HW 8 - Operator Decomposition, Characteristic and Minimal Polynomials

8.B: Operator Decomposition

6

Question

Define $N \in L (F^{5})$ by:

N (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = (2 x_{2}, 3 x_{3}, - x_{4}, 4 x_{5}, 0)

Find a square root of $I + N$ .

Proof
As a suggestion, use the Taylor series for $\sqrt{1 + x}$ as motivation (because $N$ here seems to be nilpotent):

\sqrt{1 + x} = \sum_{j = 0}^{\infty} (\binom{\frac{1}{2}}{j}) (- 1)^{j} x^{j} = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} - \frac{5}{128} x^{4} + \dots

Substitute $x := N$ , we assume and will verify:

\sqrt{I + N} = I + \frac{1}{2} N - \frac{1}{8} N^{2} + \frac{1}{16} N^{3} - \frac{5}{128} N^{4}

First, let's show that $N$ is nilpotent by showing that $N^{5} = \vec{0}$ . Notice that using $β$ as the standard basis:

M (N, β) = [\begin{matrix} 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]

Thus $M (N)$ has the form of a nilpotent operator, so it's nilpotent, thus $N^{5} = \vec{0}$ . Notice that we can verify our factor for $\sqrt{1 + N}$ works by just squaring it:

\begin{aligned} (\sqrt{I + N})^{2} & = (I + \frac{1}{2} N - \frac{1}{8} N^{2} + \frac{1}{16} N^{3} - \frac{5}{128} N^{4})^{2} \\ = I^{5} & = I \\ + 2 (\frac{1}{2}) N & = N \\ + 2 (\frac{- 1}{8}) N^{2} + {(\frac{1}{2} N)}^{2} & = \vec{0} \\ + 2 (\frac{1}{16} N^{3}) + (2 \cdot \frac{1}{2} N \cdot \frac{- 1}{8} N^{3}) & = \vec{0} \\ + 2 (\frac{- 5}{128} N^{4}) + 2 (\frac{1}{2} N \cdot \frac{1}{16} N^{3}) + {(\frac{- 1}{8} N^{2})}^{2} & = \vec{0} \\ + (terms with N^{5} or higher powers) & = \vec{0} \\ = I + N \end{aligned}

Showing it's a valid square root.
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7

Question

Suppose $V$ is a complex vector space. Prove that every invertible operator on $V$ has a cube root.

Proof
Let $T \in L (V)$ be an invertible operator, so $\exists T^{- 1}$ . We'll actually prove that every invertible operator on $V$ has some $k$ -th root instead (here $k \in Z^{+}$ , so $\exists \sqrt[k]{T}$ which proves the lemma for $k = 3$ .

Following the proof structure showing the existence of square roots, let $λ_{1}, . . ., λ_{m}$ be the distinct eigenvalues of $T$ . For each $j$ there exists a nilpotent operator $N_{j} \in L (G (λ_{j}, T))$ such that $T |_{G (λ_{j}, T)} = λ_{j} I + N_{j}$ via the description of operators on complex vector spaces. Because $T$ is invertible then $λ_{j} \neq 0$ for all $j$ , so then:

T |_{G (λ_{j}, T)} = λ_{j} (I + \frac{N_{j}}{λ_{j}})

for each $j$ .

Cool. Now we need to prove an important lemma:

Lemma

Suppose $N \in L (V)$ is nilpotent. Then $I + N$ has a $k$ -th root, where $k \in Z^{+}$ .

Proof
Say $N^{m} = \vec{0}$ . Consider the choice:

\sqrt[k]{I + N} \overset{?}{=} \sum_{n = 0}^{m} (\binom{\frac{1}{k}}{n}) N^{n}

The Taylor Series of $1 + x$ is the same sum as above, using $x := N$ . If you believe the multiplication of the power series works out the same, then they will equal.
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As such then $k = 3$ implies that since $\frac{N_{j}}{λ_{j}}$ is nilpotent, then $I + \frac{N_{j}}{λ_{j}}$ has a cube root using the lemma. Therefore we obtain a cube root $R_{j}$ of $T |_{G (λ_{j}, T)}$ .

A typical vector $v \in V$ can be written as:

v = u_{1} + \dots + u_{m}

where each $u_{j} \in G (λ_{j}, T)$ . Define $R \in L (V)$ by:

R v = R_{1} u_{1} + \dots + R_{m} u_{m}

Where $R$ is a cube root of $T$ because:

\begin{aligned} R^{3} v & = R^{2} \sum_{i = 1}^{m} R_{i} u_{i} \\ = R \sum_{i = 1}^{m} R R_{i} u_{i} \\ = R \sum_{i = 1}^{m} R_{i}^{2} u_{i} \\ = \sum_{i = 1}^{m} R_{i}^{3} u_{i} \\ = \sum_{i = 1}^{m} {(\sqrt[3]{λ_{j}} \sqrt[3]{I + \frac{N_{j}}{λ_{j}}})}^{3} u_{j} \\ = \sum_{i = 1}^{m} λ_{j} (I + \frac{N_{j}}{λ_{j}}) u_{j} \\ = \sum_{i = 1}^{m} (I λ_{j} + N_{j}) u_{j} \\ = \sum_{i = 1}^{m} T |_{G (λ_{j}, T)} u_{j} \\ = T v \end{aligned}

Thus $R$ is a valid cube root for $T$ .
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8

Question

Suppose $T \in L (V)$ and $3, 8$ are eigenvalues of $T$ . Let $n = \dim (V)$ . Prove that $V = (null (T^{n - 2})) \oplus (range (T^{n - 2}))$ .

Proof
Let's show the intersection is trivial, so suppose $v \in null (T^{n - 2}) \cap range (T^{n - 2})$ . So then $T^{n - 2} v = \vec{0}$ and $\exists u$ s.t. $T^{n - 2} u = v$ . Apply $T^{n}$ to both sides to get:

T^{n} (v) = T^{n} (T^{n - 2} u) = T^{2 n - 2} u = \vec{0} = T^{n - 2} (T^{n} u)

because null spaces have to stop growing and eventually encompass the whole $V$ . Notice we have to consider that when $2 n - 2 \geq n \Rightarrow n \geq 2$ so that our power of $T$ is over $n$ , but notice this is true since $3, 8$ are eigenvalues of $T$ , then $\dim (G (3, T)) \geq 1$ and $\dim (G (8, T)) \geq 1$ so then $\dim (V) = n \geq \dim (G (3, T)) + \dim (G (8, T)) \geq 2$ as desired.

So then $T^{n - 2} u = v = \vec{0}$ showing the intersection is trivial, so the sum is a direct sum.

Also:

\dim (null (T^{n - 2}) \oplus range (T^{n - 2})) = \dim (null (T^{n - 2})) + \dim (range (T^{n - 2})) = \dim (V)

Using a combination of the fact that a direct sum's dimensions add up, as well as the FTOLM.
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9

Question

Suppose $A, B$ are block diagonal matrices of the form:

A = [\begin{matrix} A_{1} & 0 \\ ⋱ \\ 0 & A_{m} \end{matrix}], B = [\begin{matrix} B_{1} & 0 \\ ⋱ \\ 0 & B_{m} \end{matrix}]

where $A_{j}, B_{j}$ are the same size for $j = 1, . . ., m$ . Show that $A B$ is a block diagonal matrix of the form:

A B = [\begin{matrix} A_{1} B_{1} & 0 \\ ⋱ \\ 0 & A_{m} B_{m} \end{matrix}]

Proof
First consider the whole $A, B$ matrices. Since $A, B$ are the same size, say $n$ , then consider $V$ as the vector space of a column vector of that size $n$ . We can always choose any basis $β$ of $V$ and then define $T_{A}, T_{B} \in L (V)$ in a way such that $M (T_{A}, β) = A$ and $M (T_{B}, β) = B$ . Therefore, $M (T_{A} T_{B}) = A B$ .

Now for the "sub-matrices". Let $d_{j}$ be the size of $A_{j}$ which is the same as $B_{j}$ . Consider the list of the first $d_{1}$ vectors from $β$ . $A$ and $B$ show that the span of these vectors are invariant under $T_{A}, T_{B}$ (because $A, B$ are block diagonal, then all $A_{j}, B_{j}$ are UT, showing invariance)

Similarly, the span of the next $d_{2}$ vectors in $β$ are also invariant. Continuing, we see that there are $m$ distinct lists of consecutive vectors with no intersections.

Let $U_{1}, . . ., U_{m}$ denote such spans, namely $U_{i} = span (v_{\sum_{j = 1}^{i - 1} d_{j} + 1}, . . ., v_{\sum_{j = 1}^{i + 1} d_{j} - 1})$ . Clearly $M (T_{A} |_{U_{j}}) = A_{j}$ and likewise $M (T_{B} |_{U_{j}}) = B_{j}$ for each $j$ , so $M (T_{A} |_{U_{j}} T_{B} |_{U_{j}}) = A_{j} B_{j}$ as desired.

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10

Question

Suppose $F = C$ and $T \in L (V)$ . Prove that there exist $D, N \in L (V)$ such that $T = D + N$ where $D$ is diagonalizable and $N$ is nilpotent and $D N = N D$ .

Proof
Let $λ_{1}, . . ., λ_{m}$ be the distinct eigenvalues of $T$ . Using the description of operators on complex vector spaces, then there exists a nilpotent operator $N_{j} \in L (G (λ_{j}, T))$ such that $T |_{G (λ_{j}, T)} = λ_{j} I + N_{j}$ for all $j$ . Notice though that $λ_{j} I$ is a diagonalizable operator since $\dim (G (λ_{j}, T)) = k$ for some $k \in Z^{+}$ , so then there's a basis $β_{j} = {v_{1}, . . ., v_{k}}$ spanning $G (λ_{j}, T)$ . Since all the $v_{i}$ 's are generalized eigenvectors, then we can create the 1-dimensional subspaces $U_{i}$ where $U_{i} = span (v_{i})$ for each $1 \leq i \leq k$ . Notice that these are invariant under $λ_{j} I$ clearly, so then by the conditions equivalent to diagonalizability, then $λ_{j} I$ is diagonalizable. This makes even more sense since:

M (λ_{j} I, β_{j}) = diag (λ_{j}, . . ., λ_{j})

As such, then we've done the construction we want. Now we show that $λ_{j} I N_{j} = N_{j} λ_{j} I$ for all $j$ . But that's easy since:

λ_{j} N_{j} v = N_{j} (λ_{j} v) = (N_{j} λ_{j}) v

as desired.

Now we make $D, N$ . For all $v \in V$ we can set them as sums of vectors from each generalized eigenspace:

v = u_{1} + \dots + u_{m}

where each $u_{i} \in G (λ_{i}, T)$ . Create the operator $D$ defined by:

D v = λ_{1} u_{1} + \dots + λ_{m} u_{m}

Each $u_{j} \in G (λ_{j}, T)$ which has its own basis $β_{j} = {u_{j, 1}, . . ., u_{j, \dim (G (λ_{j}, T)}}$ . Notice that:

D_{j} u_{j} = D_{j} \sum_{k = 1}^{\dim (G (λ_{j}, T))} u_{j, k} = \sum_{k = 1}^{\dim (G (λ_{j}, T))} λ_{j} u_{j, k}

So each $u_{j, k}$ is an eigenvector of our subspace $G (λ_{j}, T)$ with eigenvalue $λ_{j}$ , so as a result each $D_{j}$ is diagonalizable where $M (D_{j}, β_{j}) = diag (λ_{j}, . . ., λ_{j})$ . As a result, for each $u_{j}$ then:

D (u_{j}) = D_{j} u_{j} = \sum_{k = 1}^{\dim (G (λ_{j}, T))} λ_{j} u_{j, k} = λ_{j} \sum_{k = 1}^{\dim (G (λ_{j}, T))} u_{j, k} = λ_{j} u_{j}

So if we use a basis $β = ⋃_{j = 1}^{m} β_{j}$ (in the same order) implies that:

M (D, β) = diag (\overset{\dim (G (λ_{1}, T)) times}{\overset{⏞}{λ_{1}, . . .}}, \overset{\dim (G (λ_{2}, T)) times}{\overset{⏞}{λ_{2}, . . .}}, . . ., \dots, \overset{\dim (G (λ_{m}, T)) times}{\overset{⏞}{λ_{m}, . . .}})

which is a diagonal matrix.

Similarly we define $N$ by:

N v = \sum_{j = 1}^{m} N_{j} u_{j}

Clearly $N$ is nilpotent since each $N_{j}$ is nilpotent, so just take the max number $k$ for the power of the highest-degree nilpotent operator, and take $N^{k} = \vec{0}$ as a result:

\begin{aligned} N^{k} v & = {(\sum_{j = 1}^{m} N_{j} u_{j})}^{k} \\ = \sum_{j = 1}^{m} N_{j}^{k} u_{j} & (N_{j} u_{\neq j} = \vec{0}) so no combinations of different N_{j} ’s allowed \\ = \sum_{j = 1}^{m} \vec{0} u_{j} & k is the max power by definition for all N_{j} ’s \\ = \vec{0} \end{aligned}

Now we just show $N D = D N$ but that's easy since:

\begin{aligned} N D v & = N (\sum_{j = 1}^{m} \sum_{k = 1}^{\dim (G (λ_{j}, T))} λ_{j} u_{j, k}) \\ = \sum_{j = 1}^{m} \sum_{k = 1}^{\dim (G (λ_{j}, T))} λ_{j} N_{j} u_{j, k} \\ = \sum_{k = 1}^{\dim (G (λ_{j}, T))} \sum_{j = 1}^{m} λ_{j} N_{j} u_{j, k} \\ = D (\sum_{j = 1}^{m} N_{j} u_{j, k}) \\ = D N v \end{aligned}

Thus $N D = D N$ .
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11

Question

Suppose $T \in L (V)$ and $λ \in F$ . Prove that for every basis of $V$ with respect to which $T$ has an upper triangular matrix, the number of times that $λ$ appears on the diagonal of the matrix of $T$ equals the multiplicity of $λ$ as an eigenvalue of $T$ .

Proof
Let $β = {v_{1}, . . ., v_{n}}$ be a basis where $M (T, β)$ is an upper triangular matrix. As such, we know that the $λ$ 's have to appear on the diagonal of $M (T, β)$ .

Let $λ \in F$ be arbitrary, and say it appears on the $j$ -th diagonal entry of $M (T, β)$ . Then:

T v_{j} = a_{1} v_{1} + \dots + a_{j - 1} v_{j - 1} + λ v_{j}

for $a_{i} \in F$ . Thus:

(T - λ I) v_{j} = \sum_{i = 1}^{j - 1} a_{i} v_{i} \in span (v_{1}, . . ., v_{j - 1})

We suppose that we started looking at the last entry of the same $λ$ on the diagonal. Say that $n$ is the number of times it appears on that diagonal. We want to show that $\dim G (λ, T) = n$ as a result. Note that in a similar vein of logic:

\begin{aligned} T v_{j - 1} & = b_{1} v_{1} + \dots + b_{j - 2} v_{j - 2} + λ v_{j - 1} \\ \Leftrightarrow (T - λ I) v_{j - 1} = \sum_{i = 1}^{j - 2} b_{i} v_{i} \in span (v_{1}, . . ., v_{j - 2}) \end{aligned}

for some $b_{1}, . . ., b_{j - 1} \in F$ . We can repeat this process over an over again, and show a few important things:

$span (v_{1}) \subset span (v_{1}, v_{2}) \subset \dots \subset span (v_{1}, . . ., v_{j - 1})$
$v_{j} \in span (v_{1}, . . ., v_{j - 1}), . . ., v_{1} \in span (v_{1})$

Now are these span strict subsets or not? Well we just have to show that the $v_{1}, . . ., v_{j - 1}$ are LI, which they are because they come from a subspace of the basis $β$ . Consequently, we can treat each subspace $span (v_{1}, . . ., v_{i})$ as a vector space, and thus since the subspaces are strict then the dimensions have to increase. Therefore:

1 = \dim (span (v_{1})) < \underset{j - 1 times}{\underset{⏟}{\dots}} < \dim (span (v_{1}, . . ., v_{j - 1}))

Thus then $j < \dim (span (v_{1}, . . ., v_{j - 1}))$ but we know that we only can go $n$ entries, so clearly $n < j$ so $n < \dim (span (v_{1}, . . ., v_{j - 1}))$ . But since all $v_{1}, . . ., v_{j - 1}$ are eigenvectors, then $span (v_{1}, . . ., v_{j - 1}) \subseteq G (λ, T)$ so then clearly $\dim (G (λ, T)) \geq n$ .

Clearly $\dim (G (λ, T) \leq n$ because we have $\dim (V)$ diagonal entries and each one adds at least 1 to the dimension of each generalized eigenspace $G (λ_{j}, T)$ for all $1 \leq j \leq m$ .

Thus $\dim (G (λ, T)) = n$ for all $λ$ completing the proof.
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8.C: Characteristic & Minimal Polynomials

1

Question

Suppose $T \in L (C^{4})$ is such that the eigenvalues of $T$ are 3,5, and 8. Prove that $(T - 3 I)^{2} (T - 5 I)^{2} (T - 8 I)^{2} = \vec{0}$ .

Proof
Notice that $\dim (V) = 4$ in this case. Since we have 3 distinct eigenvalues, then at least 1 generalized eigenspace is of dimension 2, while the others are 1. For instance, if $\dim (G (3, T)) = 2$ then $\dim (G (5, T)) = \dim (G (8, T)) = 1$ . As such, then the characteristic polynomial for $T$ would be $p (z)$ given by:

p (z) = (z - 3)^{d_{1}} (z - 5)^{d_{2}} (z - 8)^{d_{3}}

where at least one $d_{i} = 2$ while the others are equal to 1. Notice then that by the Cayley-Hamilton Theorem that $p (T) = \vec{0}$ . Furthermore, we can increase the degree of $p$ such that we have:

q (z) = (z - 3)^{2} (z - 5)^{2} (z - 8)^{2} = s (z) p (z)

where $s$ contains the other factors (which we won't care about here, without loss of generality). If we show that $q (T) = \vec{0}$ here then we are done:

q (T) = s (T) p (T) = s (T) \vec{0} = \vec{0}

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2

Question

Suppose $V$ is a complex vector space. Suppose $T \in L (V)$ is such that $5, 6$ are eigenvalues of $T$ and $T$ has no other eigenvalues. Prove that $(T - 5 I)^{n - 1} (T - 6 I)^{n - 1} = \vec{0}$ where $n = \dim (V)$ .

Proof
This takes a similar process to 8.C.1 done earlier. Namely, we know that since $5, 6$ are the only eigenvalues of $T$ then:

V = G (5, T) \oplus G (6, T) \Leftrightarrow n = \dim (G (5, T)) + \dim (G (6, T))

Thus then the characteristic polynomial of $T$ would be:

p (z) = (z - 5)^{d_{1}} (z - 6)^{d_{2}}

where we require that since $\dim (G (5, T)) = d_{1}$ and $\dim (G (6, T)) = d_{2}$ then $d_{1} + d_{2} = n$ as a result.

First note that we must have each $d_{i} \geq 1$ . Why? Because if $d_{i} 0$ then $\dim (G (λ_{i}, T)) = 0$ so $λ_{i}$ is not an eigenvalue which is a contradiction to the given. As such, then at minimum for example $d_{1} \geq 1 \Rightarrow d_{2} \leq n - 1$ and vice versa (without loss of generality). Clearly then $d_{i} \leq n - 1$ for similar reasoning. Therefore, consider the polynomial $q$ defined by:

q (z) = (z - 5)^{n - 1} (z - 6)^{n - 1}

Clearly since $d_{i} \leq n - 1$ then there is some $s$ where:

q (z) = s (z) p (z)

Now if we show that $q (T) = \vec{0}$ then we are done. Notice that $p (T) = \vec{0}$ by the Cayley-Hamilton Theorem. Thus:

q (T) = s (T) p (T) = s (T) \vec{0} = \vec{0}

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3

Question

Give an example of an operator on $C^{4}$ whose characteristic polynomial equals $(z - 7)^{2} (z - 8)^{2}$ .

Proof
Consider $T \in L (C^{4})$ given by:

T (z_{1}, z_{2}, z_{3}, z_{4}) = (7 z_{1}, 7 z_{2}, 8 z_{3}, 8 z_{4})

Notice that clearly $λ_{1} = 7$ with $d_{1} = 2$ since $G (λ_{1}, T) = span (e_{1}, e_{2})$ (using the standard basis $β = {e_{1}, e_{2}, e_{3}, e_{4}}$ ). Similarly $λ_{2} = 8$ with $d_{2} = 2$ since $G (λ_{2}, T) = span (e_{3}, e_{4})$ . Thus the characteristic polynomial for $T$ is:

p (z) = (z - 7)^{2} (z - 8)^{2}

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4

Question

Give an example of an operator on $C^{4}$ whose characteristic polynomial equals $(z - 1) (z - 5)^{3}$ and whose minimal polynomial equals $(z - 1) (z - 5)^{2}$ .

Proof

Consider $T \in L (C^{4})$ defined by its matrix representation:

M (T, β) = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 5 & 1 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \end{matrix}]

Notice that using the standard basis $β = {e_{1}, e_{2}, e_{3}, e_{4}}$ , that $e_{1}$ is an eigenvector with $λ_{1} = 1$ and $e_{2}, e_{3}, e_{4}$ are eigenvectors with $λ_{2} = 5$ given by the diagonals of the matrix. As such, then the characteristic polynomial is:

p_{T} (z) = (z - 1) (z - 5)^{3}

We'll show that the minimal polynomial is $(z - 1) (z - 5)^{2}$ by showing that $(T - I) (T - 5 I)^{2} = \vec{0}$ while $(T - I) (T - 5 I) \neq \vec{0}$ :

M (T^{2}) = M (T)^{2} = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 25 & 10 & 0 \\ 0 & 0 & 25 & 0 \\ 0 & 0 & 0 & 25 \end{matrix}]

M (T^{3}) = M (T)^{3} = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 125 & 75 & 0 \\ 0 & 0 & 125 & 0 \\ 0 & 0 & 0 & 125 \end{matrix}]

\begin{aligned} M ((T - I) (T - 5 I)^{2}) & = M (T^{3} - 11 T^{2} + 35 T - 25 I) \\ = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 125 & 75 & 0 \\ 0 & 0 & 125 & 0 \\ 0 & 0 & 0 & 125 \end{array}] - 11 [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 25 & 10 & 0 \\ 0 & 0 & 25 & 0 \\ 0 & 0 & 0 & 25 \end{array}] + 35 [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 5 & 1 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \end{array}] - 25 M (I) \\ = [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] \end{aligned}

Thus $(T - I) (T - 5 I)^{2} = \vec{0}$ . Further:

\begin{aligned} M ((T - I) (T - 5 I)) & = M (T^{2} - 6 T + 5 I) \\ = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 25 & 10 & 0 \\ 0 & 0 & 25 & 0 \\ 0 & 0 & 0 & 25 \end{array}] - 6 [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 5 & 1 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \end{array}] + 5 M (I) \\ = [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 9 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] & = M (\vec{0}) \end{aligned}

Therefore $(T - I) (T - 5 I) \neq \vec{0}$ .

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5

Question

Give an example of an operator on $C^{4}$ whose characteristic and minimal polynomials both equal $z (z - 1)^{2} (z - 3)$ .

Proof
Choose $T \in L (C^{4})$ where:

M (T) = [\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 3 \end{matrix}]

Notice that the diagonal tells the $λ$ 's (since it's UT), so then the characteristic polynomial comes from these $λ$ 's:

p_{T} (z) = (z - 0) (z - 1)^{2} (z - 3) = z (z - 1)^{2} (z - 3)

where notice we have a multiplicity of 2 on the $λ_{2} = 1$ since $e_{2}, e_{3}$ from the standard basis are eigenvectors.

To show that the minimal polynomial is the characteristic, we show that $T (T - 1 I) (T - 3 I) \neq \vec{0}$ :

M (T^{2}) = M (T)^{2} = [\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 9 \end{matrix}]

M (T^{3}) = M (T)^{3} = [\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 27 \end{matrix}]

\begin{aligned} M (T (T - 1) (T - 3 I)) & = M (T^{3} - 4 T^{2} + 3 T) \\ = [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 27 \end{array}] - 4 [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 9 \end{array}] + 3 [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 3 \end{array}] \\ = [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] \\ \neq M (\vec{0}) \end{aligned}

So since we can only possible reduce the power of 2 from the characteristic polynomial, then we have no other options. The characteristic polynomial must be the minimal polynomial in this case.
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6

Question

Give an example of an operator on $C^{4}$ whose characteristic polynomial equals $z (z - 1)^{2} (z - 3)$ while the minimal polynomial equals $z (z - 1) (z - 3)$ .

Proof
Do a similar process to the last problem.

Choose $T \in L (C^{4})$ where:

M (T) = [\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 3 \end{matrix}]

Notice that the diagonal tells the $λ$ 's (since it's UT), so then the characteristic polynomial comes from these $λ$ 's:

p_{T} (z) = (z - 0) (z - 1)^{2} (z - 3) = z (z - 1)^{2} (z - 3)

where notice we have a multiplicity of 2 on the $λ_{2} = 1$ since $e_{2}, e_{3}$ from the standard basis are eigenvectors.

To show that the minimal polynomial is indeed $z (z - 1) (z - 3)$ , we show that $T (T - 1 I) (T - 3 I) = \vec{0}$ :

M (T^{2}) = M (T)^{2} = [\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 9 \end{matrix}]

M (T^{3}) = M (T)^{3} = [\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 27 \end{matrix}]

\begin{aligned} M (T (T - 1) (T - 3 I)) & = M (T^{3} - 4 T^{2} + 3 T) \\ = [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 27 \end{array}] - 4 [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 9 \end{array}] + 3 [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 3 \end{array}] \\ = [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] \\ = M (\vec{0}) \end{aligned}

Thus the minimal polynomial in this case is $z (z - 1) (z - 3)$ (notice we cannot reduce any more powers, so this has to be the minimal polynomial).
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7

Question

Suppose $V$ is a complex vector space. Suppose $P \in L (V)$ is such that $P^{2} = P$ . Prove that the characteristic polynomial of $P$ is $z^{m} (z - 1)^{n}$ where $m = \dim (null (P))$ and $n = \dim (range (P))$ .

Proof
Since $P^{2} = P$ then:

\begin{aligned} P^{2} = P & \Leftrightarrow P^{2} - P = \vec{0} \\ \Leftrightarrow P (P - 1) = \vec{0} \\ \Leftrightarrow λ = 0, 1 are the only eigenvalues of P \\ \Leftrightarrow V = G (0, P) \oplus G (1, P) \\ \Leftrightarrow \dim (V) = \dim G (0, P) + \dim G (1, P) \\ \Leftrightarrow \dim (V) = \dim null (P^{\dim (V)}) + \dim null ((P - I)^{\dim (V)}) \end{aligned}

But notice an important thing. First, $\dim (V) \in Z^{+}$ so then it's either even or odd. If it's even then:

\begin{aligned} P^{\dim (V)} & = \underset{\frac{\dim V}{2} times}{\underset{⏟}{P^{2} \circ P^{2} \dots \circ P^{2}}} \\ = P \circ P \circ \dots P & (P^{2} = P) \\ = P^{\frac{\dim V}{2}} \end{aligned}

If instead it's odd then:

\begin{aligned} P^{\dim (V)} & = \underset{\frac{\dim V}{2} times}{\underset{⏟}{P^{2} \circ \dots \circ P^{2}}} \circ P \\ = P \circ \dots \circ P \circ P & (P^{2} = P) \\ = P^{\frac{\dim V}{2} + 1} \end{aligned}

In both cases, we can repeatedly apply this to $P^{\dim (V)}$ (just treat the new integer power as $\dim (V)$ ) and keep reducing it down further and further. It must stop once we hit the lowest element in $Z^{+}$ which would be $1$ so then $P^{\dim (V)} = P$ as a result.

Therefore:

\begin{aligned} \Leftrightarrow \dim (V) = \dim null (P) + \dim (null (P - I)^{\dim (V)}) \end{aligned}

This shows that $m = \dim (null (P^{\dim (V)}) = \dim (null (P))$ as we wanted in the theorem.

But also the FTOLM implies that:

\dim (V) = \dim (null (P)) + \dim range (P)

Implying that $n = \dim range (P) = \dim null (P - I)^{\dim (V)}$ . Notice that we now have the right information to build our characteristic polynomial, since $\dim G (0, P) = \dim null (P^{\dim (V)}) = m$ as we wanted, so there's going to be a factor of $z^{m}$ in the characteristic polynomial. Further, we found that $\dim G (1, P) = \dim null (P - I)^{\dim (V)} = \dim range (P) = n$ so then we expect a factor of $(z - 1)^{n}$ as well. Putting them together, since we have no other eigenvalues and thus have no other possible factors:

p_{P} (z) = z^{m} (z - 1)^{n}

as desired.
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8

Question

Suppose $T \in L (V)$ . Prove that $T$ is invertible iff the constant term in the minimal polynomial of $T$ is non-zero.

Proof
Let's prove the negated version of this, that $T$ is not invertible iff the constant term in $p_{T}$ is zero.

The constant term $c$ for $p_{T}$ is zero ( $c = 0$ ) iff $0$ is a factor of $p_{T}$ (we can show this via contradiction. If instead 0 wasn't a factor then $p_{T} (0)$ would equal the constant term which would have to be non-zero which contradicts 0 not being a factor) iff $p_{T}$ can factor out the $z$ as a multiple:

p_{T} (z) = z q_{T} (z)

where $q_{T}$ is the same polynomial of $p_{T}$ with the factor of 0 removed. This is iff $λ = 0$ is an eigenvalue of $T$ iff $T - 0 I = T$ is not invertible.
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9

Question

Suppose $T \in L (V)$ has minimal polynomial $4 + 5 z - 6 z^{2} - 7 z^{3} + 2 z^{4} + z^{5}$ . Find the minimal polynomial of $T^{- 1}$ .

Proof
We are given $p_{T}$ shown above. Since $p_{T} (T) = \vec{0}$ then:

4 I + 5 T - 6 T^{2} - 7 T^{3} + 2 T^{4} + T^{5} = \vec{0}

Since $T$ is invertible then apply $T^{- 1}$ a total of 5 times to both sides:

4 T^{- 5} + 5 T^{- 4} - 6 T^{- 3} - T^{- 2} + 2 T^{- 1} + I = T^{- 5} \vec{0} = \vec{0}

So since having $q (T^{- 1}) = \vec{0}$ implies that $q$ is a multiple of the minimal polynomial of $T^{- 1}$ then have the left side be $q_{T^{- 1}} (T^{- 1})$ , which is a multiple of the minimal polynomial. Namely:

q_{T^{- 1}} (z) = 4 z^{5} + 5 z^{4} - 6 z^{3} - z^{2} + 2 z + 1

Now all we have to do to get the actual minimal polynomial is to make the polynomial monic. Notice that we can add factors (otherwise it's not the minimal polynomial) nor remove factors (otherwise it wasn't the minimal polynomial as it was missing factors) so all we can do is multiply by a non-zero constant. As a result, to make it monic let's just divide by the coefficient of the highest term:

p_{T^{- 1}} = z^{5} + \frac{5}{4} z^{4} - \frac{6}{4} z^{3} - \frac{1}{4} z^{2} + \frac{2}{4} z + \frac{1}{4}

as desired.
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