition SSET says that $K ({(T - λ I_{V})}^{m}) \subseteq U$ . It is always the case that $K ({(T |_{U} - λ I_{U})}^{m}) \subseteq K ({(T - λ I_{V})}^{m})$ . However, we can also see that in this case, the opposite set inclusion is true as well. By Definition SE we have $K ({(T |_{U} - λ I_{U})}^{m}) = K ({(T - λ I_{V})}^{m})$ . Then

\begin{array}{l} #_{T} (λ) & = #_{T |_{U}} (λ) \\ \geq dim (G_{T |_{U}} (λ)) & Induction Hypothesis \\ = dim (K ({(T |_{U} - λ I_{U})}^{m - 1})) & Theorem GEK \\ = dim (K ({(T |_{U} - λ I_{U})}^{m})) & Theorem KPLT \\ = dim (K ({(T - λ I_{V})}^{m})) \\ = dim (G_{T} (λ)) & Theorem GEK \end{array}

The second case is that $λ = λ_{m}$ . Notice then that $λ$ occurs one more time along the diagonal in the representation of $T$ compared to the representation of $T |_{U}$ . Then

\begin{array}{l} {(T |_{U} - λ I_{U})}^{m} (u) & = {(T - λ I_{V})}^{m} (u) \\ = {(T - λ I_{V})}^{m} (u) + 0 & Property Z \\ = {(T - λ I_{V})}^{m} (u) + α {(λ_{m} - λ)}^{m} v_{m} & Theorem ZSSM \\ = {(T - λ I_{V})}^{m} (u) + α {(T - λ I_{V})}^{m} (v_{m}) & Theorem EOMP \\ = {(T - λ I_{V})}^{m} (u + α v_{m}) & Theorem LTLC \\ = {(T - λ I_{V})}^{m} (v) \\ = 0 & Definition KLT \end{array}

So $u \in K (T |_{U} - λ I_{U})$ . The vector $v$ is an arbitrary member of $K ({(T - λ I_{V})}^{m})$ and is also equal to an element of $K (T |_{U} - λ I_{U})$ ( $u$ ) plus a scalar multiple of the vector $v_{m}$ . This observation yields

\begin{array}{l} dim (K ({(T - λ I_{V})}^{m})) & \leq dim (K (T |_{U} - λ I_{U})) + 1 \end{array}

Now count eigenvalues on the diagonal,

\begin{array}{l} #_{T} (λ) & = #_{T |_{U}} (λ) + 1 \\ \geq dim (G_{T |_{U}} (λ)) + 1 & Induction Hypothesis \\ = dim (K ({(T |_{U} - λ I_{U})}^{m - 1})) + 1 & Theorem GEK \\ = dim (K ({(T |_{U} - λ I_{U})}^{m})) + 1 & Theorem KPLT \\ \geq dim (K ({(T - λ I_{V})}^{m})) \\ = dim (G_{T} (λ)) & Theorem GEK \end{array}

In Theorem UTMR we constructed an upper-triangular matrix represntation of $T$ where each eigenvalue occurred $α_{T} (λ)$ times on the diagonal. So

\begin{array}{l} α_{T} (λ_{i}) & = #_{T} (λ_{i}) & Theorem UTMR \\ \geq dim (G_{T} (λ_{i})) \\ \geq dim (U_{i}) & Theorem PSSD \\ = α_{T} (λ_{i}) & Theorem PSSD \end{array}

Thus, $dim (G_{T} (λ_{i})) = α_{T} (λ_{i})$ and by Theorem EDYES, $U_{i} = G_{T} (λ_{i})$ and we can write

\begin{array}{l} V & = U_{1} \oplus U_{2} \oplus U_{3} \oplus \dots \oplus U_{m} \\ = G_{T} (λ_{1}) \oplus G_{T} (λ_{2}) \oplus G_{T} (λ_{3}) \oplus \dots \oplus G_{T} (λ_{m}) \end{array}

■

Besides a nice decomposition into invariant subspaces, this proof has a bonus for us.

Theorem DGES
Dimension of Generalized Eigenspaces
Suppose $T : V \mapsto V$ is a linear transformation with eigenvalue $λ$ . Then the dimension of the generalized eigenspace for $λ$ is the algebraic multiplicity of $λ$ , $dim (G_{T} (λ_{i})) = α_{T} (λ_{i})$ . $□$

Proof At the very end of the proof of Theorem GESD we obtain the inequalities

\begin{array}{l} α_{T} (λ_{i}) & \leq dim (G_{T} (λ_{i})) \leq α_{T} (λ_{i}) \end{array}

which establishes the desired equality. $■$

Subsection JCF: Jordan Canonical Form

Now we are in a position to define what we (and others) regard as an especially nice matrix representation. The word “canonical” has at its root, the word “canon,” which has various meanings. One is the set of laws established by a church council. Another is a set of writings that are authentic, important or representative. Here we take to to mean the accepted, or best, representative among a variety of choices. Every linear transformation admits a variety of representations, and will declare one as the best. Hopefully you will agree.

Definition JCF
Jordan Canonical Form
A square matrix is in Jordan canonical form if it meets the following requirements:

The matrix is block diagonal.
Each block is a Jordan block.
If $ρ < λ$ then the block $J_{k} (ρ)$ occupies rows with indices greater than the indices of the rows occupied by $J_{ℓ} (ρ)$ .
If $ρ = λ$ and $ℓ < k$ , then the block $J_{k} (λ)$ occupies rows with indices greater than the indices of the rows occupied by $J_{ℓ} (λ)$ .

△

Theorem JCFLT
Jordan Canonical Form for a Linear Transformation
Suppose $T : V \mapsto V$ is a linear transformation. Then there is a basis $B$ for $V$ such that the matrix representation of $T$ with the following properties:

The matrix representation is in Jordan canonical form.
If $J_{k} (λ)$ is one of the Jordan blocks, then $λ$ is an eigenvalue of $T$ .
For a fixed value of $λ$ , the largest block of the form $J_{k} (λ)$ has size equal to the index of $λ$ , $ι_{T} (λ)$ .
For a fixed value of $λ$ , the number of blocks of the form $J_{k} (λ)$ is the geometric multiplicity of $λ$ , $γ_{T} (λ)$ .
For a fixed value of $λ$ , the number of rows occupied by blocks of the form $J_{k} (λ)$ is the algebraic multiplicity of $λ$ , $α_{T} (λ)$ .

□

Proof This theorem is really just the consequence of applying to $T$ , consecutively Theorem GESD, Theorem MRRGE and Theorem CFNLT.

Theorem GESD gives us a decomposition of $V$ into generalized eigenspaces, one for each distinct eigenvalue. Since these generalized eigenspaces ar invariant relative to $T$ , this provides a block diagonal matrix representation where each block is the matrix representation of the restriction of $T$ to the generalized eigenspace.

Restricting $T$ to a generalized eigenspace results in a “nearly nilpotent” linear transformation, as stated more precisely in Theorem RGEN. We unravel Theorem RGEN in the proof of Theorem MRRGE so that we can apply Theorem CFNLT about representations of nilpotent linear transformations.

We know the dimension of a generalized eigenspace is the algebraic multiplicity of the eigenvalue (Theorem DGES), so the blocks associated with the generalized eigenspaces are square with a size equal to the algebraic multiplicity. In refining the basis for this block, and producing Jordan blocks the results of Theorem CFNLT apply. The total number of blocks will be the nullity of $T |_{G_{T} (λ)} - λ I_{G_{T} (λ)}$ , which is the geometric multiplicity of $λ$ as an eigenvalue of $T$ (Definition GME). The largest of the Jordan blocks will have size equal to the index of the nilpotent linear transformation $T |_{G_{T} (λ)} - λ I_{G_{T} (λ)}$ , which is exactly the definition of the index of the eigenvalue $λ$ (Definition IE). $■$

Before we do some examples of this result, notice how close Jordan canonical form is to a diagonal matrix. Or, equivalently, notice how close we have come to diagonalizing a matrix (Definition DZM). We have a matrix representation which has diagonal entries that are the eigenvalues of a matrix. Each occurs on the diagonal as many times as the algebraic multiplicity. However, when the geometric multiplicty is strictly less than the algebraic multiplicity, we have some entries in the representation just above the diagonal (the “superdiagonal”). Furthermore, we have some idea how often this happens if we know the geometric multiplicity and the index of the eigenvalue.

We now recognize just how simple a diagonalizable linear transformation really is. For each eigenvalue, the generalized eigenspace is just the regular eigenspace, and it decomposes into a direct sum of one-dimensional subspaces, each spanned by a different eigenvector chosen from a basis of eigenvectors for the eigenspace.

Some authors create matrix representations of nilpotent linear transformations where the Jordan block has the ones just below the diagonal (the “subdiagonal”). No matter, it is really the same, just different. We have also defined Jordan canonical form to place blocks for the larger eigenvalues earlier, and for blocks with the same eigenvalue, we place the bigger ones earlier. This is fairly standard, but there is no reason we couldn’t order the blocks differently. It’d be the same, just different. The reason for choosing some ordering is to be assured that there is just one canonical matrix representation for each linear transformation.

Subsection CHT: Cayley-Hamilton Theorem

Jordan was a French mathematician who was active in the late 1800’s. Cayley and Hamilton were 19th-century contemporaries of Jordan from Britian. The theorem that bears their names is perhaps one of the most celebrated in basic linear algebra. While our result applies only to vector spaces and linear transformations with scalars from the set of complex numbers, $ℂ$ , the result is equally true if we restrict our scalars to the real numbers, $ℝ^{}$ . It says that every matrix satisfies its own characteristic polynomial.

Theorem CHT
Cayley-Hamilton Theorem
Suppose $A$ is a square matrix with characteristic polynomial $p_{A} (x)$ . Then $p_{A} (A) = O$ . $□$

Proof Suppose $B$ and $C$ are similar matrices via the matrix $S$ , $B = S^{- 1} C S$ , and $q (x)$ is any polynomial. Then $q (B)$ is similar to $q (C)$ via $S$ , $q (B) = S^{- 1} q (C) S$ . (See Example HPDM for hints on how to convince yourself of this.)

By Theorem JCFLT and Theorem SCB we know $A$ is similar to a matrix, $J$ , in Jordan canonical form. Suppose $λ_{1}, λ_{2}, λ_{3}, \dots, λ_{m}$ are the distinct eigenvalues of $A$ (and are therefore the eigenvalues and diagonal entries of $J$ ). Then by Theorem EMRCP and Definition AME, we can factor the characteristic polynomial as

\begin{array}{l} p_{A} (x) & = {(x - λ_{1})}^{α_{A} (λ_{1})} {(x - λ_{2})}^{α_{A} (λ_{2})} {(x - λ_{3})}^{α_{A} (λ_{3})} \dots {(x - λ_{m})}^{α_{A} (λ_{m})} \end{array}

On substituting the matrix $J$ we have

\begin{array}{l} p_{A} (J) & = {(J - λ_{1} I)}^{α_{A} (λ_{1})} {(J - λ_{2} I)}^{α_{A} (λ_{2})} {(J - λ_{3} I)}^{α_{A} (λ_{3})} \dots {(J - λ_{m} I)}^{α_{A} (λ_{m})} \end{array}

The matrix $J - λ_{k} I$ will be block diagonal, and the block arising from the generalized eigenspace for $λ_{k}$ will have zeros along the diagonal. Suitably adjusted for matrices (rather than linear transformations), Theorem RGEN tells us this matrix is nilpotent. Since the size of this nilpotent matrix is equal to the algebraic multiplicity of $λ_{k}$ , the power ${(J - λ_{k} I)}^{α_{A} (λ_{k})}$ will be a zero matrix (Theorem KPNLT) in the location of this block.

Repeating this argument for each of the $m$ eigenvalues will place a zero block in some term of the product at every location on the diagonal. The entire product will then be zero blocks on the diagonal, and zero off the diagonal. In other words, it will be the zero matrix. Since $A$ and $J$ are similar, $p_{A} (A) = p_{A} (J) = O$ . $■$

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