Linear Algebra/Strings/Solutions

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Solutions[edit]

This exercise is recommended for all readers.
Problem 1

What is the index of nilpotency of the left-shift operator, here acting on the space of triples of reals?


(x,y,z)\mapsto(0,x,y)
Answer

Three. It is at least three because \ell^2(\,(1,1,1)\,)=(0,0,1)\neq \vec{0}. It is at most three because (x,y,z)\mapsto (0,x,y)\mapsto (0,0,x)\mapsto (0,0,0).

This exercise is recommended for all readers.
Problem 2

For each string basis state the index of nilpotency and give the dimension of the rangespace and nullspace of each iteration of the nilpotent map.

  1. 
\begin{array}{ccccccc}
\vec{\beta}_1 &\mapsto &\vec{\beta}_2 &\mapsto &\vec{0}  \\
\vec{\beta}_3 &\mapsto &\vec{\beta}_4 &\mapsto &\vec{0}
\end{array}
  2. 
\begin{array}{ccccccc}
\vec{\beta}_1 &\mapsto &\vec{\beta}_2 &\mapsto &\vec{\beta}_3
&\mapsto &\vec{0}  \\
\vec{\beta}_4 &\mapsto &\vec{0} \\
\vec{\beta}_5 &\mapsto &\vec{0} \\
\vec{\beta}_6 &\mapsto &\vec{0}
\end{array}
  3. 
\begin{array}{ccccccccc}
\vec{\beta}_1 &\mapsto &\vec{\beta}_2 &\mapsto &\vec{\beta}_3
&\mapsto &\vec{0}
\end{array}

Also give the canonical form of the matrix.

Answer
  1. The domain has dimension four. The map's action is that any vector in the space c_1\cdot \vec{\beta}_1+c_2\cdot \vec{\beta}_2
+c_3\cdot \vec{\beta}_3+c_4\cdot \vec{\beta}_4 is sent to c_1\cdot \vec{\beta}_2+c_2\cdot \vec{0}
+c_3\cdot \vec{\beta}_4+c_4\cdot \vec{0}
=c_1\cdot \vec{\beta}_3+c_3\cdot\vec{\beta}_4. The first application of the map sends two basis vectors \vec{\beta}_2 and \vec{\beta}_4 to zero, and therefore the nullspace has dimension two and the rangespace has dimension two. With a second application, all four basis vectors are sent to zero and so the nullspace of the second power has dimension four while the rangespace of the second power has dimension zero. Thus the index of nilpotency is two. This is the canonical form.
    
\begin{pmatrix}
0  &0  &0  &0  \\
1  &0  &0  &0  \\
0  &0  &0  &0  \\
0  &0  &1  &0
\end{pmatrix}
  2. The dimension of the domain of this map is six. For the first power the dimension of the nullspace is four and the dimension of the rangespace is two. For the second power the dimension of the nullspace is five and the dimension of the rangespace is one. Then the third iteration results in a nullspace of dimension six and a rangespace of dimension zero. The index of nilpotency is three, and this is the canonical form.
    
\begin{pmatrix}
0  &0  &0  &0  &0  &0  \\
1  &0  &0  &0  &0  &0  \\
0  &1  &0  &0  &0  &0  \\
0  &0  &0  &0  &0  &0  \\
0  &0  &0  &0  &0  &0  \\
0  &0  &0  &0  &0  &0
\end{pmatrix}
  3. The dimension of the domain is three, and the index of nilpotency is three. The first power's null space has dimension one and its range space has dimension two. The second power's null space has dimension two and its range space has dimension one. Finally, the third power's null space has dimension three and its range space has dimension zero. Here is the canonical form matrix.
    
\begin{pmatrix}
0  &0  &0  \\
1  &0  &0  \\
0  &1  &0
\end{pmatrix}
Problem 3

Decide which of these matrices are nilpotent.

  1. \begin{pmatrix}
-2  &4  \\
-1  &2
\end{pmatrix}
  2. \begin{pmatrix}
3  &1  \\
1  &3
\end{pmatrix}
  3. \begin{pmatrix}
-3  &2  &1  \\
-3  &2  &1  \\
-3  &2  &1
\end{pmatrix}
  4. \begin{pmatrix}
1  &1  &4  \\
3  &0  &-1 \\
5  &2  &7
\end{pmatrix}
  5. \begin{pmatrix}
45  &-22  &-19  \\
33  &-16  &-14  \\
69  &-34  &-29
\end{pmatrix}
Answer

By Lemma 1.3 the nullity has grown as large as possible by the n-th iteration where n is the dimension of the domain. Thus, for the 2 \! \times \! 2 matrices, we need only check whether the square is the zero matrix. For the 3 \! \times \! 3 matrices, we need only check the cube.

  1. Yes, this matrix is nilpotent because its square is the zero matrix.
  2. No, the square is not the zero matrix.
    
\begin{pmatrix}
3  &1  \\
1  &3
\end{pmatrix}^2
=\begin{pmatrix}
10  &6  \\
6   &10
\end{pmatrix}
  3. Yes, the cube is the zero matrix. In fact, the square is zero.
  4. No, the third power is not the zero matrix.
    
\begin{pmatrix}
1  &1  &4  \\
3  &0  &-1 \\
5  &2  &7
\end{pmatrix}^3
=\begin{pmatrix}
206  &86  &304  \\
26  &8   &26   \\
438  &180 &634
\end{pmatrix}
  5. Yes, the cube of this matrix is the zero matrix.

Another way to see that the second and fourth matrices are not nilpotent is to note that they are nonsingular.

This exercise is recommended for all readers.
Problem 4

Find the canonical form of this matrix.


\begin{pmatrix}
0  &1  &1  &0  &1  \\
0  &0  &1  &1  &1  \\
0  &0  &0  &0  &0  \\
0  &0  &0  &0  &0  \\
0  &0  &0  &0  &0
\end{pmatrix}
Answer

The table os calculations


\begin{array}{c|cc}
 p   & N^p   & \mathcal{N}(N^p)     \\  \hline
 1 
&\begin{pmatrix}
0  &1  &1  &0  &1  \\
0  &0  &1  &1  &1  \\
0  &0  &0  &0  &0  \\
0  &0  &0  &0  &0  \\
0  &0  &0  &0  &0
\end{pmatrix}  
& \{\begin{pmatrix} r \\ u \\ -u-v \\ u \\ v \end{pmatrix}
\,\big|\, r,u,v\in\mathbb{C}\}   \\
 2 
&\begin{pmatrix}
0  &0  &1  &1  &1  \\
0  &0  &0  &0  &0  \\
0  &0  &0  &0  &0  \\
0  &0  &0  &0  &0  \\
0  &0  &0  &0  &0
\end{pmatrix}  
& \{\begin{pmatrix} r \\ s \\ -u-v \\ u \\ v \end{pmatrix}
\,\big|\, r,s,u,v\in\mathbb{C}\}   \\
 2 
&\textit{--zero matrix--}
& \mathbb{C}^5 
\end{array}

gives these requirements of the string basis: three basis vectors are sent directly to zero, one more basis vector is sent to zero by a second application, and the final basis vector is sent to zero by a third application. Thus, the string basis has this form.


\begin{array}{ccccccc}
\vec{\beta}_1 &\mapsto &\vec{\beta}_2 &\mapsto
&\vec{\beta}_3 &\mapsto &\vec{0}               \\
\vec{\beta}_4 &\mapsto &\vec{0}                  \\
\vec{\beta}_5 &\mapsto &\vec{0}
\end{array}

From that the canonical form is immediate.


\begin{pmatrix}
0  &0  &0  &0  &0  \\
1  &0  &0  &0  &0  \\
0  &1  &0  &0  &0  \\
0  &0  &0  &0  &0  \\
0  &0  &0  &0  &0
\end{pmatrix}
This exercise is recommended for all readers.
Problem 5

Consider the matrix from Example 2.16.

  1. Use the action of the map on the string basis to give the canonical form.
  2. Find the change of basis matrices that bring the matrix to canonical form.
  3. Use the answer in the prior item to check the answer in the first item.
Answer
  1. The canonical form has a 3 \! \times \! 3 block and a 2 \! \times \! 2 block
    
\left(\begin{array}{ccc|cc}
0  &0  &0  &0  &0  \\
1  &0  &0  &0  &0  \\
0  &1  &0  &0  &0  \\ \hline
0  &0  &0  &0  &0  \\
0  &0  &0  &1  &0  \\
\end{array}\right)
    corresponding to the length three string and the length two string in the basis.
  2. Assume that N is the representation of the underlying map with respect to the standard basis. Let B be the basis to which we will change. By the similarity diagram
    Linalg similarity cd 2.png
    we have that the canonical form matrix is PNP^{-1} where
    
P^{-1}
={\rm Rep}_{B,\mathcal{E}_5}(\mbox{id})
=\begin{pmatrix}
1 &0 &0 &0 &0 \\
0 &1 &0 &1 &0 \\
1 &0 &1 &0 &0 \\
0 &0 &1 &1 &1 \\
0 &0 &0 &0 &1
\end{pmatrix}
    and P is the inverse of that.
    
P={\rm Rep}_{\mathcal{E}_5,B}(\mbox{id})
=(P^{-1})^{-1}
=\begin{pmatrix}
1 &0 &0 &0 &0 \\
-1 &1 &1 &-1&1 \\
-1 &0 &1 &0 &0 \\
1 &0 &-1&1 &-1\\
0 &0 &0 &0 &1
\end{pmatrix}
  3. The calculation to check this is routine.
This exercise is recommended for all readers.
Problem 6

Each of these matrices is nilpotent.

  1. 
\begin{pmatrix}
1/2  &-1/2  \\
1/2  &-1/2
\end{pmatrix}
  2. 
\begin{pmatrix}
0  &0  &0  \\
0  &-1 &1  \\
0  &-1 &1
\end{pmatrix}
  3. 
\begin{pmatrix}
-1  &1  &-1 \\
1  &0  &1  \\
1  &-1 &1
\end{pmatrix}

Put each in canonical form.

Answer
  1. The calculation
    
\begin{array}{c|cc}
 p   & N^p   & \mathcal{N}(N^p)     \\  \hline
 1 
&\begin{pmatrix}
1/2  &-1/2  \\
1/2  &-1/2
\end{pmatrix}  
& \{\begin{pmatrix} u \\ u \end{pmatrix}
\,\big|\, u\in\mathbb{C}\}   \\
 2 
&\textit{--zero matrix--}
& \mathbb{C}^2 
\end{array}

    shows that any map represented by the matrix must act on the string basis in this way

    
\begin{array}{ccccccc}
\vec{\beta}_1 &\mapsto &\vec{\beta}_2  &\mapsto &\vec{0}
\end{array}

    because the nullspace after one application has dimension one and exactly one basis vector, \vec{\beta}_2, is sent to zero. Therefore, this representation with respect to \langle \vec{\beta}_1,\vec{\beta}_2 \rangle is the canonical form.

    
\begin{pmatrix}
0    &0   \\
1    &0
\end{pmatrix}
  2. The calculation here is similar to the prior one.
    
\begin{array}{c|cc}
 p   & N^p   & \mathcal{N}(N^p)     \\  \hline
 1 
&\begin{pmatrix}
0  &0  &0  \\
0  &-1 &1  \\
0  &-1 &1
\end{pmatrix}  
& \{\begin{pmatrix} u \\ v \\ v \end{pmatrix}
\,\big|\, u,v\in\mathbb{C}\}   \\
 2 
&\textit{--zero matrix--}
& \mathbb{C}^3 
\end{array}

    The table shows that the string basis is of the form

    
\begin{array}{ccccccc}
\vec{\beta}_1 &\mapsto &\vec{\beta}_2 &\mapsto &\vec{0}  \\
\vec{\beta}_3 &\mapsto &\vec{0}
\end{array}

    because the nullspace after one application of the map has dimension two— \vec{\beta}_2 and \vec{\beta}_3 are both sent to zero— and one more iteration results in the additional vector being brought to zero.

  3. The calculation
    
\begin{array}{c|cc}
 p   & N^p   & \mathcal{N}(N^p)     \\  \hline
 1 
&\begin{pmatrix}
-1  &1  &-1  \\
1  &0  &1   \\
1  &-1 &1
\end{pmatrix}  
& \{\begin{pmatrix} u \\ 0 \\ -u \end{pmatrix}
\,\big|\, u\in\mathbb{C}\}   \\
 2 
&\begin{pmatrix}
1  &0  &1   \\
0  &0  &0   \\
-1  &0  &-1
\end{pmatrix}  
& \{\begin{pmatrix} u \\ v \\ -u \end{pmatrix}
\,\big|\, u,v\in\mathbb{C}\}   \\
 3 
&\textit{--zero matrix--}
& \mathbb{C}^3 
\end{array}

    shows that any map represented by this basis must act on a string basis in this way.

    
\begin{array}{ccccccc}
\vec{\beta}_1 &\mapsto &\vec{\beta}_2 &\mapsto
&\vec{\beta}_3 &\mapsto &\vec{0}
\end{array}

    Therefore, this is the canonical form.

    
\begin{pmatrix}
0    &0   &0   \\
1    &0   &0   \\
0    &1   &0
\end{pmatrix}
Problem 7

Describe the effect of left or right multiplication by a matrix that is in the canonical form for nilpotent matrices.

Answer

A couple of examples


\begin{pmatrix}
0  &0  \\
1  &0
\end{pmatrix}
\begin{pmatrix}
a  &b  \\
c  &d
\end{pmatrix}
=
\begin{pmatrix}
0  &0  \\
a  &b
\end{pmatrix}
\qquad
\begin{pmatrix}
0  &0  &0 \\
1  &0  &0 \\
0  &1  &0
\end{pmatrix}
\begin{pmatrix}
a  &b  &c \\
d  &e  &f \\
g  &h  &i
\end{pmatrix}
=
\begin{pmatrix}
0  &0  &0 \\
a  &b  &c \\
d  &e  &f
\end{pmatrix}

suggest that left multiplication by a block of subdiagonal ones shifts the rows of a matrix downward. Distinct blocks


\begin{pmatrix}
0  &0  &0  &0  \\
1  &0  &0  &0  \\
0  &0  &0  &0  \\
0  &0  &1  &0
\end{pmatrix}
\begin{pmatrix}
a  &b  &c  &d  \\
e  &f  &g  &h  \\
i  &j  &k  &l  \\
m  &n  &o  &p
\end{pmatrix}
=
\begin{pmatrix}
0  &0  &0  &0  \\
a  &b  &c  &d  \\
0  &0  &0  &0  \\
i  &j  &k  &l
\end{pmatrix}

act to shift down distinct parts of the matrix.

Right multiplication does an analgous thing to columns. See Problem 1.

Problem 8

Is nilpotence invariant under similarity? That is, must a matrix similar to a nilpotent matrix also be nilpotent? If so, with the same index?

Answer

Yes. Generalize the last sentence in Example 2.9. As to the index, that same last sentence shows that the index of the new matrix is less than or equal to the index of \hat{N}, and reversing the roles of the two matrices gives inequality in the other direction.

Another answer to this question is to show that a matrix is nilpotent if and only if any associated map is nilpotent, and with the same index. Then, because similar matrices represent the same map, the conclusion follows. This is Exercise 14 below.

This exercise is recommended for all readers.
Problem 9

Show that the only eigenvalue of a nilpotent matrix is zero.

Answer

Observe that a canonical form nilpotent matrix has only zero eigenvalues; e.g., the determinant of this lower-triangular matrix


\begin{pmatrix}
-x  &0  &0  \\
1  &-x &0  \\
0  &1  &-x
\end{pmatrix}

is  (-x)^3 , the only root of which is zero. But similar matrices have the same eigenvalues and every nilpotent matrix is similar to one in canonical form.

Another way to see this is to observe that a nilpotent matrix sends all vectors to zero after some number of iterations, but that conflicts with an action on an eigenspace \vec{v}\mapsto \lambda\vec{v} unless \lambda is zero.

Problem 10

Is there a nilpotent transformation of index three on a two-dimensional space?

Answer

No, by Lemma 1.3 for a map on a two-dimensional space, the nullity has grown as large as possible by the second iteration.

Problem 11

In the proof of Theorem 2.13, why isn't the proof's base case that the index of nilpotency is zero?

Answer

The index of nilpotency of a transformation can be zero only when the vector starting the string must be \vec{0}, that is, only when V is a trivial space.

This exercise is recommended for all readers.
Problem 12

Let  t:V\to V be a linear transformation and suppose  \vec{v}\in V is such that  t^k(\vec{v})=\vec{0} but  t^{k-1}(\vec{v})\neq\vec{0} . Consider the t-string \langle \vec{v},t(\vec{v}),\dots,t^{k-1}(\vec{v}) \rangle .

  1. Prove that  t is a transformation on the span of the set of vectors in the string, that is, prove that  t restricted to the span has a range that is a subset of the span. We say that the span is a  t -invariant subspace.
  2. Prove that the restriction is nilpotent.
  3. Prove that the t-string is linearly independent and so is a basis for its span.
  4. Represent the restriction map with respect to the t-string basis.
Answer
  1. Any member \vec{w} of the span can be written as a linear combination \vec{w}=c_0\cdot \vec{v}+c_1\cdot t(\vec{v})+\dots
+c_{k-1}\cdot t^{k-1}(\vec{v}). But then, by the linearity of the map, t(\vec{w})=c_0\cdot t(\vec{v})+c_1\cdot t^2(\vec{v})+\dots
+c_{k-2}\cdot t^{k-1}(\vec{v})+c_{k-1}\cdot \vec{0} is also in the span.
  2. The operation in the prior item, when iterated k times, will result in a linear combination of zeros.
  3. If  \vec{v}=\vec{0} then the set is empty and so is linearly independent by definition. Otherwise write  c_1\vec{v}+\dots+c_{k-1}t^{k-1}(\vec{v})=\vec{0} and apply  t^{k-1} to both sides. The right side gives  \vec{0} while the left side gives  c_1t^{k-1}(\vec{v}) ; conclude that  c_1=0 . Continue in this way by applying  t^{k-2} to both sides, etc.
  4. Of course, t acts on the span by acting on this basis as a single, k-long, t-string.
    
\begin{pmatrix}
0 &0 &0 &0 &\ldots &0 &0 \\
1 &0 &0 &0 &\ldots &0 &0 \\
0 &1 &0 &0 &\ldots &0 &0 \\
0 &0 &1 &0 &       &0 &0 \\
&  &  &\ddots          \\
0 &0 &0 &0 &       &1 &0 \\
\end{pmatrix}
Problem 13

Finish the proof of Theorem 2.13.

Answer

We must check that  B\cup\hat{C}\cup\{\vec{v}_1,\dots,\vec{v}_j\} is linearly independent where  B is a  t -string basis for  \mathcal{R}(t) , where  \hat{C} is a basis for  \mathcal{N}(t) , and where  t(\vec{v}_1)=\vec{\beta}_1,\dots,t(\vec{v}_i=\vec{\beta}_i . Write


\vec{0}=c_{1,-1}\vec{v}_1+c_{1,0}\vec{\beta}_1
+c_{1,1}t(\vec{\beta}_1)+\dots+
c_{1,{h_1}}t^{h_1}(\vec{\vec{\beta}}_1)
+c_{2,-1}\vec{v}_2+\dots+c_{j,h_i}t^{h_i}(\vec{\beta_i})

and apply  t .


\vec{0}=c_{1,-1}\vec{\beta}_1+c_{1,0}t(\vec{\beta}_1)
+\dots+
c_{1,h_1-1}t^{h_1}(\vec{\vec{\beta}}_1)+c_{1,h_1}\vec{0}
+c_{2,-1}\vec{\beta}_2+\cdots+c_{i,h_i-1}t^{h_i}(\vec{\beta_i})
+c_{i,h_i}\vec{0}

Conclude that the coefficients  c_{1,-1},\dots,c_{1,h_i-1}, c_{2,-1},\dots,c_{i,h_i-1} are all zero as  B\cup\hat{C} is a basis. Substitute back into the first displayed equation to conclude that the remaining coefficients are zero also.

Problem 14

Show that the terms "nilpotent transformation" and "nilpotent matrix", as given in Definition 2.6, fit with each other: a map is nilpotent if and only if it is represented by a nilpotent matrix. (Is it that a transformation is nilpotent if an only if there is a basis such that the map's representation with respect to that basis is a nilpotent matrix, or that any representation is a nilpotent matrix?)

Answer

For any basis B, a transformation n is nilpotent if and only if N={\rm Rep}_{B,B}(n) is a nilpotent matrix. This is because only the zero matrix represents the zero map and so  n^j is the zero map if and only if  N^j is the zero matrix.

Problem 15

Let  T be nilpotent of index four. How big can the rangespace of  T^3 be?

Answer

It can be of any size greater than or equal to one. To have a transformation that is nilpotent of index four, whose cube has rangespace of dimension k, take a vector space, a basis for that space, and a transformation that acts on that basis in this way.


\begin{array}{ccccccccc}
\vec{\beta}_1 &\mapsto &\vec{\beta}_2 &\mapsto &\vec{\beta}_3
&\mapsto &\vec{\beta}_4 &\mapsto &\vec{0}  \\
\vec{\beta}_5 &\mapsto &\vec{\beta}_6 &\mapsto &\vec{\beta}_7
&\mapsto &\vec{\beta}_8 &\mapsto &\vec{0}  \\
&\vdots                                    \\
\vec{\beta}_{4k-3} &\mapsto &\vec{\beta}_{4k-2}
&\mapsto &\vec{\beta}_{4k-1}
&\mapsto &\vec{\beta}_{4k} &\mapsto &\vec{0}  \\
&\vdots 
\end{array}


--possibly other, shorter, strings--

So the dimension of the rangespace of T^3 can be as large as desired. The smallest that it can be is one— there must be at least one string or else the map's index of nilpotency would not be four.

Problem 16

Recall that similar matrices have the same eigenvalues. Show that the converse does not hold.

Answer

These two have only zero for eigenvalues


\begin{pmatrix}
0  &0  \\
0  &0
\end{pmatrix}
\qquad
\begin{pmatrix}
0  &0  \\
1  &0
\end{pmatrix}

but are not similar (they have different canonical representatives, namely, themselves).

Problem 17

Prove a nilpotent matrix is similar to one that is all zeros except for blocks of super-diagonal ones.

Answer

A simple reordering of the string basis will do. For instance, a map that is assoicated with this string basis


\begin{array}{ccccccc}
\vec{\beta}_1 &\mapsto &\vec{\beta}_2 &\mapsto &\vec{0}
\end{array}

is represented with respect to B=\langle \vec{\beta}_1,\vec{\beta}_2 \rangle by this matrix


\begin{pmatrix}
0  &0  \\
1  &0
\end{pmatrix}

but is represented with respect to B=\langle \vec{\beta}_2,\vec{\beta}_1 \rangle in this way.


\begin{pmatrix}
0  &1  \\
0  &0
\end{pmatrix}
This exercise is recommended for all readers.
Problem 18

Prove that if a transformation has the same rangespace as nullspace. then the dimension of its domain is even.

Answer

Let t:V\to V be the transformation. If  \mathop{\mbox{rank}} (t)=\text{nullity}\, (t) then the equation  \mathop{\mbox{rank}}(t)+\text{nullity}\,(t)=\dim (V) shows that  \dim (V) is even.

Problem 19

Prove that if two nilpotent matrices commute then their product and sum are also nilpotent.

Answer

For the matrices to be nilpotent they must be square. For them to commute they must be the same size. Thus their product and sum are defined.

Call the matrices  A and  B . To see that  AB is nilpotent, multiply 
(AB)^2=ABAB=AABB=A^2B^2, and (AB)^3=A^3B^3, etc., and, as  A is nilpotent, that product is eventually zero.

The sum is similar; use the Binomial Theorem.

Problem 20

Consider the transformation of  \mathcal{M}_{n \! \times \! n} given by  t_S(T)=ST-TS where  S is an  n \! \times \! n matrix. Prove that if  S is nilpotent then so is  t_S .

Answer

Some experimentation gives the idea for the proof. Expansion of the second power


t^2_S(T)=S(ST-TS)-(ST-TS)S=S^2-2STS+TS^2

the third power

\begin{array}{rl}
t^3_S(T)
&=S(S^2-2STS+TS^2)-(S^2-2STS+TS^2)S  \\
&=S^3T-3S^2TS+3STS^2-TS^3
\end{array}

and the fourth power

\begin{array}{rl}
t^4_S(T)
&=S(S^3T-3S^2TS+3STS^2-TS^3)-(S^3T-3S^2TS+3STS^2-TS^3)S  \\
&=S^4T-4S^3TS+6S^2TS^2-4STS^3+TS^4
\end{array}

suggest that the expansions follow the Binomial Theorem. Verifying this by induction on the power of t_S is routine. This answers the question because, where the index of nilpotency of S is k, in the expansion of t^{2k}_S


t^{2k}_S(T)=\sum_{0\leq i\leq 2k}(-1)^i\binom{2k}{i} S^iTS^{2k-i}

for any i at least one of the S^i and S^{2k-i} has a power higher than k, and so the term gives the zero matrix.

Problem 21

Show that if  N is nilpotent then  I-N is invertible. Is that "only if" also?

Answer

Use the geometric series: 
I-N^{k+1}=(I-N)(N^k+N^{k-1}+\cdots+I)
. If  N^{k+1} is the zero matrix then we have a right inverse for  I-N . It is also a left inverse.

This statement is not "only if" since


\begin{pmatrix}
1  &0  \\
0  &1
\end{pmatrix}
-\begin{pmatrix}
-1  &0  \\
0  &-1
\end{pmatrix}

is invertible.