1. Solution: By 8.21 (a), $V = G(0, N)$. Since $G(0, N) = \operatorname{null} N^{\operatorname{dim} V}$ (see 8.11), it follows that $N^{\operatorname{dim} V} = 0$ and so $N$ is nilpotent.

2. Solution: Define $T \in \mathcal{L}(\mathbb{R}^3)$ by

$$ T(x, y, z) = (0, -z, y). $$ That is, $T$ squashes vectors onto the $yz$ plane and rotates them counterclockwise by $\pi/2$ radians. So all eigenvectors of $T$ are contained in the $x$-axis and correspond to the eigenvalue $0$. $T$ obviously is not nilpotent.

3. Solution: Suppose $\lambda$ is an eigenvalue of $T$ and $v \in V$ a corresponding eigenvalue. $S$ is surject, so there exists $u \in V$ such that $Su = v$. Then

$$ S^{-1}TSu = S^{-1}Tv = \lambda S^{-1}v = \lambda u, $$ which shows that $\lambda$ is an eigenvalue of $S^{-1}TS$. Hence every eigenvalue of $T$ is an eigenvalue of $S^{-1}TS$. We will prove these eigenvalues have the same multiplicity and it will follow that $S^{-1}TS$ cannot have other eigenvalues (by 8.26).

Suppose $\lambda_1, \dots, \lambda_m$ are the distinct eigenvalues of $T$. Fix $k \in \{1, \dots, m\}$. Let $v_1, \dots, v_d$ be a basis of $G(\lambda_k, T)$. There exist $u_1, \dots, u_d \in V$ such that $Su_j = v_j$ for each $j = 1, \dots, d$. It easy to check that the $u$’s are linearly independent. We have

$$ G(\lambda_k, S^{-1}TS) = \operatorname{null} (S^{-1}TS – \lambda_k I)^{\operatorname{dim} V} = \operatorname{null} S^{-1}(T – \lambda_k I)^{\operatorname{dim} V}S, $$ where the first equality comes from 8.11 and the second from Exercise 5 in section 5B. For each $j$, we have

$$ S^{-1}(T – \lambda_k I)^{\operatorname{dim} V}Su_j = S^{-1}(T – \lambda_k I)^{\operatorname{dim} V}v_j = 0, $$ where the second equality follows because $v_j \in G(\lambda_k, T)$. This shows that $u_1, \dots, u_d \in G(\lambda_k, S^{-1}TS)$. Hence

$$ \operatorname{dim} G(\lambda_k, S^{-1}TS) \ge d = \operatorname{dim} G(\lambda_k, T). \tag{1} $$ By 8.26, we must have

$$ \operatorname{dim} G(\lambda_1, T) + \dots + \operatorname{dim} G(\lambda_m, T) = \operatorname{dim} V \tag{2} $$ and

$$ \operatorname{dim} G(\lambda_1, S^{-1}TS) + \dots + \operatorname{dim} G(\lambda_m, S^{-1}TS) \le \operatorname{dim} V. \tag{3} $$ $(1)$ and $(2)$ imply that $(3)$ is only possible if $$\operatorname{dim} G(\lambda_k, S^{-1}TS) = \operatorname{dim} G(\lambda_k, T).$$ Hence, their multiplicieties are the same and $S^{-1}TS$ cannot have other generalized eigenspaces (the ones shown here already eat up the dimension of $V$).

4. Solution: By the same reasoning used in the proof of 8.4, it follows that $\operatorname{dim} \operatorname{null} T^{n-1} \ge n – 1$. But $\operatorname{null} T^{n-1} \subset \operatorname{null} T^n = G(0, T)$ (see 8.2 and 8.11). Thus $\operatorname{dim} G(0, T) \ge n – 1$ and $0$ is an eigenvalue of $T$.

If $\operatorname{dim} G(0, T) = n$, 8.26 shows that $0$ is the only eigenvalue of $T$. If $\operatorname{dim} G(0, T) = n – 1$, there is only space for one more eigenvalue with multiplicty $1$.

5. Solution: Every eigenvector is also a generalized eigenvector, so in the forward direction we can make a similar argument to that of Exercise 3, where the dimensions only fit if each eigenspace equals its corresponding generalized one (because the former is a subset of the latter). The other direction is obvious from 8.23.

6. Solution: The formula for $N$ doesn’t really matter here. We only care about the dimension of $\mathbb{F}^{5}$, which is $5$. Using the same reasoning from the proof of 8.31, and because $N^j = 0$ for $j \ge 5$, we have

$$ \begin{aligned} I + N &= (I + a_1N + a_2N^2 + a_3N^3 + a_4N^4)\\ &= I + 2a_1N + (2a_2 + a_1^2)N^2 + (2a_3 + 2a_1a_2)N^3 + (2a_4 + 2a_1a_3 + a_2^2)N^4 \end{aligned} $$ for some $a_1, a_2, a_3, a_4 \mathbb{F}$.

Choose

$$ a_1 = \frac{1}{2},\quad a_2 = \frac{-1}{8},\quad a_3 = \frac{1}{16}\quad a_4 = \frac{-5}{128}$$ and the terms on the second line will collapse to $N + I$. Hence

$$ I + \frac{1}{2}N – \frac{1}{8}N^2 + \frac{1}{16}N^3 – \frac{5}{128}N^4 $$ is a square root of $N + I$.

7. Solution: One can use the same strategy as in the proof of 8.31 to show that $I + N$ has a cube root for any nilpotent $N \in \mathcal{L}(V)$ and the rest of the proof will be same as the proof of 8.33.

8. Solution: If $0$ is not an eigenvalue of $T$, then $T^j$ is injective and surjective for all integers $j$, which gives the desired result (take $j = n-2$).

Suppose $0$ is an eigenvalue of $T$. Since $3$ and $8$ are also eigenvalues of $T$, by 8.26 the multiplicity of $0$, namely $\operatorname{dim} G(0, T)$ which equals $\operatorname{dim} \operatorname{null} T^n$, is at most $n – 2$. By the same reasoning used in the proof of 8.4, we have $\operatorname{null} T^{n-2} = \operatorname{null} T^n$ (because the $\operatorname{null} T^{n-2} \subset \operatorname{null} T^n$). Exercise 19 of section 8A implies that $\operatorname{range} T^{n-2} = \operatorname{range} T$. Now 8.5 completes the proof.

9. Solution: Keep in mind that when we mention the size of an $n$-by-$n$ matrix here we mean $n$ and not $n$ times $n$.

Let $V$ be vector space whose dimension equals the size $A$ (or $B$, since they’re the same). Choose a basis of $V$ and define $S, T \in \mathcal{L}(V)$ such that $\mathcal{M}(S) = A$ and $\mathcal{M}(T) = B$. Then $\mathcal{M}(ST) = AB$.

Let $d_j$ equal the size of $A_j$ (or $B_j$, because they’re the same). Consider the list consisting of the first $d_1$ vectors in the chosen basis. $A$ and $B$ show that the span of these vectors are invariant under $S$ and $T$.

Similarly, the span of the next $d_2$ vectors after this list is also invariant under $T$. Continuing in this fashion, we see that there are $m$ distinct lists of consecutive vectors, with no intersections, in the chosen basis whose spans are invariant under $S$ and $T$.

Let $U_1, \dots, U_m$ denote such spans. Clearly $\mathcal{M}(S|_{U_j}) = A_j$ and $\mathcal{M}(T|_{U_j}) = B_j$ for each $j$. Hence $\mathcal{M}(S|_{U_j}T|_{U_j}) = A_jB_j$ and so it easy to see that $\mathcal{M}(ST)$ (which equals $AB$) has the desired form.

10. Solution: Let $v_1, \dots, v_n$ denote a basis of $V$ consisting of generalized eigenvectors of $T$ (which exists by 8.23). Define $\langle \cdot, \cdot \rangle: V \times V \to \mathbb{C}$ by

$$ \langle a_1 v_1 + \dots + a_n v_n, b_1 v_1 + \dots + b_n v_n \rangle = a_1\overline{b_1} + \dots + a_n\overline{b_n}, $$ where the $a$’s and $b$’s are complex numbers. You can check that $\langle \cdot, \cdot \rangle$ is a well defined inner product on $V$. Thus $v_1, \dots, v_n$ is an orthonormal basis of $V$. Moreover, the generalized eigenspaces of $T$ are orthogonal to each other. This implies that, if $v \in G(\beta, T)$, then

$$ P_{G(\alpha, T)} v = \begin{cases} v, \text{ if } \alpha = \beta\\ 0, \text{ if } \alpha \neq \beta \end{cases} \tag{$*$} $$ where $P_{G(\alpha, T)}$ is the orthogonal projection of $V$ onto $G(\alpha, T)$.

Let $\lambda_1, \dots, \lambda_m$ denote the distinct eigenvalues of $T$. We have

$$ T = T|_{G(\lambda_1, T)}P_{G(\lambda_1, T)} + \dots + T|_{G(\lambda_m, T)}P_{G(\lambda_m, T)}. $$ For each $j = 1, \dots, m$, we can write $T|_{G(\lambda_j, T)} = \lambda_j I + N_j$ where $N_j$ is a nilpotent operator under which $G(\lambda_j, T)$ is invariant (see 8.21 (c)). Therefore

$$ \begin{aligned} T &= (\lambda_1 I + N_1)P_{G(\lambda_1, T)} + \dots + (\lambda_m I + N_m)P_{G(\lambda_m, T)}\\ &= \underbrace{\lambda_1 P_{G(\lambda_1, T)} + \dots + \lambda_m P_{G(\lambda_m, T)}}_\text{(4)} + \underbrace{N_1P_{G(\lambda_1, T)} + \dots + N_mP_{G(\lambda_m, T)}}_\text{(5)}. \end{aligned} $$ Fix $k \in \{1, \dots, n\}$. Then $v_k \in G(\lambda_j, T)$ for some $j \in \{1, \dots m\}$. $(*)$ shows that $(4)$ maps $v_k$ to $\lambda_j v_k$. Hence $v_1, \dots, v_n$ is a basis of eigenvectors of $(4)$ and so $(4)$ is diagonalizable. $(*)$ also shows that $(5)$ maps $v_k$ to $N_j v_k$. But $G(\lambda_j, T)$ is invariant under $N_j$, so $(*)$ actually implies that $(5)$ raised to the power of $\operatorname{dim} V$ maps $v_k$ to $N_j^{\dim V}v_k$ which equals $0$. Therefore $(5)$ is nilpotent. It is easy to see that $(4)$ and $(5)$ commute (they map $v_k$ to $\lambda_j N_j v_k$, no matter the order), which completes the proof.

11. Solution: Suppose $T$ has an upper-triangular matrix with respect to the basis $v_1, \dots, v_n$. Suppose also that $\lambda$ appears on the $j$-th diagonal entry of $\mathcal{M}(T)$. Then

$$ Tv_j = a_1 v_1 + \dots + a_{j-1} v_{j-1} + \lambda v_j $$ for some $a_1, \dots, a_{j-1} \in \mathbb{F}$, and so

$$ (T – \lambda I)v_j = a_1 v_1 + \dots + a_{j-1} v_{j-1} \in \operatorname{span}(v_1, \dots, v_{j-1}). $$ We have

$$ (T – \lambda I)v_{j-1} = c_1 v_1 + \dots + (c_{j-1} – \lambda) v_{j-1} $$ for some $c_1, \dots, c_{j-1} \in \mathbb{F}$. If $c_{j-1} – \lambda = 0$, then $(T – \lambda I)^2v_j \in \operatorname{span}(v_1, \dots, v_{j-2})$. If $c_{j-1} – \lambda \neq 0$, then $$ (T – \lambda I)\left(v_j – \frac{a_{j-1}}{c_{j-1} – \lambda}v_{j-1}\right) \in \operatorname{span}(v_1, \dots, v_{j-2}). $$ We go on, either squaring by squaring $(T – \lambda I)$ or subtracting a vector $u \in \operatorname{span}(v_1, \dots, v_{j-1})$ from $v_j$ in the argument of $(T – \lambda I)$, and we will have $(T – \lambda I)^2(v_j – u)$ in the span of the first $j – 3$ vectors of the basis, then in span of the first $n – 4$, an so on until it will be in the span of an empty list, that is, $\{0\}$. This means that $v_j – u \in G(\lambda, T)$ for some $u \in \operatorname{span}(v_1, \dots, v_{j-1})$.

Let $\nu_1, \dots, \nu_d$ denote the vectors of the chosen basis of $V$ that correspond to the columns of $\mathcal{M}(T)$ in which $\lambda$ appears and in the order that they appear on the basis. We can repeat the previous process and find $u_1, \dots, u_d$ such

$$ \nu_1 – u_1, \dots, \nu_d – u_d \in G(\lambda, T), \tag{6} $$ where each $u_k$ is in the span of the basis vectors that come before $\nu_k$. We claim this list is linearly independent. To see this, fix $k \in \{1, \dots, d\}$. Suppose $\nu_k$ is the $j$-th basis vector, i.e. $\nu_k = v_j$. This means that

$$ \operatorname{span}(\nu_1 – u_1, \dots, \nu_{k-1} – u_{k-1}) \subset \operatorname{span}(v_1, \dots, v_{j-1}). $$ Therefore, we can’t have $\nu_k – u_k \in \operatorname{span}(\nu_1 – u_1, \dots, \nu_{k-1} – u_{k-1})$, because that would imply that

$$ v_j \in \operatorname{span}(v_1, \dots, v_{j-1}), $$ since $u_k \in \operatorname{span}(v_1, \dots, v_{j-1})$. This argument can be repeated for each $k$. Therefore no vector in $(6)$ is in the span of the previous ones. It follows that the list in $(6)$ is linearly independent.

Hence $\operatorname{dim} G(\lambda, T) \ge d$. By 8.26, the dimension of $G(\lambda, T)$ cannot be greater $d$, because we have $\operatorname{dim} V$ diagonal entries and each one adds at least $1$ to the dimension of some generalized eigenspace. Thus $\operatorname{dim} G(\lambda, T) = d$, completing the proof.

## Russell Sharpe

5 Apr 2021The proof given here for 8.B.8 uses 8.26, which requires V to be complex. We can dispense with this condition by arguing as follows.

Since T has nonzero eigenvalues, it cannot be nilpotent. So by Exercise (8.A.10) V is the direct sum of null T^(n-1) and range T^(n-1) and the sequence stabilizes either at j = n-1 or at some j <=n-2.

If it stabilizes at j = n-1 then by a similar argument to (8.4), either (a) dim null T^(n-1) = n or (b) dim null T^(n-1) = n-1. But both of these lead to contradiction as follows.

Case (a): dim null T^(n-1) = n: in this case T^(n-1)=0 and T is nilpotent: contradiction.

Case (b): dim null T^(n-1) = n-1. Then range T^(n-1) = 1, so range T^(n-1) = span for some v != 0. Since range T^(n-1) is invariant under T (and is not mapped to zero, since null T^n = null T^(n-1) != V in this case) we have Tv = {lambda}.v for some {lambda} != 0. So span is the eigenspace for eigenvalue {lambda}. V is then the direct sum of null T^(n-1) and range T^(n-1) = span ie the direct sum of G(0,T) and G({lambda},T): there is only room for *one* nonzero eigenvalue. But both 3 and 8 are eigenvalues by assumption: contradiction)

So we conclude that the sequence stabilises after all at some j <= n-2, whereupon we conclude that V is the direct sum of null T^(n-2) and range T^(n-2).

## Russell Sharpe

5 Apr 2021There are some subtleties in the text processing here which have obscured my post. Everything in angle brackets became invisible. I'll try now with normal parentheses instead.

"span" in the above should read "span(v)", and reference to "the sequence stabilizing" should be "the sequence (null T^j) stabilizing"

## Phi

27 Nov 2020A sketch of an alternative proof of 11:

Lemma 1. Suppose 𝔽∈{ℝ,ℂ}, V is a finite dimensional vector space over 𝔽, T is a linear operator on V, $b_1, \dots, b_n$ is a basis of V, the matrix of T w.r.t. $b_1, \dots, b_n$ is upper triangular, its m-th diagonal element is $\lambda_1$, its (m+1)-st diagonal element is $\lambda_2 \neq \lambda_1$. Then there exists a vector c such that $b_1, \dots, b_{m-1}, c, b_m, b_{m+2}, \dots, b_n$ is a basis of V (we replaced $b_{m+1}$ with $c$ and also changed order of 2 vectors), the matrix of T w.r.t. this new basis is upper triangular, and its m-th and (m+1)-st diagonal elements are swapped.

In other words, lemma 1 says that given an operator with an upper triangular matrix, if in some place there are two different diagonal elements, we can swap them.

So, if we have T with some matrix which contains k occurences of some $\lambda$ on the diagonal, applying lemma 1 to T many times, we can find a basis of V so that in the matrix of T w.r.t. this new basis, on the diagonal, all occurences of $\lambda$ go before the occurences of all other eigenvalues, and this matrix is upper triangular. Therefore, $\operatorname{dim} G(\lambda, T) \geq k$.

Do this for each distinct eigenvalue, we get that the dimension of each generalized eigenspace is at least its count on the main diagonal. Because sum of their dimensions can't exceed n, their dimensions must be exactly their counts on the diagonal. QED

## Larry Baker

13 Jun 2022Thanks for this sketch of a simpler proof.

I specifically determined a change of bases (not unique) to swap diagonal entries as your lemma suggests. In my case, the vectors b(m) and b(m+1) are replaced by a linear combination of themselves.

Having established this lemma, I proceeded slightly differently. As you say, you can transform the original basis in such a way that the upper triangular block at the top has all one eigenvalue, say lambda, and the upper triangular block at the bottom has eigenvalues that are all different from lambda. If you then subtract lambda times the identity matrix from this matrix, you’ll have an upper triangular block with all zeroes on the diagonal, which is nilpotent by Exercise 8.A.12, and an upper triangular block at the bottom without any zeroes on the diagonal. Raising the full matrix to the power of the dimension of V results in a new matrix with the upper triangular block at the top all zeros and the upper triangular block at the bottom without any zeroes on the diagonal. The dimension of the nullspace of this new matrix is precisely the number of diagonal entries equal to lambda obviating the need to make a counting argument.

## Larry Baker

24 Jun 2022Professor Axler has a proof for this exercise in the second edition of LADR: Theorem 8.10 on pages 169-171. It's a fairly complicated proof by induction so it's doubtful that there's a simple solution to this exercise.

## Phi

25 Nov 2020Am I correct thinking that the theorem 8.31 holds not only for ℂ, but for all fields?

## Phi

25 Nov 2020Actually, nevermind - it doesn't work in ℤ/(2), because 2=1+1=0 has no multiplicative inverse in that field.

## Jonathan Sharir-Smith

22 Aug 2020For equation (3) in your solution to question 3, why is there a (non-strict) inequality there. Why can't we write this as an equality (which seems to me to follow from 8.21 a) because we already showed that there is a bijective correspondence between the eigenvalues of $T$ and $S^-1TS$?

## Jonathan Sharir-Smith

22 Aug 2020Edit: It should be $S^{-1}TS$ at the bottom of the comment.

## David

20 Aug 2020For #6, I think you meant to raise the right hand side of the first line to power of 2.

## Jonathan Sharir-Smith

22 Aug 2020Agreed, there is a mistake there. Although I think he meant to write $\sqrt{I+N}$ on the left-hand side of the equality, rather than just $I+N$.

## David

26 Aug 2020No, I think Linearity mean to square right-hand side so it would equal the second line.

## Aditya

10 May 2020Q11, I feel there might be another approach, using Q3. T' be the upper triangular matrix, S be the change of basis matrix such that S^-1T'S has the block diagonal form with \lambda_i having multiplicity d_i. Since these two transformations (I use matrices and transformations interchangeably as Q3 can be extended to hold for matrices) are conjugate, dim null(T'-\lambda_i)^n = d_i.

Now let's take an eigenvalue \lambda. It must appear on diagonal of T'. (T'-\lambda) has zeroes, say, k places. We must now show k = dim null(T'-\lambda)^n which is the multiplicity. To see that, observe, (T'-\lambda)^n is also upper triangular and has diagonal elements raised to the power n. So it has 0 at k places, Now we try to find the null space of this matrix. If we multiply it with (x_1,...,x_n) then we get a system of linear equations. Since it is an upper triangular matrix, it is in Row Echelon form. We see leading entries of row = 0 at k places. Hence k = dim null(T'-\lambda)^n.

Hope this makes sense. I found it much easier to come up with. Btw, your website is a life saver! Thank you!

## Larry Baker

13 Jun 2022There's a flaw at the end of this argument. It is true that both the original matrix (T'-\lambda) and (T'-\lambda)^n have 0 at k places. But an upper triangular matrix can have k zeros on its main diagonal, but its null space may not have dimension k. For instance, consider the 2x2 matrix whose first row is {0,1} and whose second row is {0,0}. The dimension of its null space is 1, not 2. I believe that one needs to do the additional work described by Phi to get all the same eigenvalues in the upper triangular block at the top and then appeal to Exercise 8A12.

## Brian

18 Oct 20198B.8) null T^(n-2) = null T^(n)

is not true. 8.4) states that n = dim V; null T^n = null T^(n+1) and doesn't statisfy null T^(n-1) = null T^(n)

## Lazlo P

18 Sep 2019For 8B.2, I think you meant to write T(x, y, z) = (0, -z, y)

## Linearity

21 Sep 2019Yes, thanks.

## Chi Yuan Lau

9 Aug 2019For 10, I think if we consider in the matrix version , then we can decomposite it directly by theorem 8.29

## Linearity

9 Aug 2019Yes, for $\mathbb C$ it is easy. This is a special case of Jordan-Chevalley decomposition Theorem for algebraically closed field.

## Chi Yuan Lau

9 Aug 2019so it follows from 8B.8 :

If $T$ has at least 2 distinct egeinvalues , then $$V=\mathrm{range} T^{n-2}\oplus\mathrm{null}~ T^{n-2}.$$Or replace the $2$ by any $k$, the conclusion will also hold.

## Marcel Ackermann

21 Sep 2017Hey, I would appreciate a lot a solution for 8B8 :)

## Ying Kit Hui

27 May 2018Hint: prove that nullT^(n-2) = nullT^(n-1)

## Marcel Ackermann

22 Aug 20178B1) according to 8.21a: $V=G(0,N)$

so $G(0,N)=\mathrm{null}(N-0I)^{\dim V}=null(N)^{\dim V}=V$

according to 8.21c: $(N-0I)|_{G(0,N) = N|V = N}, so N is nilpotent