# Selected Exercises #

## 5.A #

12. Define $T \in \mathcal L(\mathcal P_4(\mathbf{R}))$ by

$$(Tp)(x) = xp’(x)$$

for all $x \in \mathbf{R}$. Find all eigenvalues and eigenvectors of $T$.

Observe that, if $p = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$, then $$x p’(x) = a_1 x + 2 a_2 x^2 + 3 a_3 x^3 + 4 a_4 x^4.$$

This immediately tells us that eigenvectors will have to include only one term.

In light of this, the eigenvalues of $T$ are $0, 1, 2, 3, 4$ and the corresponding eigenvectors are represented by $a_0, a_1 x, a_2 x^2, a_3 x^3, a_4 x^4$. Note that this fits with the theorem that a $\dim 5$ vector space can have at most $5$ eigenvalues.

15. Suppose $T \in \mathcal L(V)$. Suppose $S \in \mathcal L(V)$ is invertible.
(a) Prove that $T$ and $S^{-1}TS$ have the same eigenvalues.
(b) What is the relationship between the eigenvectors of $T$ and the eigenvectors of $S^{-1}TS$?

Regarding (a), suppose $T$ has eigenvalue $\lambda$. Then $Tv = \lambda v$ for some $v \in V$. $S$ is surjective (since it’s invertible). So, there exists some $u \in V$ such that $Su = v$. In this case, $$(S^{-1}TS)(u) = S^{-1}(T(S(u))) = S^{-1}(Tv) = S^{-1}(\lambda v) = \lambda S^{-1}(v) = \lambda u.$$

Hence, $\lambda$ is also an eigenvalue of $S^{-1}TS$.

Now for (b). In (a), $v$ was an eigenvector of $T$ and $S^{-1}(v)$ was an eigenvector of $S^{-1}TS$. This relationship generalizes, since, for each eigenvalue of the two linear maps, $v$ is an eigenvector of $T$ implies that $S^{-1}(v)$ is an eigenvector of $S^{-1}TS$.

18. Show that the operator $T \in \mathcal L(C^{\infty})$ defined by $$T(z_1, z_2, \dots) = (0, z_1, z_2, \dots)$$ has no eigenvalues.

Assume the operator has an eigenvalue. Then

$$T(z_1, z_2, \dots) = (\lambda z_1, \lambda z_2, \lambda z_3, \dots)$$

for some $\lambda \in F$ and eigenvector $v \in V$. This implies that

$$(0, z_1, z_2, \dots) = (\lambda z_1, \lambda z_2, \lambda z_3, \dots).$$

If $\lambda z_1 = 0$, then either $\lambda = 0$ or $z_1 = 0$. If $\lambda = 0$, then all $\lambda z_i = 0$ implying that all $z_i = 0$. This can’t be true since $z_1, \dots$ is supposed to be an eigenvector. If $z_1 = 0$, then by the latter elements in the two sides of this equation, $\lambda z_2 = z_1 = 0$, and so on for each $\lambda z_i = z_{i-1} = 0$. We already showed what would happen if $\lambda$ was 0, so this implies that $z_i = 0$ for all $i \in 1, \dots$. However, this gives us a contradiction, since $z_1, \dots$ is supposed to be an eigenvector and therefore is non-zero.

20. Find all eigenvalues and eigenvectors of the backward shift operator $T \in \mathcal L(F^{\infty})$ defined by $$T(z_1, z_2, z_3, \dots) = (z_2, z_3, \dots).$$

In order for $T(z_1, z_2, z_3, \dots) = (z_2, z_3, \dots) = \lambda(z_1, z_2, z_3, \dots)$, $\lambda z_i = z_{i+1}$ for all $i \in 1, \dots$. This means that $\lambda$ can be any positive value in $F$, but to be an eigenvector, $z$ must obey the recurrence $z_i = \lambda^{i-1} z_{i-1}$ for all $i \in 1, 2, \dots$.

22. Suppose $T \in \mathcal L(V)$ and there exist nonzero vectors $v$ and $w$ in $V$ such that $$Tv = 3w \text{ and } Tw = 3v.$$

Prove that $3$ or $-3$ is an eigenvalue of $T$.

Observe that $Tv, 3w$ and $Tw, 3v$ are linearly dependent lists in $V$. So, $$Tw - 3v = 0 = Tv - 3w.$$

By algebra, $$Tw - Tv = 3v - 3w$$ and, by homogeneity, $$T(w-v) = 3(v-w).$$

Hence, $3$ is an eigenvalue of $T$ for the eigenvector $v-w$. Note that we can show the same thing for $-3$ if we just flip $v-w$ to $w-v$ and pull out the resulting negative sign.

30. Suppose $T \in \mathcal L(R^3)$ and $-4$, $5$, and $\sqrt{7}$ are eigenvalues of $T$. Prove that there exists $x \in R^{3}$ such that $Tx - 9x = (-4, 5, \sqrt 7)$.

To start, $Tx \neq 9x$ for all $x \in R^3$, since that would imply that $9$ is an eigenvalue of $T$, contradicting 5.13. Thus, $T-9I$ is surjective, meaning there exists some $x$ such that $$(T-9I)(x) = (-4, 5, \sqrt 7).$$

32. Suppose $\lambda_1, \dots, \lambda_n$ is a list of distinct real numbers. Prove that the list $e^{\lambda_1 x}, \dots, e^{\lambda_n x}$ is linearly independent in the vector space of real-valued functions on $\mathbf{R}$.
Hint: Let $V = \operatorname{span}(e^{\lambda_1 x}, \dots, e^{\lambda_n x})$ and define an operator $T \in \mathcal L(V)$ by $Tf = f’$. Find eigenvalues and eigenvectors of $T$.

Following the hint, if we define $T$ as described, then $f$ will be an eigenvector of $T$ when $$f’ = (a_1 \lambda_1 e^{\lambda_1 x} + \cdots + a_n \lambda_n e^{\lambda_n x}) = \tau (a_1 e^{\lambda_1 x} + \cdots + a_n e^{\lambda_n x})$$ for some $\tau \in F$. Since $\lambda_1, \dots, \lambda_n$ are distinct, finding a $\tau$ is only possible when $a_i \neq 0$ for one $i$ and $a_j = 0$ for all $j \neq i \in 1, \dots, n$. Taking that constraint, the eigenvalues of $T$ are $\lambda_1, \dots, \lambda_n$ and the eigenvectors are $a_1 e^{\lambda_1 x}, \dots, a_n e^{\lambda_n x}$ and all $a_1, \dots, a_n \in F$. Thus, by 5.10, if we set $a_1 = \cdots = a_n = 1$, we get that $e^{\lambda_1 x}, \dots, e^{\lambda_n x}$ are linearly independent in the vector space of real-valued functions on $\mathbf{R}$.

## 5.B #

1. Suppose $T \in \mathcal L(V)$ and there exists a positive integer $n$ such that $T^n = 0$.
(a) Prove that $I - T$ is invertible and that $$(I-T)^{-1] = I + T + \cdots + T^{n-1}. (b) Explain how you would guess the formula above. For (a), we use the strategy of showing that  (I-T)(I + T + \cdots + T^{n-1}) = I  and  (I + T + \cdots + T^{n-1})(I-T) = I . Observe that$$ (I-T)(I + T + \cdots + T^{n-1}) = I + T + \cdots + T^{n-1} - (T + T^2 + \cdots + T^n). $$Since  T^n  is  0 , we have$$ I + T + \cdots + T^{n-1} - (T + T^2 + \cdots + T^n) = I + T + \cdots + T^{n-1} - (T + T^2 + \cdots + T^{n-1}) = I. $$Also observe that$$ (I + T + \cdots + T^{n-1})(I-T) = I(I-T) + T(I-T) + \cdots + T^{n-1}(I-T) = I - T + T - T^2 + \cdots - T^{n-1} + T^{n-1} - T^n. $$Again, since  T^n  is  0 , we get$$ I - T + T - T^2 + \cdots - T^{n-1} + T^{n-1} - T^n = I. $$Hence, (I-T) is invertible and (I-T)^{-1} = I + T + \cdots + T^{n-1} . For (b), I wouldn’t have guessed this but apparently the way to guess this is to know that$$ 1-x^n = (1-x)(1+x+\cdots+x^{n-1}). $$2. Suppose  T \in \mathcal L(V)  and  (T-2I)(T-3I)(T-4I) = 0 . Suppose  \lambda  is an eigenvalue of  T . Prove that  \lambda = 2  or  \lambda = 3  or \lambda = 4. Suppose  v  is the eigenvector for  \lambda . Then$$ \begin{eqnarray} (T-2I)(T-3I)(T-4I)v & = & (\lambda v - 2v)(\lambda v - 3v)(\lambda v - 4v) \\\\ &=& (\lambda - 2)(\lambda - 3)(\lambda - 4)v = 0. \end{eqnarray} $$So, because  v  isn’t  0  by the definition of an eigenvector, one of  \lambda - 4, \lambda -3, \lambda -2  must equal 0. Hence  \lambda  is one of 2, 3, or 4. ## 5.C # 1. Suppose  T \in \mathcal L(V)  is diagonalizable. Prove that  V = \operatorname{null} T \oplus \operatorname{range} T . Going to solve this in two ways. (First way.) If  T  is invertible, then we’re done.  T  is injective and therefore  operatorname{null} T = {0}  and  operatorname{range} T = V , so  operatorname{null} T \intersect operatorname{range} T = {0}  and  \operatorname{null} T \oplus \operatorname{range} T = V . If  T  is not invertible, since  T  is diagonalizable, at least one eigenvalue must equal  0 , by 5.30 and 5.32. Let  0, \lambda_1, \dots, \lambda_n  be the eigenvalues of  T  where  \lambda_i = 0  for  i = 1, \dots, n . Then, by 5.41(d), we have$$ V = E(0, T) \oplus E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_n, T). $$E(0,T) = \operatorname{null} T because if  v \in E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_n, T) , then Tv \noteq 0  by definition. It remains to show that  E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_n, T) = \operatorname{range} T . Assume that  v \in E(\lambda_i, T) . Then,  T(\frac{1}{\lambda_i} v) = v , so  E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_n, T) \subset \operatorname{range} T . Now assume that  u \in \operatorname{range} T . It follows that$$ u = T(v) = T(v_0 + v_1 + \cdots + v_n) = \lambda_1 v_1 + \cdots + \lambda_n v_n \in E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_n, T) $$where  v_0, \dots, v_n  are elements of  E(\lambda_0, T), \dots, E(\lambda_n, T)  respectively. This implies that  \operatorname{range} T \subset E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_n, T) , so \operatorname{range} T = E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_n, T) . Taken together, this means that  \operatorname{null} T \oplus \operatorname{range} T = V . 2. Prove the converse of the statement in the exercise above or give a counterexample to the converse. The converse of the statement is false. When  V = C^2  and  T(w, z) = (z, w) \in \mathcal L(V) , T has only eigenvalue,  \lambda = 1  and therefore is not diagonalizable, since having only one eigenvalue violates 5.41(d) and 5.41(e). 8. Suppose  T \in \mathcal L(F^5)  and  \dim E(8, T) = 4 . Prove that  T-2I  or  T - 6I  is invertible. By 5.38, we know that$$ \dim V \geq \dim E(\lambda_1, T) + \cdots \dim E(\lambda_m, T).  So, $4 + \dim E(\lambda_1, T) + \cdots \dim E(\lambda_n, T) \leq 5$, meaning $\dim E(\lambda_1, T) + \cdots \dim E(\lambda_n, T) \leq 1$. Since $\dim E(\lambda_i, T) \gt 0$ iff $\lambda_i$ is an eigenvalue, then at most one of $6$ and $2$ can be an eigenvalue. By 5.6, for whichever $n$ of $6$ and $2$ is not an eigenvalue, $T - nI$ is invertible. Hence, one of $T-2I$, $T-6I$ is invertible.

14. Find $T \in \mathcal L(C^3)$ such that $6$ and $7$ are eigenvalues of $T$ and such that $T$ does not have a diagonal matrix with respect to any basis of $C^3$.
TODO, I have a solution but am feeling a lot of resistance to writing it up.