Chapter 9 - Operators on Real Vector Spaces

In Chapter 8 we covered the overall structure of an operator on a finite-dimensional complex vector space. In this chapter, we'll try to extend it to real vector spaces, which requires a lot of caveats.

9.A: Complexification

Complexification of a Vector Space

As we'll see, a real vector space $V$ can be embedded in a complex vector space called the complexification of $V$ . This is similar to the idea that since $R \subset C$ that any value $x \in R$ can really be though of as a value $x \in C$ and treated like a complex value. We do the same idea with vector spaces $V$ .

complexification of $V$ , $V_{C}$

Suppose $V$ is a real vector space.

The complexification of $V$ , denoted $V_{C}$ , equals $V \times V$ . An element of $V_{C}$ is an ordered pair $(u, v)$ where $u, v \in V$ , but we'll write this as $u + i v$ .
Addition on $V_{C}$ is defined by:

(u_{1} + i v_{1}) + (u_{2} + i v_{2}) = (u_{1} + u_{2}) + i (v_{1} + v_{2})

for $u_{1}, v_{1}, u_{2}, v_{2} \in V$ .

Complex scalar multiplication on $V_{C}$ is defined by:

(a + b i) (u + i v) = (a u - b v) + i (a v + b u)

for $a, b \in R$ and $u, v \in V$ .

We think of $V \subseteq V_{C}$ since $u \in V$ is $u + i \vec{0}$ . The idea of constructing $V_{C}$ from $V$ is similar to how we can construct $C ≃ R \times R$ and extend that for $C^{n} ≃ R^{n} \times R^{n}$ .

V_{C}

is a complex vector space.

Suppose $V$ is a real vector space. Then with the definitions of addition and scalar multiplication as above, then $V_{C}$ is a complex vector space.

Proof
Trivial. It's a vector space proof, it's clear that there's all the properties but we don't really need to go through it here. Proof by "trust me bro".
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Note that the additive identity for $V_{C}$ is $\vec{0} + i \vec{0}$ which we just write as $\vec{0}$ .

Basis of

V

is a basis of

V_{C}

Suppose $V$ is a real vector space.

If $v_{1}, . . ., v_{n}$ is a basis of $V$ (real vector space), then $v_{1}, . . ., v_{n}$ is a basis of $V_{C}$ (complex vector space)
The dimension of $V_{C}$ equals the dimension of $V$ .

Proof
(a): Suppose $v_{1}, . . ., v_{n}$ is a basis of $V$ . Then $span (v_{1}, . . ., v_{n})$ in $V_{C}$ contains all the vectors:

v_{1}, . . ., v_{n}, i v_{1}, . . ., i v_{n}

Thus $v_{1}, . . ., v_{n}$ spans the complex vector space $V_{C}$ .

To show LI, suppose $λ_{1}, . . ., λ_{n} \in C$ and consider when:

λ_{1} v_{1} + \dots + λ_{n} v_{n} = \vec{0}

The equation above, using our definitions, implies that:

R (λ_{1}) v_{1} + \dots + R (λ_{n}) v_{n} = \vec{0}

and:

I (λ_{1}) v_{1} + \dots + I (λ_{n}) v_{n} = \vec{0}

Because $v_{1}, . . ., v_{n}$ is LI in $V$ , the equations above imply that $R (λ_{i}) = 0$ and $I (λ_{i}) = 0$ for all $1 \leq i \leq n$ , so then $λ_{i} = 0$ for all $i$ , so $v_{1}, . . ., v_{n}$ are LI in $V_{C}$ .

(b): Since the two bases in (a) are the same size, both $V$ and $V_{C}$ are of the same dimension.
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Complexification of an Operator

complexification of $T$ , $T_{C}$

Suppose $V$ is a real vector space and $T \in L (V)$ . The complexification of $T$ , denoted $T_{C}$ , is the operator $T_{C} \in L (V_{C})$ defined by:

T_{C} (u + i v) = T u + i T v

for $u, v \in V$ .

See HW 9 - Jordon Form, Complexification#2 for why $T_{C} \in L (V_{C})$ . The important thing is that specifically for scalar multiplication, that:

T_{C} (λ (u + i v)) = λ T_{C} (u + i v)

for all $u, v \in V$ and importantly all complex $λ \in C$ .

Example

Suppose $A$ is an $n \times n$ matrix of real numbers. Define $T \in L (R^{n})$ by $T x = A x$ , where elements of $R^{n}$ are though of as $n \times 1$ column vectors. Identifying the complexification of $R^{n}$ with $C^{n}$ , we then have $T_{C} z = A z$ for $z \in C^{n}$ , where again elements of $C^{n}$ are $n \times 1$ column vectors.

In other words, if $T$ is an operator of matrix multiplication by $A$ on $R^{n}$ , then the complexification $T_{C}$ is also a matrix multiplication by $A$ but now acting on the larger domain $C^{n}$ .

Matrix of

T_{C}

equals matrix of

T

Suppose $V$ is a real vector space with basis $v_{1}, . . ., v_{n}$ and $T \in L (V)$ . Then $M (T) = M (T_{C})$ , where both matrices are with respect to the same basis $v_{1}, . . ., v_{n}$ .

Proof
Using a basis of $V$ as a basis of $V_{C}$ , we get that both are bases of the same spaces, then that implies that each matrix element, which is determined from $T v_{i} = \dots$ will be the same for $T_{C} v_{i}$ in $V_{C}$ .
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Looking back at the guarunteed existence of eigenvalues on complex vector spaces, is it the same for real vectors spaces? Sadly no, because the transformation $T (w, z) = (- z, w)$ on $V = R^{2}$ (a CCW rotation by 90 degrees) has no eigenvalues (since normally they're always complex).

However, looking at that example, it's on a nonzero finite-dimensional real vector space with no eigenvalues and thus no 1-dimensional invariant subspaces. That implies that maybe a dimension of 1 or 2 always exists, similar to how all polynomials can be factored into products of 2nd-degree polynomials (without having to factor out complex roots).

As such, we get the following:

Every operator has an invariant subspace of dimension 1 or 2.

Every operator on a nonzero finite-dimensional vector space has an invariant subspace of dimension 1 or 2.

Proof
Every operator on a nonzero finite-dimensional complex vector space has an eigenvalue, and thus has a 1-dimensional invariant subspace.

Hence, assume $V$ is a real vector space and $T \in L (V)$ . The complexification $T_{C}$ has an eigenvalue $a + b i$ , where $a, b \in R$ . Thus $\exists u, v \in V$ , not both 0, where $T_{C} (u + i v) = (a + b i) (u + i v)$ . Using the definition of $T_{C}$ the last equation can be written as:

T_{C} (u + i v) = (a + b i) (u + i v) = T u + i T v = (a u - b v) + (a v + b u) i

Thus:

T u = a u - b v, T v = a v + b u

Let $U = span (u, v)$ in $V$ . Then $U$ is a subspace of $V$ with dimension 1 or 2. The equation above show that $U$ is invariant under $T$ , completing the proof.
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The Minimal Polynomial of the Complexification

Suppose $V$ is a real vector space and $T \in L (V)$ . Repeated application of the definition of $T_{C}$ shows that:

(T_{C})^{n} (u + i v) = T^{n} u + i T^{n} v

$\forall n \in Z^{+}, u, v \in V$ .

Notice that the next result implies that the minimal polynomial of $T_{C}$ has real coefficients:

Minimal polynomial of

T_{C}

equals minimal polynomial of

T

Suppose $V$ is a real vector space and $T \in L (V)$ . Then the minimal polynomial of $T_{C}$ equals the minimal polynomial of $T$ .

Proof
Let $p \in P (R)$ denote the minimal polynomial of $T$ . Using the equation above for $(T_{C})^{n}$ , it's easy to see that $p (T_{C}) = (p (T))_{C}$ and thus $p (T_{C}) = \vec{0}$ .

Now suppose $q \in P (C)$ is a monic polynomial such that $q (T_{C}) = \vec{0}$ . Then $(q (T_{C})) u = \vec{0}$ for all $u \in V$ . Letting $r$ denote the polynomial whose $j$ -th coefficient is the real part of the $j$ -th coefficient of $q$ , we see that $r$ is a monic polynomial and $r (T) = \vec{0}$ . Thus $\deg q = \deg r \geq \deg p$ .

Thus, the two findings in the previous paragraphs show that $p$ is the minimal polynomial of $T_{C}$ as desired.
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Eigenvalues of the Complexification

We'll show that the real eigenvalues of $T_{C}$ are the same eigenvalues of $T$ . We give two proofs:

Real eigenvalues of

T_{C}

Suppose $V$ is a real vector space, $T \in L (V)$ and $λ \in R$ . Then $λ$ is an eigenvalue of $T_{C}$ iff $λ$ is an eigenvalue of $T$ .

Proof (1)
( $\leftarrow$ ): First suppose $λ$ is an eigenvalue of $T$ , so $\exists v \in V$ where $λ v = T v$ and $v \neq \vec{0}$ . Thus $T_{C} v = λ v$ which shows that $λ$ is an eigenvalue of $T_{C}$ .

( $\to$ ): Suppose $λ$ is an eigenvalue of $T_{C}$ . Then $\exists u, v \in V$ where $u + i v \neq \vec{0}$ where:

T_{C} (u + i v) = λ (u + i v)

Then applying $T_{C}$ implies that $T u = λ u$ and $T v = λ v$ . Because either $u \neq \vec{0}$ or $v \neq \vec{0}$ then that implies $λ$ is an eigenvalue of $T$ .
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Proof (2)
The real eigenvalues of $T$ are the real zeroes of the minimal polynomial for $T$ . The real eigenvalues of $T_{C}$ are the real zeroes of the minimal polynomial of $T_{C}$ using the same argument. These two minimal polynomials are the same, so the eigenvalues of $T$ are the real eigenvalues of $T_{C}$ as desired.
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T_{C} - λ I

and

T_{C} - \overset{―}{λ} I

Suppose $V$ is a real vector space, $T \in L (V)$ , $λ \in C$ , and $j \in Z^{\geq 0}$ and $u, v \in V$ . Then:

(T_{C} - λ I)^{j} (u + i v) = \vec{0} \Leftrightarrow (T_{C} - \overset{―}{λ} I)^{j} (u - i v) = \vec{0}

Proof
We'll do an induction on $j$ . The base case $j = 0$ is (since any operator to the 0 is just the identity) claims that $u + i v = \vec{0} \Leftrightarrow u - i v = \vec{0}$ which is clearly true.

Now suppose that $j \geq 1$ and the desired result holds for the $j - 1$ case. Suppose $(T_{C} - λ I)^{j} (u + i v) = \vec{0}$ . Then:

(T_{C} - λ I)^{j - 1} ((T_{C} - λ I) (u + i v)) = \vec{0}

Writing $λ = a + b i$ where $a, b \in R$ then:

(T_{C} - λ I) (u + i v) = (T u - a u + b v) + i (T v - a v - b u)

(T_{C} - \overset{―}{λ} I) (u - i v) = (T u - a u + b v) - i (T v - a v - b u)

Using the inductive hypothesis above with the top equation gives that:

(T_{C} - \overset{―}{λ} I)^{j - 1} ((T u - a u + b v) - i (T v - a v - b u)) = \vec{0}

The equation above and the second equation in the block above implies that $(T_{C} - \overset{―}{λ} I)^{j} (u - i v) = \vec{0}$ . This does the proof for $(\to)$ , but the proof for $(\leftarrow)$ is the same, replacing $λ$ with $\overset{―}{λ}$ .
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Nonreal eigenvalues of

T_{C}

come in pairs

Suppose $V$ is a real vector space, $T \in L (V)$ and $λ \in C$ . Then $λ$ is an eigenvalue of $T_{C}$ iff $\overset{―}{λ}$ is an eigenvalue of $T_{C}$ .

Proof
Take $j = 1$ in Chapter 9 - Operators on Real Vector Spaces#^914b01.
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Multiplicity of

λ

equals the multiplicity of

\overset{―}{λ}

Suppose $V$ is a real vector space, $T \in L (V)$ and $λ \in C$ is an eigenvalue of $T_{C}$ . Then the multiplicity of $λ$ as an eigenvalue of $T_{C}$ equals the multiplicity of $\overset{―}{λ}$ as an eigenvalue of $T_{C}$ .

Proof
Suppose $u_{1} + i v_{1}, . . ., u_{m} + i v_{m} \in V_{C}$ is a basis of the generalized eigenspace $G (λ, T_{C})$ , where $u_{j}, v_{j} \in V$ . Then using Chapter 9 - Operators on Real Vector Spaces#^914b01, then clearly $u_{1} - i v_{1}, . . ., u_{m} - i v_{m}$ is a basis of $G (\overset{―}{λ}, T_{C})$ . Thus both $λ, \overset{―}{λ}$ have multiplicity $m$ as eigenvalues of $T_{C}$ .
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Example

Suppose $T \in L (R^{3})$ is defined by:

T (x_{1}, x_{2}, x_{3}) = (2 x_{1}, x_{2} - x_{3}, x_{2} + x_{3})

Then:

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

You can verify that $2$ is an eigenvalue of $T$ with multiplicity 1 and $T$ has no other eigenvalues.

If we identity the complexification of $R^{3}$ with $C^{3}$ then the matrix of $T_{C}$ with respect to the standard basis of $C^{3}$ is the matrix above. The eigenvalues of $T_{C}$ are $2, 1 + i, 1 - i$ , each with multiplicity 1. Thus the nonreal eigenvalues of $T_{C}$ come as a pair, with each the complex conjugate of the other and with the same multiplicity, as we expect.

Operator on odd-dimensional vector space has an eigenvalue.

Every operator on an odd-dimensional real vector space has an eigenvalue.

Proof
Suppose $V$ is a real vector space with odd dimension and $T \in L (V)$ . Because the nonreal eigenvalues of $T_{C}$ come in pairs with equal multiplicity, the sum of the multiplicities of all the nonreal eigenvalues of $T_{C}$ is an even number.

Because the sum of the multiplicities of all eigenvalues of $T_{C}$ equals the (complex) dimension of $V_{C}$ , the conclusion of the paragraph above implies that $T_{C}$ has a real eigenvalue. Every real eigenvalue of $T_{C}$ is an eigenvalue of $T$ , giving the desired result.
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Characteristic Polynomial of the Complexification

In Chapter 8 - Operators on Complex Vector Spaces#^df8684 we defined the characteristic polynomial for complex vector spaces. The next result generalizes it for finite-dimensional real vector spaces.

Characteristic polynomial of

T_{C}

Suppose $V$ is a real vector space and $T \in L (V)$ . Then the coefficients of the characteristic polynomial of $T_{C}$ are all real.

Proof
Suppose $λ$ is a nonreal eigenvalue of $T_{C}$ with multiplicity $m$ . Then $\overset{―}{λ}$ is also an eigenvalue of $T_{C}$ with multiplicity $m$ . Thus the characteristic polynomial of $T_{C}$ includes the factor $(z - λ)^{m}$ and $(z - \overset{―}{λ})^{m}$ , where multipliying these two factors gives:

(z - λ)^{m} (z - \overset{―}{λ})^{m} = (z^{2} + 2 R (λ) z + | λ |^{2})^{m}

The polynomial on the right only has real coefficients.

The characteristic polynomial of $T_{C}$ is the product of terms of th form above and the terms of the form $(z - t)^{d}$ where $t$ is a real eigenvalue of $T_{C}$ with multiplicity $d$ . Thus the coefficients of the characteristic polynomial of $T_{C}$ are all real.
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characteristic polynomial (real vector spaces)

Suppose $V$ is a real vector space and $T \in L (V)$ . Then the characteristic polynomial of $T$ is defined by the characteristic polynomial of $T_{C}$ .

Example

See:

Example

Suppose $T \in L (R^{3})$ is defined by:
$T (x_{1}, x_{2}, x_{3}) = (2 x_{1}, x_{2} - x_{3}, x_{2} + x_{3})$
Then:
$M (T) = [\begin{matrix} 2 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & 1 & 1 \end{matrix}]$
You can verify that $2$ is an eigenvalue of $T$ with multiplicity 1 and $T$ has no other eigenvalues.

We said the eigenvalues of $T_{C}$ are $2, 1 + i, 1 - i$ each with multiplicity 1. Thus the characteristic polynomial for the complexification $T_{C}$ is:
$(z - 2) (z - (1 + i)) (z - (1 - i)) = z^{3} - 4 z^{2} + 6 z - 4$
Which is the same as the characteristic polynomial of $T$ .

Degree and zeros of characteristic polynomial

Suppose $V$ is a real vector space and $T \in L (V)$ . Then:

the coefficients of the characteristic polynomial of $T$ are all real.
the characteristic polynomial of $T$ has degree $\dim (V)$
The eigenvalues of $T$ are precisely the real zeros of the characteristic polynomial of $T$ .

Proof
(a): Holds from

Characteristic polynomial of

T_{C}

Suppose $V$ is a real vector space and $T \in L (V)$ . Then the coefficients of the characteristic polynomial of $T_{C}$ are all real.

(b): Follows from

Degree and zeros of characteristic polynomial

Suppose $V$ is a complex vector space and $T \in L (V)$ . Then:

the characteristic polynomial of $T$ has degree $\dim (V)$
the zeroes of the characteristic polynomial of $T$ are the eigenvalues of $T$ .

(c): The real zeroes of the characteristic polynomial of $T$ are the real eigenvalues of $T_{C}$ (see (a)), which are the eigenvalues of $T$ .
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Cayley-Hamilton Theorem (Real

V

)

Suppose $T \in L (V)$ . Let $q$ denote the characteristic polynomial of $T$ . Then $q (T) = \vec{0}$ .

Proof
We have already proved this result for $V$ as a complex vector space, so suppose $V$ is a real vector space.

The Cayley-Hamilton Theorem for complex vector spaces implies that $q (T_{C}) = \vec{0}$ . Thus we also have $q (T) = \vec{0}$ as desired (just consider changing $V_{C} \to V$ and $T_{C} \to T$ ).
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Example

Refer to:

Example

See:

Example

We said the eigenvalues of $T_{C}$ are $2, 1 + i, 1 - i$ each with multiplicity 1. Thus the characteristic polynomial for the complexification $T_{C}$ is:
$(z - 2) (z - (1 + i)) (z - (1 - i)) = z^{3} - 4 z^{2} + 6 z - 4$
Which is the same as the characteristic polynomial of $T$ .

The Cayley-Hamilton Theorem says that $T^{3} - 4 T^{2} + 6 T - 4 I = \vec{0}$ , which is easily verified.

Characteristic polynomial is a multiple of minimal polynomial

Suppose $T \in L (V)$ . Then:

The degree of the minimal polynomial of $T$ is at most $\dim (V)$
The characteristic polynomial of $T$ is a polynomial multiple of the minimal polynomial of $T$ .

Proof
(a): Follows immediately from Chapter 9 - Operators on Real Vector Spaces#^b355e7.

(b): Follows from the same theorem and Chapter 8 - Operators on Complex Vector Spaces#^ea692e.
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9.B: Operators on Real Inner Product Spaces

Let's switch to focus on inner product spaces. We'll need to use the fact that all real vector spaces have an invariant subspace with dimension 1 or 2.

Normal Operators on Real Inner Product Spaces

The Complex Spectral Theorem gives a complete description of normal operators on complex inner product spaces.