1. Solution: $V_\mathbb{C}$ is clearly closed under addition. We can write each complex number in the form $a + bi$ for some $a, b \in \mathbb{R}$ and we have

$$ (a + bi)(u + iv) = (au – bv) + i(av + bu) = (au – bv, av + bu) \in V \times V = V_{\mathbb{C}}, $$ where the first equality follows from the definition and the second because the vectors inside both parentheses are in $V$. Thus $V_{\mathbb{C}}$ is closed under complex scalar multiplication. If $0$ is the additive identity on $V$, then $0 + i0$ is the additive identity on $V_\mathbb{C}$ because

$$ (u + iv) + (0 + i0) = (u + 0) + i(v + 0) = u + iv. $$ One easily checks that the rest of the properties listed in 1.19 are satisfied by $V_\mathbb{C}$. Therefore $V_\mathbb{C}$ is a complex vector space.

2. Solution: Let $u_1, u_2, v_1, v_2 \in V$. Then

$$ \begin{aligned} T_\mathbb{C} \big((u_1 + iv_1) + (u_2 + iv_2)\big) &= T_\mathbb{C} \big((u_1 + u_2) + i(v_1 + v_2)\big)\\ &= T(u_1 + u_2) + iT(v_1 + v_2)\\ &= (Tu_1 + iTv_1) + (Tu_2 + iTv_2)\\ &= T_\mathbb{C}(u_1 + iv_1) + T_\mathbb{C}(u_2 + iv_2). \end{aligned} $$ Hence $T_\mathbb{C}$ satisfies the additivity property of linear maps. Now let $u, v \in V$ and $a, b \in \mathbb{R}$. Then

$$ \begin{aligned} T_\mathbb{C}\big((a + bi)(u + iv)\big) &= T_\mathbb{C}\big((au – bv) + i(av + bu)\big)\\ &= T(au – bv) + iT(av + bu)\\ &= (aTu – bTv) + i(aTv + bTu)\\ &= (a + bi)(Tu + iTv)\\ &= (a + bi)T_\mathbb{C}(u + iv) \end{aligned} $$ where the fourth line follows from the definition of complex scalar multiplication on $V_\mathbb{C}$. Hence $T_\mathbb{C}$ satisfies the homogeneity property of linear maps. Therefore $T_\mathbb{C}$ is a linear map.

3. Solution: The forward direction is obvious, because $\mathbb{R} \subset \mathbb{C}$. For the other direction we can just repeat the same argument used in the proof of 9.4 to show the linear independence of $v_1, \dots, v_n$.

4. Solution: Suppose $v_1, \dots, v_m$ spans $V_\mathbb{C}$. Let $v \in V$. Then $v + i0 \in V_\mathbb{C}$ and we can write

$$ v + i0 = \lambda_1 v_1 + \dots + \lambda_m v_m = (\operatorname{Re} \lambda_1 v_1 + \dots + \operatorname{Re} \lambda_m v_m) + i(\operatorname{Im} \lambda_1 v_1 + \dots + \operatorname{Im} \lambda_m v_m) $$ for some $\lambda_1, \dots, \lambda_m \in \mathbb{C}$. The equation above implies that

$$ \operatorname{Re} \lambda_1 v_1 + \dots + \operatorname{Re} \lambda_m v_m = v $$ Therefore $v \in \operatorname{span}(v_1, \dots, v_m)$. Hence $v_1, \dots, v_m$ spans $V$.

Conversely, suppose $v_1, \dots, v_m$ spans $V$. Then we can reduce this list to a basis of $V$. But a basis of $V$ is also a basis $V_\mathbb{C}$. Therefore we can reduce $v_1, \dots, v_m$ to a basis of $V_\mathbb{C}$ and this implies that it spans $V_\mathbb{C}$.

5. Solution: Suppose $u, v \in V$. Then

$$ \begin{aligned} (S + T)_\mathbb{C}(u + iv) &= (S + T)u + i(S + T)v\\ &= (Su + iSv) + (Tu + iTv)\\ &= S_\mathbb{C}(u + iv) + T_\mathbb{C}(u + iv). \end{aligned} $$ Thus $(S + T)_\mathbb{C} = S_\mathbb{C} + T_\mathbb{C}$. Now suppose $\lambda \in \mathbb{R}$. Then

$$ (\lambda T)_\mathbb{C}(u + iv) = \lambda Tu + \lambda iTv = \lambda(Tu + iTv) = \lambda T_\mathbb{C}(u + iv). $$ Therefore $(\lambda T)_\mathbb{C} = \lambda T_\mathbb{C}$.

6. Solution: Suppose $T_\mathbb{C}$ is invertible. Then, because $T_\mathbb{C}$ is surjective, for every $w \in V$ there exist $u, v \in V$ such that $T_\mathbb{C}(u + iv) = w + i0$. This means that $$ Tu + iTv = w + i0. $$ Thus $Tu = w$. Hence $T$ is surjective and therefore invertible.

Conversely, suppose $T$ is invertible. Let $u, v \in V$. By surjectivity of $T$, there exist $\hat{u}, \hat{v} \in V$ such that $T\hat{u} = u$ and $T\hat{v} = v$. Thus

$$ T_\mathbb{C}(\hat{u} + i\hat{v}) = T\hat{u} + iT\hat{v} = u + iv. $$ Since $u$ and $v$ were arbitrary, it follows that $T_\mathbb{C}$ is surjective and therefore invertible.

7. Solution: Suppose $\mathbb{N}_C$ is nilpotent. By 8.18, for any $v \in V$, we have

$$ 0 + i0 = (N_\mathbb{C})^{\dim V}(v + i0) = N^{\dim V}v + i0 $$ where the second equality follows from 9.9. This implies that $N^{\dim V}v = 0$. Thus $N$ is nilpotent.

The other direction is obvious from 9.9 and the definitions.

8. Solution: If $T_\mathbb{C}$ had a nonreal eigenvalue $\lambda$, then by 9.11 and 9.16 it would have $4$ eigenvalues, namely $5, 7, \lambda, \overline{\lambda}$, which is a contradiction because the dimension of $(\mathbb{R}^3)_\mathbb{C}$ equals $3$ (see 9.4 (b)) and $\mathbb{T}_C$ has at most $3$ eigenvalues (see 5.13).

9. Solution: Suppose by contradiction that $T \in \mathcal{L}(\mathbb{R}^7)$ is such that $T^2 + T + I$ is nilpotent. Then by Exercise 7 $(T^2 + T + I)_\mathbb{C}$ is also nilpotent and its minimal polynomial is of the form $z^j$ for some positive integer $j$ (because $0$ is its only eigenvalue, see Exercise 7 in section 8A and 8.49). We have $$ z^2 + z + 1 = \left(z – \frac{-1 + i\sqrt{3}}{2}\right)\left(z – \frac{-1 – i\sqrt{3}}{2}\right). $$ Define $p \in \mathcal{P}(\mathbb{R})$ by

$$ p(z) = (z^2 + z + 1)^j. $$ Then $p$ has no real roots and $p(T_\mathbb{C}) = 0$. This is a contradiction, because $p$ must be a polynomial multiple of the minimal polynomial of $T_\mathbb{C}$ (by 8.46) and the minimal polynomial of $T_\mathbb{C}$ has at least one real root (see 9.19, 9.11 and 8.49).

10. Solution: Choose $\lambda \in \mathbb{C}$ such that $\lambda^2 + \lambda + 1 = 0$ and define $T \in \mathcal{L}(\mathbb{C}^7)$ by

$$ Tv = \lambda v $$ for all $v \in \mathbb{C}^7$. Then

$$ (T^2 + T + I)v = T^2v + Tv + Iv = (\lambda^2 + \lambda + 1)v = 0. $$ Therefore $T^2 + T + I$ is nilpotent.

11. Solution: From the definitions, it is easy to see that

$$ (T_\mathbb{C}^2 + bT_\mathbb{C} + cI) = 0. $$ Therefore $z^2 + bz + c$ is polynomial multiple of the minimal polynomial of $T_\mathbb{C}$ (see 8.46). Note that, because this polynomial has real coefficients, it either has two real roots or two nonreal roots. Thus $T$ has an eigenvalue if and only if $T_\mathbb{C}$ has a real root (see 9.10), which happens if and only if $z^2 + bz + c$ has a real root (by the previous reasoning and 8.49), which happens if and only if $b^2 \ge 4c$.

12. Solution: By Exercise 3 in section 8D, the minimal polynomial of $T^2 + bT + cI$ is of the form $z^m$ for some positive integer $m$. Let the $p$ be the polynomial defined by $$ p(z) = (z^2 + bz + c)^m. $$ Then $p$ has no real roots (because $z^2 + bz + c$ does not) and $p(T_\mathbb{C}) = 0$. Thus 8.46 and 8.49 imply that $T_\mathbb{C}$ has no real eigenvalues. Now 9.10 implies that $T$ has no eigenvalues.

13. Solution: We have

$$ z^2 + bz + c = (z – \lambda)(z – \overline{\lambda}) $$ for some nonreal $\lambda \in \mathbb{C}$. Suppose $v \in \operatorname{null}(T_\mathbb{C}^2 + bT_\mathbb{C} + cI)^j$. Then

$$ 0 = (T_\mathbb{C}^2 + bT_\mathbb{C} + cI)^j v = (T_\mathbb{C} – \lambda I)^j(T_\mathbb{C} – \overline{\lambda}I)^jv. $$ It is easy to check that $v$ is of the form $v = u + w$ where $u \in G(\lambda, T_\mathbb{C})$ and $w \in G(\overline{\lambda}, T_\mathbb{C})$. We have

$$ 0 = (T_\mathbb{C} – \lambda I)^j(T_\mathbb{C} – \overline{\lambda}I)^jv = (T_\mathbb{C} – \lambda I)^j(T_\mathbb{C} – \overline{\lambda}I)u + (T_\mathbb{C} – \lambda I)^j(T_\mathbb{C} – \overline{\lambda}I)^jw. $$ Since each generalized eigenspace is invariant under $T_\mathbb{C}$ and every vector in $V$ (the left side of the equation) can be written uniquely as linear combination of generalized eigenvectors that correspond to distinct eigenvalues, the equation above implies that

$$ (T_\mathbb{C} – \lambda I)^j(T_\mathbb{C} – \overline{\lambda}I)^ju = 0 \text{ and } (T_\mathbb{C} – \lambda I)^j(T_\mathbb{C} – \overline{\lambda}I)^jw = 0. $$ Hence $u \in \operatorname{null} (T_\mathbb{C} – \lambda I)^j$ and $w \in \operatorname{null} (T_\mathbb{C} – \overline{\lambda} I)^j$. Therefore

$$ \operatorname{null}(T_\mathbb{C}^2 + bT_\mathbb{C} + cI)^j \subset \operatorname{null} (T_\mathbb{C} – \lambda I)^j \oplus \operatorname{null} (T_\mathbb{C} – \overline{\lambda} I)^j. $$ The inclusion in the other direction is obvious. Thus

$$ \operatorname{null}(T_\mathbb{C}^2 + bT_\mathbb{C} + cI)^j = \operatorname{null} (T_\mathbb{C} – \lambda I)^j \oplus \operatorname{null} (T_\mathbb{C} – \overline{\lambda} I)^j, $$ which gives us that

$$ \dim \operatorname{null}(T_\mathbb{C}^2 + bT_\mathbb{C} + cI)^j = \dim \operatorname{null} (T_\mathbb{C} – \lambda I)^j + \dim \operatorname{null} (T_\mathbb{C} – \overline{\lambda} I)^j. $$ By 9.12, the two dimensions on the right side of the equation above are equal. Therefore the left side is even. One easily checks that

$$ (\operatorname{null} (T^2 + bT + cI))_\mathbb{C} = \operatorname{null} (T_\mathbb{C}^2 + bT_\mathbb{C} + Ic). $$ Now 9.4 (b) yields the desired result.

14. Solution: Because it is nilpotent, the minimal polynomial of $T_\mathbb{C}^2 + T_\mathbb{C} + I$ is of the form $z^m$ for some positive integer $m$. We have

$$ z^2 + z + 1 = (z – \lambda)(z – \overline{\lambda}) $$ for some nonreal $\lambda \in \mathbb{C}$. Thus

$$ 0 = (T_\mathbb{C}^2 + T_\mathbb{C} + I)^m = (T_\mathbb{C} – \lambda)^m(T_\mathbb{C} – \overline{\lambda})^m. $$ This, together with 9.16, implies that the eigenvalues of $T_\mathbb{C}$ are $\lambda$ and $\overline{\lambda}$, with equal multiplicities, namely $4$. The characteristic polynomial of $T_\mathbb{C}$, and of $T$ as well by definition, is therefore $$ (z – \lambda)^4(z – \overline{\lambda})^4. $$ which equals $(z^2 + z + 1)^4$. By the Cayley-Hamilton Theorem (9.24), it follows that $$ (T^2 + T + I)^4 = 0. $$

15. Solution: Suppose $U$ is a subspace of $V$ invariant under $T$. Because $T|_U$ doesn’t have an eigenvalue, 9.19 implies that $U$ has even dimension.

16. Solution: Suppose $T$ is such that $T^2 = -I$. Then clearly $T$ does not have an eigenvalue. Then 9.19 implies that $\dim V$ is even.

Conversely, suppose $V$ has even dimension. Let $v_1, \dots, v_n, u_1, \dots, u_n$ be a basis of $V$. Define $T \in \mathcal{L}(V)$ by

$$ Tv_j = -u_j, Tu_j = v_j $$ for each $j = 1, \dots, n$. We have

$$ T^2v_j = -Tu_j = -v_j \text{ and } T^2u_j = Tv_j = -u_j. $$ Thus $T^2 = -I$.

17. Solution:

(a) Obviously $V$ closed under addition and complex scalar multiplication. One can easily check that the other properties in 1.19 are satisfied.

(b) Let $n$ be the integer such that $\dim V = 2n$ and consider the following process.

- Step $1$: Choose a nonzero $v_1 \in V$. Then $v_1, Tv_1$ is linearly independent in $V$ as real vector space because $v_1$ is not an eigenvector of $T$ (because $T$ has no eigenvectors). Set $U_1 = \operatorname{span}(v_1, Tv_1)$. Then $\dim U_1 = 2$ and $U_1$ is invariant under $T$.
- Step $j$: If $j = n + 1$, stop the process. We have that

$$ \dim U_{j-1} = 2(j-1) \le 2n – 2 < \dim V, $$ $U_{j-1}$ is invariant under $T$ and

$$ v_1, Tv_1, \dots, v_{j-1}, Tv_{j-1} $$ is a basis of $U_{j-1}$. Hence there exists a nonzero $w \in V$ such that $w \notin U_{j-1}$. Since $T$ is surjective, there exists $v_j \in V$ such that $Tv_j = w$. Thus $v_j \notin U_{j-1}$, because $U_{j-1}$ being invariant under $T$ would imply $w \in U_{j-1}$. Moreover, the list

$$ v_1, Tv_1, \dots, v_j, Tv_j \tag{1} $$ is linearly independent. To see this, let $a_1, \dots, a_j, c_1, \dots, c_j \in \mathbb{R}$ such that

$$ a_1v_1 + c_1Tv_1 + \dots + a_jv_j + c_jTv_j = 0 $$ We already know that $v_j$ is not in the span of the previous vectors, so $c_j = 0$ implies $a_j = 0$ which implies that the rest of the $a$’s and $c$’s is $0$. Assume by contradiction $c_j \neq 0$. We can write the equation above as

$$ a_1v_1 + c_1Tv_1 + \dots + a_{j-1}v_{j-1} + c_{j-1}Tv_{j-1} = – a_jv_j – c_jTv_j. \tag{2} $$ Applying $T$ to both sides we get

$$ a_1Tv_1 – c_1v_1 + \dots + a_{j-1}Tv_{j-1} – c_{j-1}v_{j-1} = – a_jTv_j + c_jv_j. \tag{3} $$ Multiplying $(3)$ by $\frac{a_j}{c_j}$ and summing with $(2)$ shows us that

$$ \left(\frac{-a_j^2}{c_j} – c_j\right)Tv_j \in U_{j-1}. $$ Hence the parentheses above equals $0$. This means that $-a_j^2 = c_j^2$, which is only possible if $a_j = c_j = 0$ and contradicts our assumption that $c_j \neq 0$. Therefore the list in $(1)$ is linearly independent. Set

$$ U_j = \operatorname{span}(v_1, Tv_1, \dots, v_j, Tv_j). $$ Then $\dim U_j = 2j$ and $U_j$ is invariant under $T$.

At the end of the process, we will have constructed a subspace $U_n$ of $V$ which has a basis

$$ v_1, Tv_1, \dots, v_n, Tv_n $$ and dimension $2n$. Thus $U_n = V$ and the above list is a basis of $V$. Clearly $v_1, \dots, v_n$ spans $V$ as a complex vector space. Futhermore, it is linearly independent in $V$ as a complex vector space, because if a (complex) linear combination of it equals $0$, then a linear combination of the list above equals $0$, which implies that the coefficients are $0$. Therefore $V$ as a complex vector space has dimension $n$, completing the proof.

18. Solution: The proof that (a) $\Rightarrow$ (b) below is basically a rewrite of the proof 5.27 with a few tweakings.

Suppose (a) holds. We will prove (b) by induction on $\dim V$. First, if $\dim V = 1$, then the matrix of $T$ with respect to any basis only has one entry and thus (b) trivially holds. Assume that $\dim V > 1$ and that (a) implies (b) for all vector spaces of lower dimension.

Let $\lambda \in \mathbb{R}$ be an eigenvalue of $T$, which exists by (a) and 9.11. Define

$$ U = \operatorname{range}(T – \lambda I). $$ Because $\dim \operatorname{null}(T – \lambda I) \ge 1$, the Fundamental Theorem of Linear Maps implies that $\dim U < \dim V$. Futhermore, $U$ is invariant under $T$ and so $U_\mathbb{C}$ is invariant under $({T|_U})_\mathbb{C}$. All the eigenvalues of $({T|_U})_\mathbb{C}$ are real. To see this, suppose $\lambda \in \mathbb{C}$ is an eigenvalue of $({T|_U})_\mathbb{C}$ and $u_1 + iu_2$ a corresponding eigenvector for some $u_1, u_2 \in U$. Then

$$ \lambda(u_1 + iu_2) = ({T|_U})_\mathbb{C}(u_1 + iu_2) = (T|_U)u_1 + i(T|_U)u_2 = Tu_1 + iTu_2 = T_\mathbb{C}(u_1 + iu_2), $$ which implies that $\lambda$ is an eigenvalue of $T_\mathbb{C}$ and thus must be real. By the induction hypothesis there exists a basis $u_1, \dots, u_m$ of $U$ with respect to which the matrix of $T|_U$ is upper triangular. From 5.26, we get that

$$ Tu_j = (T|_U)u_j \in \operatorname{span}(u_1, \dots, u_j). \tag{4} $$ Extend it to a basis $u_1, \dots, u_m, v_1, \dots, v_n$ of $V$. For each $k$ we have

$$ Tv_k = (T – \lambda I)v_k + \lambda v_k. $$ The definition of $U$ shows that $(T – \lambda I)v_k \in U$. Hence, the equation above shows that

$$ Tv_k \in \operatorname{span}(u_1, \dots, u_m, v_1, \dots, v_k). \tag{5} $$ Using $(4)$, $(5)$ and 5.26, we conclude that the matrix of $T$ with respect to basis $u_1, \dots, u_m, v_1, \dots, v_n$ of $V$ is upper triangular.

9.7 and 5.32 show that (b) implies (a). Hence the equivalence between (a) and (b) is established. Now it suffices to show that (a) and (c) are equivalent.

Suppose (a) holds. Let $\lambda \in \mathbb{R}$ be an eigenvalue of $T_\mathbb{C}$ and let $d$ denote its multiplicity. Let $u_1 + iv_1, \dots, u_d + iv_d$ be a basis of $G(\lambda, T_\mathbb{C})$. For each $j = 1, \dots, d$, we have

$$ (T – \lambda)^{\dim V} u_j + i(T – \lambda)^{\dim V} v_j = ((T – \lambda)^{\dim V})_\mathbb{C}(u_j + iv_j) = (T_\mathbb{C} – \lambda)^{\dim V}(u_j + iv_j) = 0. $$ This implies that $u_1, v_1, \dots, u_d, v_d \in G(\lambda, T)$. Since $G(\lambda, T_\mathbb{C})$ is contained in the complex span of this list, it follows that the dimension of this complex span is at least $d$. 9.4 (b) now implies that the dimension of the real span of $u_1, v_1, \dots, u_d, v_d$ is at least $d$. Hence $\dim G(\lambda, T) \ge d$. Since the sum of multiplicities of the generalized eigenspaces of $T$ is at most $\dim V$, it follows that the only way the dimensions will fit is if $\dim G(\lambda, T) = d$.

Thus, summing all the generalized eigenspaces of $T$ we get a subspace of $V$ with dimension $\dim V$, namely $V$. Taking a basis of generalized eigenspace and putting this bases together gives a basis consisting of generalized eigenvectors of $V$. Thus (c) holds.

Now suppose (c) holds. Then $T_\mathbb{C}$ has real eigenvalues, whose sum of multiplicities already equal $\dim V$. Hence $T_\mathbb{C}$ cannot have a nonreal eigenvalue. Thus (a) holds.

19. Solution: By the same reasoning used in the proof 8.4, it follows that $\dim \operatorname{null} T^{n-1} \ge n – 1$. Thus, the multiplicty of $0$ as an eigenvalue of $T$, and $T_\mathbb{C}$ as well, is at least $n – 1$. Hence $T_\mathbb{C}$ has at most another eigenvalue. Since nonreal eigenvalues of $T_\mathbb{C}$ come in pairs (see 9.16), this other eigenvalue of $T_\mathbb{C}$ must be real.

## Jonathan Sharir-Smith

24 Aug 2020Perhaps I am totally missing it, but can someone explain why Q9 is true while Q10 is true (that is, I don't really see where the hypothesis that we're working with $\Bbb R$ in particular is used in Q9)?

## Jonathan Sharir-Smith

24 Aug 2020Never mind, I was mixing up the characteristic polynomial of the operator and of the polynomial of the operator. Thanks again for this great website!

## Yuheng

24 Aug 2020For 9A13, can't we use a similar method as 11 and 12?

First, because null(T^2+bT+cI)^j is invariant under T (8.20), define S as T subject to null(T^2+bT+cI)^j. Because b^2<4c, p(z) = (z^2+bz+c)^j has no real roots. Also, by definition, p(S)=0. So p(Sc)=0. So p is a polynomial multiple of minimal polynomial of Sc, which means minimal polynomial of Sc has no real roots. So Sc has no real eigenvalue. So S has no eigenvalue. So null(T^2+bT+cI)^j must have even dimension, otherwise 9.19 tells us that S has some eigenvalue.

## Yuheng

24 Aug 2020For the reverse direction of 9A11, I think it's better to argue with more detail. We have p(z)=z^2+b*z+c and p(Tc)=0, so p is a polynomial multiple of minimal polynomial of Tc. And because Tc is an operator on complex vector space, it must have some eigenvalue, which is also a root of its minimal polynomial, which must equal some root of p. Because all roots of p are real, this root must be real, so Tc has a real eigenvalue, and so is T.

## Zixiu Su

28 Jun 20209A13, I am not sure how to verify v is in the form of the sum of two vectors from those two distinct generalized eigenspaces. How do you know those two are the only generalized eigenspaces of Tc?

## Linearity

28 Jun 2020This is a particular case of the primary decomposition theorem, see Hoffman-Kunze. You can see it as follows, suppose $f(T)g(T)v=0$ and polynomials $f,g$ are relatively prime. Then $v=u+w$ such that $f(T)u=0$ and $g(T)w=0$ in a unique way.

Proof: Since $f$ and $g$ are relatively prime, there exists polynomials$h_1,h_2$ such that $$fh_1+gh_2=1.$$Then $$v=1v=f(T)h_1(T)v+g(T)h_2(T)v.$$ Let $u=g(T)h_2(T)v$ and $w=f(T)h_1(T)v$. This shows existence. Please try uniqueness yourself.

## Leo

18 Apr 2020In exercise 19, you say "the multiplicity of T, and T_C as well, is at least n-1." But I don't see where it is proved that the multiplicity of real eigenvalues of T equal those of T_C.

## Linearity

20 Apr 2020You must have some misunderstanding. That sentence means it can be shown that the multiplicity of zero as an eigenvalue of $T$ as well $T_{\mathbb C}$ is at least $n-1$.

## Marcel Ackermann

10 Feb 2019Does anybody have a solution for 9A16?