Linear Algebra/Jordan Canonical Form/Solutions

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Solutions[edit]

Problem 1

Do the check for Example 2.3.

Answer

We are required to check that


\begin{pmatrix}
3  &0  \\
1  &3
\end{pmatrix}
=
N+3I=PTP^{-1}
=
\begin{pmatrix}
1/2  &1/2  \\
-1/4 &1/4
\end{pmatrix}
\begin{pmatrix}
2  &-1  \\
1  &4
\end{pmatrix}
\begin{pmatrix}
1  &-2  \\
1  &2
\end{pmatrix}

That calculation is easy.

Problem 2

Each matrix is in Jordan form. State its characteristic polynomial and its minimal polynomial.

  1. \begin{pmatrix}
3  &0  \\
1  &3
\end{pmatrix}
  2. \begin{pmatrix}
-1  &0  \\
0  &-1
\end{pmatrix}
  3. \begin{pmatrix}
2  &0  &0  \\
1  &2  &0  \\
0  &0  &-1/2
\end{pmatrix}
  4. \begin{pmatrix}
3  &0  &0  \\
1  &3  &0  \\
0  &1  &3  \\
\end{pmatrix}
  5. \begin{pmatrix}
3  &0  &0  &0  \\
1  &3  &0  &0  \\
0  &0  &3  &0  \\
0  &0  &1  &3
\end{pmatrix}
  6. \begin{pmatrix}
4  &0  &0  &0  \\
1  &4  &0  &0  \\
0  &0  &-4 &0  \\
0  &0  &1  &-4
\end{pmatrix}
  7. \begin{pmatrix}
5  &0  &0  \\
0  &2  &0  \\
0  &0  &3
\end{pmatrix}
  8. \begin{pmatrix}
5  &0  &0  &0  \\
0  &2  &0  &0  \\
0  &0  &2  &0  \\
0  &0  &0  &3
\end{pmatrix}
  9. \begin{pmatrix}
5  &0  &0  &0  \\
0  &2  &0  &0  \\
0  &1  &2  &0  \\
0  &0  &0  &3
\end{pmatrix}
Answer
  1. The characteristic polynomial is c(x)=(x-3)^2 and the minimal polynomial is the same.
  2. The characteristic polynomial is c(x)=(x+1)^2. The minimal polynomial is m(x)=x+1.
  3. The characteristic polynomial is c(x)=(x+(1/2))(x-2)^2 and the minimal polynomial is the same.
  4. The characteristic polynomial is c(x)=(x-3)^3 The minimal polynomial is the same.
  5. The characteristic polynomial is c(x)=(x-3)^4. The minimal polynomial is m(x)=(x-3)^2.
  6. The characteristic polynomial is c(x)=(x+4)^2(x-4)^2 and the minimal polynomial is the same.
  7. The characteristic polynomial is c(x)=(x-2)(x-3)(x-5) and the minimal polynomial is the same.
  8. The characteristic polynomial is c(x)=(x-2)^2(x-3)(x-5) and the minimal polynomial is m(x)=(x-2)(x-3)(x-5).
  9. The characteristic polynomial is c(x)=(x-2)^2(x-3)(x-5) and the minimal polynomial is the same.
This exercise is recommended for all readers.
Problem 3

Find the Jordan form from the given data.

  1. The matrix  T is  5 \! \times \! 5 with the single eigenvalue 3. The nullities of the powers are:  T-3I has nullity two,  (T-3I)^2 has nullity three,  (T-3I)^3 has nullity four, and  (T-3I)^4 has nullity five.
  2. The matrix  S is  5 \! \times \! 5 with two eigenvalues. For the eigenvalue 2 the nullities are:  S-2I has nullity two, and  (S-2I)^2 has nullity four. For the eigenvalue -1 the nullities are:  S+1I has nullity one.
Answer
  1. The transformation t-3 is nilpotent (that is, \mathcal{N}_\infty(t-3) is the entire space) and it acts on a string basis via two strings, \vec{\beta}_1\mapsto\vec{\beta}_2\mapsto\vec{\beta}_3
\mapsto\vec{\beta}_4\mapsto\vec{0} and \vec{\beta}_5\mapsto\vec{0}. Consequently, t-3 can be represented in this canonical form.
    
N_3=
\begin{pmatrix}
0  &0  &0  &0  &0  \\
1  &0  &0  &0  &0  \\
0  &1  &0  &0  &0  \\
0  &0  &1  &0  &0  \\
0  &0  &0  &0  &0
\end{pmatrix}
    and therefore T is similar to this this canonical form matrix.
    
J_3=N_3+3I=
\begin{pmatrix}
3  &0  &0  &0  &0  \\
1  &3  &0  &0  &0  \\
0  &1  &3  &0  &0  \\
0  &0  &1  &3  &0  \\
0  &0  &0  &0  &3
\end{pmatrix}
  2. The restriction of the transformation s+1 is nilpotent on the subspace \mathcal{N}_\infty(s+1), and the action on a string basis is given as \vec{\beta}_1\mapsto\vec{0}. The restriction of the transformation s-2 is nilpotent on the subspace \mathcal{N}_\infty(s-2), having the action on a string basis of \vec{\beta}_2\mapsto\vec{\beta}_3\mapsto\vec{0} and \vec{\beta}_4\mapsto\vec{\beta}_5\mapsto\vec{0}. Consequently the Jordan form is this
    
\begin{pmatrix}
-1  &0  &0  &0  &0  \\
0  &2  &0  &0  &0  \\
0  &1  &2  &0  &0  \\
0  &0  &0  &2  &0  \\
0  &0  &0  &1  &2
\end{pmatrix}
    (note that the blocks are arranged with the least eigenvalue first).
Problem 4

Find the change of basis matrices for each example.

  1. Example 2.13
  2. Example 2.14
  3. Example 2.15
Answer

For each, because many choices of basis are possible, many other answers are possible. Of course, the calculation to check if an answer gives that PTP^{-1} is in Jordan form is the arbiter of what's correct.

  1. Here is the arrow diagram.
    Linalg similarity cd 4.png
    The matrix to move from the lower left to the upper left is this.
    
P^{-1}=\bigl({\rm Rep}_{\mathcal{E}_3,B}(\mbox{id})\bigr)^{-1}
={\rm Rep}_{B,\mathcal{E}_3}(\mbox{id})
=\begin{pmatrix}
1  &-2   &0 \\
1  &0   &1 \\
-2  &0   &0
\end{pmatrix}
    The matrix P to move from the upper right to the lower right is the inverse of P^{-1}.
  2. We want this matrix and its inverse.
    
P^{-1}=
\begin{pmatrix}
1  &0  &3  \\
0  &1  &4  \\
0  &-2 &0
\end{pmatrix}
  3. The concatenation of these bases for the generalized null spaces will do for the basis for the entire space.
    
B_{-1}=\langle \begin{pmatrix} -1\\ 0 \\  0 \\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} -1\\ 0 \\ -1 \\ 0 \\ 1 \end{pmatrix} \rangle
\qquad
B_3=\langle \begin{pmatrix} 1 \\ 1 \\ -1 \\ 0 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 0 \\  0 \\-2 \\ 2 \end{pmatrix},
\begin{pmatrix} -1\\-1 \\  1 \\ 2 \\ 0 \end{pmatrix} \rangle
    The change of basis matrices are this one and its inverse.
    
P^{-1}=
\begin{pmatrix}
-1  &-1  &1  &0  &-1  \\
0   &0   &1  &0  &-1  \\
0   &-1  &-1 &0  &1   \\
1   &0   &0  &-2 &2   \\
0   &1   &0  &2  &0   \\
\end{pmatrix}
This exercise is recommended for all readers.
Problem 5

Find the Jordan form and a Jordan basis for each matrix.

  1. 
\begin{pmatrix}
-10  &4  \\
-25  &10
\end{pmatrix}
  2. 
\begin{pmatrix}
5   &-4 \\
9   &-7
\end{pmatrix}
  3. 
\begin{pmatrix}
4   &0    &0  \\
2   &1    &3  \\
5   &0    &4
\end{pmatrix}
  4. 
\begin{pmatrix}
5   &4    &3  \\
-1   &0    &-3 \\
1   &-2   &1
\end{pmatrix}
  5. 
\begin{pmatrix}
9   &7    &3  \\
-9   &-7   &-4 \\
4   &4    &4
\end{pmatrix}
  6. 
\begin{pmatrix}
2   &2    &-1 \\
-1   &-1   &1  \\
-1   &-2   &2
\end{pmatrix}
  7. 
\begin{pmatrix}
7   &1    &2   &2 \\
1   &4    &-1  &-1\\
-2   &1    &5   &-1\\
1   &1    &2   &8
\end{pmatrix}
Answer

The general procedure is to factor the characteristic polynomial c(x)=(x-\lambda_1)^{p_1}(x-\lambda_2)^{p_2}\cdots to get the eigenvalues \lambda_1, \lambda_2, etc. Then, for each \lambda_i we find a string basis for the action of the transformation t-\lambda_i when restricted to \mathcal{N}_\infty(t-\lambda_i), by computing the powers of the matrix T-\lambda_iI and finding the associated null spaces, until these null spaces settle down (do not change), at which point we have the generalized null space. The dimensions of those null spaces (the nullities) tell us the action of t-\lambda_i on a string basis for the generalized null space, and so we can write the pattern of subdiagonal ones to have N_{\lambda_i}. From this matrix, the Jordan block J_{\lambda_i} associated with \lambda_i is immediate J_{\lambda_i}=N_{\lambda_i}+\lambda_iI. Finally, after we have done this for each eigenvalue, we put them together into the canonical form.

  1. The characteristic polynomial of this matrix is c(x)=(-10-x)(10-x)+100=x^2, so it has only the single eigenvalue \lambda=0.

    
\begin{array}{r|ccc}
\textit{power} p &(T+0\cdot I)^p &\mathcal{N}((t-0)^p)
&\textit{nullity}                 \\
\hline
1
&\begin{pmatrix}
-10  &4  \\
-25  &10
\end{pmatrix}
&\{\begin{pmatrix} 2y/5 \\ y \end{pmatrix}\,\big|\,
y\in\mathbb{C}\}
&1                                       \\
2
&\begin{pmatrix}
0  &0  \\
0  &0
\end{pmatrix}
&\mathbb{C}^2
&2
\end{array}

    (Thus, this transformation is nilpotent: \mathcal{N}_\infty(t-0) is the entire space). From the nullities we know that t's action on a string basis is \vec{\beta}_1\mapsto\vec{\beta}_2\mapsto\vec{0}. This is the canonical form matrix for the action of t-0 on \mathcal{N}_\infty(t-0)=\mathbb{C}^2

    
N_0=
\begin{pmatrix}
0  &0  \\
1  &0
\end{pmatrix}

    and this is the Jordan form of the matrix.

    
J_0=N_0+0\cdot I=
\begin{pmatrix}
0  &0  \\
1  &0
\end{pmatrix}

    Note that if a matrix is nilpotent then its canonical form equals its Jordan form.

    We can find such a string basis using the techniques of the prior section.

    
B=\langle \begin{pmatrix} 1 \\ 0 \end{pmatrix},
\begin{pmatrix} -10 \\ -25 \end{pmatrix} \rangle

    The first basis vector has been taken so that it is in the null space of t^2 but is not in the null space of t. The second basis vector is the image of the first under t.

  2. The characteristic polynomial of this matrix is  c(x)=(x+1)^2 , so it is a single-eigenvalue matrix. (That is, the generalized null space of t+1 is the entire space.) We have
    
\mathcal{N}(t+1)=\{\begin{pmatrix} 2y/3 \\ y \end{pmatrix}\,\big|\,
y\in\mathbb{C}\}
\qquad
\mathcal{N}((t+1)^2)=\mathbb{C}^2
    and so the action of t+1 on an associated string basis is \vec{\beta}_1\mapsto\vec{\beta}_2\mapsto\vec{0}. Thus,
    
N_{-1}
=
\begin{pmatrix}
0  &0  \\
1  &0
\end{pmatrix}
    the Jordan form of T is
    
J_{-1}=N_{-1}+-1\cdot I
=
\begin{pmatrix}
-1  &0  \\
1  &-1
\end{pmatrix}
    and choosing vectors from the above null spaces gives this string basis (many other choices are possible).
    
B=\langle \begin{pmatrix} 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 6 \\ 9 \end{pmatrix} \rangle
  3. The characteristic polynomial  c(x)=(1-x)(4-x)^2=-1\cdot (x-1)(x-4)^2 has two roots and they are the eigenvalues \lambda_1=1 and \lambda_2=4. We handle the two eigenvalues separately. For \lambda_1, the calculation of the powers of T-1I yields
    
\mathcal{N}(t-1)=\{\begin{pmatrix} 0 \\ y \\ 0 \end{pmatrix}
\,\big|\, y\in\mathbb{C}\}
    and the null space of (t-1)^2 is the same. Thus this set is the generalized null space \mathcal{N}_\infty(t-1). The nullities show that the action of the restriction of t-1 to the generalized null space on a string basis is \vec{\beta}_1\mapsto\vec{0}. A similar calculation for \lambda_2=4 gives these null spaces.
    
\mathcal{N}(t-4)=\{\begin{pmatrix} 0 \\ z \\ z \end{pmatrix}
\,\big|\, z\in\mathbb{C}\}
\qquad
\mathcal{N}((t-4)^2)=\{\begin{pmatrix} y-z \\ y \\ z \end{pmatrix}
\,\big|\, y,z\in\mathbb{C}\}
    (The null space of (t-4)^3 is the same, as it must be because the power of the term associated with \lambda_2=4 in the characteristic polynomial is two, and so the restriction of t-2 to the generalized null space \mathcal{N}_\infty(t-2) is nilpotent of index at most two— it takes at most two applications of t-2 for the null space to settle down.) The pattern of how the nullities rise tells us that the action of t-4 on an associated string basis for \mathcal{N}_\infty(t-4) is \vec{\beta}_2\mapsto\vec{\beta}_3\mapsto\vec{0}. Putting the information for the two eigenvalues together gives the Jordan form of the transformation t.
    
\begin{pmatrix}
1  &0  &0  \\
0  &4  &0  \\
0  &1  &4
\end{pmatrix}
    We can take elements of the nullspaces to get an appropriate basis.
    
B=B_{1}\!\mathbin{{}^\frown}\!B_4=
\langle \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix},
\begin{pmatrix} 0 \\ 5 \\ 5 \end{pmatrix} \rangle
  4. The characteristic polynomial is  c(x)=(-2-x)(4-x)^2=-1\cdot (x+2)(x-4)^2 . For the eigenvalue \lambda_{-2}, calculation of the powers of T+2I yields this.
    
\mathcal{N}(t+2)=\{\begin{pmatrix} z \\ z \\ z \end{pmatrix}
\,\big|\, z\in\mathbb{C}\}
    The null space of (t+2)^2 is the same, and so this is the generalized null space \mathcal{N}_\infty(t+2). Thus the action of the restriction of t+2 to \mathcal{N}_\infty(t+2) on an associated string basis is \vec{\beta}_1\mapsto\vec{0}. For \lambda_2=4, computing the powers of T-4I yields
    
\mathcal{N}(t-4)=\{\begin{pmatrix} z \\ -z \\ z \end{pmatrix}
\,\big|\, z\in\mathbb{C}\}
\qquad
\mathcal{N}((t-4)^2)=\{\begin{pmatrix} x \\ -z \\ z \end{pmatrix}
\,\big|\, x,z\in\mathbb{C}\}
    and so the action of t-4 on a string basis for \mathcal{N}_\infty(t-4) is \vec{\beta}_2\mapsto\vec{\beta}_3\mapsto\vec{0}. Therefore the Jordan form is
    
\begin{pmatrix}
-2  &0  &0  \\
0  &4  &0  \\
0  &1  &4
\end{pmatrix}
    and a suitable basis is this.
    
B=B_{-2}\!\mathbin{{}^\frown}\!B_4=
\langle \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},
\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix},
\begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix} \rangle
  5. The characteristic polynomial of this matrix is  c(x)=(2-x)^3=-1\cdot (x-2)^3 . This matrix has only a single eigenvalue, \lambda=2. By finding the powers of T-2I we have
    
\mathcal{N}(t-2)=\{\begin{pmatrix} -y \\ y \\ 0 \end{pmatrix}
\,\big|\, y\in\mathbb{C}\}
\qquad
\mathcal{N}((t-2)^2)=\{\begin{pmatrix} -y-(1/2)z \\ y \\ z \end{pmatrix}
\,\big|\, y,z\in\mathbb{C}\}
\qquad
\mathcal{N}((t-2)^3)=\mathbb{C}^3
    and so the action of t-2 on an associated string basis is \vec{\beta}_1\mapsto\vec{\beta}_2\mapsto
\vec{\beta}_3\mapsto\vec{0}. The Jordan form is this
    
\begin{pmatrix}
2  &0  &0  \\
1  &2  &0  \\
0  &1  &2
\end{pmatrix}
    and one choice of basis is this.
    
B=\langle \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 7 \\ -9 \\ 4 \end{pmatrix},
\begin{pmatrix} -2 \\ 2 \\ 0 \end{pmatrix} \rangle
  6. The characteristic polynomial  c(x)=(1-x)^3=-(x-1)^3 has only a single root, so the matrix has only a single eigenvalue \lambda=1. Finding the powers of T-1I and calculating the null spaces
    
\mathcal{N}(t-1)=\{\begin{pmatrix} -2y+z \\ y \\ z \end{pmatrix}
\,\big|\, y,z\in\mathbb{C}\}
\qquad
\mathcal{N}((t-1)^2)=\mathbb{C}^3
    shows that the action of the nilpotent map t-1 on a string basis is \vec{\beta}_1\mapsto\vec{\beta}_2\mapsto\vec{0} and \vec{\beta}_3\mapsto\vec{0}. Therefore the Jordan form is
    
J=
\begin{pmatrix}
1  &0  &0  \\
1  &1  &0  \\
0  &0  &1
\end{pmatrix}
    and an appropriate basis (a string basis associated with t-1) is this.
    
B=\langle \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix},
\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \rangle
  7. The characteristic polynomial is a bit large for by-hand calculation, but just manageable  c(x)=x^4-24x^3+216x^2-864x+1296=(x-6)^4 . This is a single-eigenvalue map, so the transformation t-6 is nilpotent. The null spaces
    
\mathcal{N}(t-6)=\{\begin{pmatrix} -z-w \\ -z-w \\ z \\ w \end{pmatrix}
\,\big|\, z,w\in\mathbb{C}\}
\quad
\mathcal{N}((t-6)^2)=\{\begin{pmatrix} x \\ -z-w \\ z \\ w \end{pmatrix}
\,\big|\, x,z,w\in\mathbb{C}\}
\quad
\mathcal{N}((t-6)^3)=\mathbb{C}^4
    and the nullities show that the action of t-6 on a string basis is \vec{\beta}_1\mapsto\vec{\beta}_2\mapsto
\vec{\beta}_3\mapsto\vec{0} and \vec{\beta}_4\mapsto\vec{0}. The Jordan form is
    
\begin{pmatrix}
6  &0  &0  &0  \\
1  &6  &0  &0  \\
0  &1  &6  &0  \\
0  &0  &0  &6  \\
\end{pmatrix}
    and finding a suitable string basis is routine.
    
B=\langle \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix},
\begin{pmatrix} 2 \\ -1 \\ -1 \\ 2 \end{pmatrix},
\begin{pmatrix} 3 \\ 3 \\ -6 \\ 3 \end{pmatrix},
\begin{pmatrix} -1 \\ -1 \\ 1 \\ 0 \end{pmatrix} \rangle
This exercise is recommended for all readers.
Problem 6

Find all possible Jordan forms of a transformation with characteristic polynomial  (x-1)^2(x+2)^2  .

Answer

There are two eigenvalues, \lambda_1=-2 and \lambda_2=1. The restriction of t+2 to \mathcal{N}_\infty(t+2) could have either of these actions on an associated string basis.


\vec{\beta}_1\mapsto\vec{\beta}_2\mapsto\vec{0}
\qquad
\begin{array}[t]{l}
\vec{\beta}_1\mapsto\vec{0}  \\
\vec{\beta}_2\mapsto\vec{0}
\end{array}

The restriction of t-1 to \mathcal{N}_\infty(t-1) could have either of these actions on an associated string basis.


\vec{\beta}_3\mapsto\vec{\beta}_4\mapsto\vec{0}
\qquad
\begin{array}[t]{l}
\vec{\beta}_3\mapsto\vec{0}  \\
\vec{\beta}_4\mapsto\vec{0}
\end{array}

In combination, that makes four possible Jordan forms, the two first actions, the second and first, the first and second, and the two second actions.


\begin{pmatrix}
-2  &0  &0  &0  \\
1  &-2 &0  &0  \\
0  &0  &1  &0  \\
0  &0  &1  &1
\end{pmatrix}
\quad
\begin{pmatrix}
-2  &0  &0  &0  \\
0  &-2 &0  &0  \\
0  &0  &1  &0  \\
0  &0  &1  &1
\end{pmatrix}
\quad
\begin{pmatrix}
-2  &0  &0  &0  \\
1  &-2 &0  &0  \\
0  &0  &1  &0  \\
0  &0  &0  &1
\end{pmatrix}
\quad
\begin{pmatrix}
-2  &0  &0  &0  \\
0  &-2 &0  &0  \\
0  &0  &1  &0  \\
0  &0  &0  &1
\end{pmatrix}
Problem 7

Find all possible Jordan forms of a transformation with characteristic polynomial  (x-1)^3(x+2) .

Answer

The restriction of t+2 to \mathcal{N}_\infty(t+2) can have only the action \vec{\beta}_1\mapsto\vec{0}. The restriction of t-1 to \mathcal{N}_\infty(t-1) could have any of these three actions on an associated string basis.


\vec{\beta}_2\mapsto\vec{\beta}_3\mapsto\vec{\beta}_4\mapsto\vec{0}
\qquad
\begin{array}[t]{l}
\vec{\beta}_2\mapsto\vec{\beta}_3\mapsto\vec{0}  \\
\vec{\beta}_4\mapsto\vec{0}
\end{array}
\qquad
\begin{array}[t]{l}
\vec{\beta}_2\mapsto\vec{0}  \\
\vec{\beta}_3\mapsto\vec{0}  \\
\vec{\beta}_4\mapsto\vec{0}
\end{array}

Taken together there are three possible Jordan forms, the one arising from the first action by t-1 (along with the only action from t+2), the one arising from the second action, and the one arising from the third action.


\begin{pmatrix}
-2  &0  &0  &0  \\
0  &1  &0  &0  \\
0  &1  &1  &0  \\
0  &0  &1  &1
\end{pmatrix}
\quad
\begin{pmatrix}
-2  &0  &0  &0  \\
0  &1  &0  &0  \\
0  &1  &1  &0  \\
0  &0  &0  &1
\end{pmatrix}
\quad
\begin{pmatrix}
-2  &0  &0  &0  \\
0  &1  &0  &0  \\
0  &0  &1  &0  \\
0  &0  &0  &1
\end{pmatrix}
This exercise is recommended for all readers.
Problem 8

Find all possible Jordan forms of a transformation with characteristic polynomial  (x-2)^3(x+1) and minimal polynomial  (x-2)^2(x+1) .

Answer

The action of t+1 on a string basis for \mathcal{N}_\infty(t+1) must be \vec{\beta}_1\mapsto\vec{0}. Because of the power of  x-2 in the minimal polynomial, a string basis for t-2 has length two and so the action of  t-2 on  \mathcal{N}_\infty(t-2) must be of this form.


\begin{array}[t]{l}
\vec{\beta}_2\mapsto\vec{\beta}_3\mapsto\vec{0}  \\
\vec{\beta}_4\mapsto\vec{0}
\end{array}

Therefore there is only one Jordan form that is possible.


\begin{pmatrix}
-1  &0  &0  &0  \\
0  &2  &0  &0  \\
0  &1  &2  &0  \\
0  &0  &0  &2
\end{pmatrix}
Problem 9

Find all possible Jordan forms of a transformation with characteristic polynomial  (x-2)^4(x+1) and minimal polynomial  (x-2)^2(x+1) .

Answer

There are two possible Jordan forms. The action of t+1 on a string basis for \mathcal{N}_\infty(t+1) must be \vec{\beta}_1\mapsto\vec{0}. There are two actions for t-2 on a string basis for \mathcal{N}_\infty(t-2) that are possible with this characteristic polynomial and minimal polynomial.


\begin{array}[t]{l}
\vec{\beta}_2\mapsto\vec{\beta}_3\mapsto\vec{0}  \\
\vec{\beta}_4\mapsto\vec{\beta}_5\mapsto\vec{0}
\end{array}
\qquad
\begin{array}[t]{l}
\vec{\beta}_2\mapsto\vec{\beta}_3\mapsto\vec{0}  \\
\vec{\beta}_4\mapsto\vec{0}                      \\
\vec{\beta}_5\mapsto\vec{0}
\end{array}

The resulting Jordan form matrics are these.


\begin{pmatrix}
-1  &0  &0  &0  &0  \\
0  &2  &0  &0  &0  \\
0  &1  &2  &0  &0  \\
0  &0  &0  &2  &0  \\
0  &0  &0  &1  &2
\end{pmatrix}
\qquad
\begin{pmatrix}
-1  &0  &0  &0  &0  \\
0  &2  &0  &0  &0  \\
0  &1  &2  &0  &0  \\
0  &0  &0  &2  &0  \\
0  &0  &0  &0  &2
\end{pmatrix}
This exercise is recommended for all readers.
Problem 10
Diagonalize these.
  1.  \begin{pmatrix}
1  &1  \\
0  &0
\end{pmatrix}
  2.  \begin{pmatrix}
0  &1  \\
1  &0
\end{pmatrix}
Answer
  1. The characteristic polynomial is  c(x)=x(x-1) . For \lambda_1=0 we have
    
\mathcal{N}(t-0)=\{\begin{pmatrix} -y \\ y \end{pmatrix}
\,\big|\, y\in\mathbb{C} \}
    (of course, the null space of t^2 is the same). For \lambda_2=1,
    
\mathcal{N}(t-1)=\{\begin{pmatrix} x \\ 0 \end{pmatrix}
\,\big|\, x\in\mathbb{C} \}
    (and the null space of (t-1)^2 is the same). We can take this basis
    
B=\langle \begin{pmatrix} 1 \\ -1 \end{pmatrix},\begin{pmatrix} 1 \\ 0 \end{pmatrix} \rangle
    to get the diagonalization.
    
\begin{pmatrix}
1  &1  \\
-1  &0
\end{pmatrix}^{-1}
\begin{pmatrix}
1  &1  \\
0  &0
\end{pmatrix}
\begin{pmatrix}
1  &1  \\
-1  &0
\end{pmatrix}
=
\begin{pmatrix}
0  &0  \\
0  &1
\end{pmatrix}
  2. The characteristic polynomial is  c(x)=x^2-1=(x+1)(x-1) . For \lambda_1=-1,
    
\mathcal{N}(t+1)=\{\begin{pmatrix} -y \\ y \end{pmatrix}
\,\big|\, y\in\mathbb{C} \}
    and the null space of (t+1)^2 is the same. For \lambda_2=1
    
\mathcal{N}(t-1)=\{\begin{pmatrix} y \\ y \end{pmatrix}
\,\big|\, y\in\mathbb{C} \}
    and the null space of (t-1)^2 is the same. We can take this basis
    
B=\langle \begin{pmatrix} 1 \\ -1 \end{pmatrix},\begin{pmatrix} 1 \\ 1 \end{pmatrix} \rangle
    to get a diagonalization.
    
\begin{pmatrix}
1  &1  \\
1  &-1
\end{pmatrix}^{-1}
\begin{pmatrix}
0  &1  \\
1  &0
\end{pmatrix}
\begin{pmatrix}
1   &1  \\
-1  &1
\end{pmatrix}
=
\begin{pmatrix}
-1  &0  \\
0   &1
\end{pmatrix}
This exercise is recommended for all readers.
Problem 11

Find the Jordan matrix representing the differentiation operator on  \mathcal{P}_3 .

Answer

The transformation d/dx:\mathcal{P}_3\to \mathcal{P}_3 is nilpotent. Its action on  B=\langle x^3,3x^2,6x,6 \rangle  is x^3\mapsto 3x^2\mapsto 6x\mapsto 6\mapsto 0. Its Jordan form is its canonical form as a nilpotent matrix.


J=
\begin{pmatrix}
0  &0  &0  &0  \\
1  &0  &0  &0  \\
0  &1  &0  &0  \\
0  &0  &1  &0
\end{pmatrix}
This exercise is recommended for all readers.
Problem 12

Decide if these two are similar.


\begin{pmatrix}
1  &-1 \\
4  &-3 \\
\end{pmatrix}
\qquad
\begin{pmatrix}
-1  &0  \\
1  &-1 \\
\end{pmatrix}
Answer

Yes. Each has the characteristic polynomial (x+1)^2. Calculations of the powers of T_1+1\cdot I and T_2+1\cdot I gives these two.


\mathcal{N}(t_1+1)=\{\begin{pmatrix} y/2 \\ y \end{pmatrix} \,\big|\, y\in\mathbb{C}\}
\qquad
\mathcal{N}(t_2+1)=\{\begin{pmatrix} 0 \\ y \end{pmatrix} \,\big|\, y\in\mathbb{C}\}

(Of course, for each the null space of the square is the entire space.) The way that the nullities rise shows that each is similar to this Jordan form matrix


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

and they are therefore similar to each other.

Problem 13

Find the Jordan form of this matrix.


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

Also give a Jordan basis.

Answer

Its characteristic polynomial is  c(x)=x^2+1 which has complex roots  x^2+1=(x+i)(x-i) . Because the roots are distinct, the matrix is diagonalizable and its Jordan form is that diagonal matrix.


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

To find an associated basis we compute the null spaces.


\mathcal{N}(t+i)=\{\begin{pmatrix} -iy \\ y \end{pmatrix}
\,\big|\, y\in\mathbb{C}\}
\qquad
\mathcal{N}(t-i)=\{\begin{pmatrix} iy \\ y \end{pmatrix}
\,\big|\, y\in\mathbb{C}\}

For instance,


T+i\cdot I=
\begin{pmatrix}
i  &-1  \\
1  &i
\end{pmatrix}

and so we get a description of the null space of t+i by solving this linear system.


\begin{array}{*{2}{rc}r}
ix  &-  &y  &=  &0  \\
x  &+  &iy &=  &0
\end{array}
\;\xrightarrow[]{i\rho_1+\rho_2}\;
\begin{array}{*{2}{rc}r}
ix  &-  &y  &=  &0  \\
&   &0  &=  &0
\end{array}

(To change the relation ix=y so that the leading variable x is expressed in terms of the free variable y, we can multiply both sides by -i.)

As a result, one such basis is this.


B=\langle \begin{pmatrix} -i \\ 1 \end{pmatrix},
\begin{pmatrix} i \\ 1 \end{pmatrix} \rangle
Problem 14

How many similarity classes are there for  3 \! \times \! 3 matrices whose only eigenvalues are  -3 and  4 ?

Answer

We can count the possible classes by counting the possible canonical representatives, that is, the possible Jordan form matrices. The characteristic polynomial must be either c_1(x)=(x+3)^2(x-4) or c_2(x)=(x+3)(x-4)^2. In the c_1 case there are two possible actions of t+3 on a string basis for \mathcal{N}_\infty(t+3).


\vec{\beta}_1\mapsto\vec{\beta}_2\mapsto \vec{0}
\qquad
\begin{array}[t]{l}
\vec{\beta}_1\mapsto\vec{0} \\
\vec{\beta}_2\mapsto\vec{0}
\end{array}

There are two associated Jordan form matrices.


\begin{pmatrix}
-3  &0  &0  \\
1  &-3 &0  \\
0  &0  &4
\end{pmatrix}
\qquad
\begin{pmatrix}
-3  &0  &0  \\
0  &-3 &0  \\
0  &0  &4
\end{pmatrix}

Similarly there are two Jordan form matrices that could arise out of c_2.


\begin{pmatrix}
-3  &0  &0  \\
0  &4  &0  \\
0  &1  &4
\end{pmatrix}
\qquad
\begin{pmatrix}
-3  &0  &0  \\
0  &4  &0  \\
0  &0  &4
\end{pmatrix}

So in total there are four possible Jordan forms.

This exercise is recommended for all readers.
Problem 15

Prove that a matrix is diagonalizable if and only if its minimal polynomial has only linear factors.

Answer

Jordan form is unique. A diagonal matrix is in Jordan form. Thus the Jordan form of a diagonalizable matrix is its diagonalization. If the minimal polynomial has factors to some power higher than one then the Jordan form has subdiagonal  1 's, and so is not diagonal.

Problem 16

Give an example of a linear transformation on a vector space that has no non-trivial invariant subspaces.

Answer

One example is the transformation of  \mathbb{C} that sends  x to  -x .

Problem 17

Show that a subspace is  t-\lambda_1 invariant if and only if it is  t-\lambda_2 invariant.

Answer

Apply Lemma 2.7 twice; the subspace is t-\lambda_1 invariant if and only if it is t invariant, which in turn holds if and only if it is t-\lambda_2 invariant.

Problem 18

Prove or disprove: two  n \! \times \! n matrices are similar if and only if they have the same characteristic and minimal polynomials.

Answer

False; these two 4 \! \times \! 4 matrices each have c(x)=(x-3)^4 and m(x)=(x-3)^2.


\begin{pmatrix}
3  &0  &0  &0  \\
1  &3  &0  &0  \\
0  &0  &3  &0  \\
0  &0  &1  &3
\end{pmatrix}
\quad
\begin{pmatrix}
3  &0  &0  &0  \\
1  &3  &0  &0  \\
0  &0  &3  &0  \\
0  &0  &0  &3
\end{pmatrix}
Problem 19

The trace of a square matrix is the sum of its diagonal entries.

  1. Find the formula for the characteristic polynomial of a 2 \! \times \! 2 matrix.
  2. Show that trace is invariant under similarity, and so we can sensibly speak of the "trace of a map". (Hint: see the prior item.)
  3. Is trace invariant under matrix equivalence?
  4. Show that the trace of a map is the sum of its eigenvalues (counting multiplicities).
  5. Show that the trace of a nilpotent map is zero. Does the converse hold?
Answer
  1. The characteristic polynomial is this.
    
\begin{vmatrix}
a-x  &b  \\
c  &d-x
\end{vmatrix}
=(a-x)(d-x)-bc=ad-(a+d)x+x^2-bc
=x^2-(a+d)x+(ad-bc)
    Note that the determinant appears as the constant term.
  2. Recall that the characteristic polynomial  \left|T-xI\right| is invariant under similarity. Use the permutation expansion formula to show that the trace is the negative of the coefficient of  x^{n-1} .
  3. No, there are matrices T and S that are equivalent S=PTQ (for some nonsingular P and Q) but that have different traces. An easy example is this.
    
PTQ=
\begin{pmatrix}
2  &0  \\
0  &1
\end{pmatrix}
\begin{pmatrix}
1  &0  \\
0  &1
\end{pmatrix}
\begin{pmatrix}
1  &0  \\
0  &1
\end{pmatrix}
=
\begin{pmatrix}
2  &0  \\
0  &1
\end{pmatrix}
    Even easier examples using 1 \! \times \! 1 matrices are possible.
  4. Put the matrix in Jordan form. By the first item, the trace is unchanged.
  5. The first part is easy; use the third item. The converse does not hold: this matrix
    
\begin{pmatrix}
1  &0  \\
0  &-1
\end{pmatrix}
    has a trace of zero but is not nilpotent.
Problem 20

To use Definition 2.6 to check whether a subspace is t invariant, we seemingly have to check all of the infinitely many vectors in a (nontrivial) subspace to see if they satisfy the condition. Prove that a subspace is  t invariant if and only if its subbasis has the property that for all of its elements, t(\vec{\beta}) is in the subspace.

Answer

Suppose that  B_M is a basis for a subspace  M of some vector space. Implication one way is clear; if  M is  t invariant then in particular, if  \vec{m}\in B_M then  t(\vec{m})\in M . For the other implication, let  B_M=\langle \vec{\beta}_1,\dots,\vec{\beta}_q \rangle  and note that  t(\vec{m})=t(m_1\vec{\beta}_1+\dots+m_q\vec{\beta}_q)
=m_1t(\vec{\beta}_1)+\dots+m_qt(\vec{\beta}_q) is in  M as any subspace is closed under linear combinations.

This exercise is recommended for all readers.
Problem 21

Is  t invariance preserved under intersection? Under union? Complementation? Sums of subspaces?

Answer

Yes, the intersection of t invariant subspaces is t invariant. Assume that  M and  N are  t invariant. If  \vec{v}\in M\cap N then  t(\vec{v})\in M by the invariance of  M and  t(\vec{v})\in N by the invariance of  N .

Of course, the union of two subspaces need not be a subspace (remember that the x- and y-axes are subspaces of the plane \mathbb{R}^2 but the union of the two axes fails to be closed under vector addition, for instance it does not contain \vec{e}_1+\vec{e}_2.) However, the union of invariant subsets is an invariant subset; if  \vec{v}\in M\cup N then  \vec{v}\in M or  \vec{v}\in N so  t(\vec{v})\in M or  t(\vec{v})\in N .

No, the complement of an invariant subspace need not be invariant. Consider the subspace


\{\begin{pmatrix} x \\ 0 \end{pmatrix}\,\big|\, x\in\mathbb{C}\}

of  \mathbb{C}^2 under the zero transformation.

Yes, the sum of two invariant subspaces is invariant. The check is easy.

Problem 22

Give a way to order the Jordan blocks if some of the eigenvalues are complex numbers. That is, suggest a reasonable ordering for the complex numbers.

Answer

One such ordering is the dictionary ordering. Order by the real component first, then by the coefficient of  i . For instance,  3+2i<4+1i but  4+1i<4+2i .

Problem 23

Let  \mathcal{P}_j(\mathbb{R}) be the vector space over the reals of degree  j polynomials. Show that if  j\le k then  \mathcal{P}_j(\mathbb{R}) is an invariant subspace of  \mathcal{P}_k(\mathbb{R}) under the differentiation operator. In  \mathcal{P}_7(\mathbb{R}) , does any of  \mathcal{P}_0(\mathbb{R}) , ...,  \mathcal{P}_6(\mathbb{R}) have an invariant complement?

Answer

The first half is easy— the derivative of any real polynomial is a real polynomial of lower degree. The answer to the second half is "no"; any complement of  \mathcal{P}_j(\mathbb{R}) must include a polynomial of degree  j+1 , and the derivative of that polynomial is in  \mathcal{P}_j(\mathbb{R}).

Problem 24

In  \mathcal{P}_n(\mathbb{R}) , the vector space (over the reals) of degree  n polynomials,


\mathcal{E}=
\{p(x)\in\mathcal{P}_n(\mathbb{R})\,\big|\, p(-x)=p(x) \text{ for all }x\}

and


\mathcal{O}=
\{p(x)\in\mathcal{P}_n(\mathbb{R})\,\big|\, p(-x)=-p(x) \text{ for all }x\}

are the even and the odd polynomials;  p(x)=x^2 is even while  p(x)=x^3 is odd. Show that they are subspaces. Are they complementary? Are they invariant under the differentiation transformation?

Answer

For the first half, show that each is a subspace and then observe that any polynomial can be uniquely written as the sum of even-powered and odd-powered terms (the zero polynomial is both). The answer to the second half is "no":  x^2 is even while  2x is odd.

Problem 25

Lemma 2.8 says that if  M and  N are invariant complements then  t has a representation in the given block form (with respect to the same ending as starting basis, of course). Does the implication reverse?

Answer

Yes. If  {\rm Rep}_{B,B}(t) has the given block form, take  B_M to be the first  j vectors of  B , where  J is the  j \! \times \! j upper left submatrix. Take  B_N to be the remaining  k vectors in  B . Let  M and  N be the spans of  B_M and  B_N . Clearly  M and  N are complementary. To see  M is invariant ( N works the same way), represent any  \vec{m}\in M with respect to  B , note the last  k components are zeroes, and multiply by the given block matrix. The final  k components of the result are zeroes, so that result is again in  M .

Problem 26

A matrix  S is the square root of another  T if  S^2=T . Show that any nonsingular matrix has a square root.

Answer

Put the matrix in Jordan form. By non-singularity, there are no zero eigenvalues on the diagonal. Ape this example:


\begin{pmatrix}
9  &0  &0 \\
1  &9  &0 \\
0  &0  &4
\end{pmatrix}
=
\begin{pmatrix}
3  &0  &0 \\
1/6 &3  &0 \\
0  &0  &2
\end{pmatrix}^2

to construct a square root. Show that it holds up under similarity: if  S^2=T then  (PSP^{-1})(PSP^{-1})=PTP^{-1} .