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Some review questions that were created by the professor and provided (in the form of handouts) in-class on several occasions. The answers are student-created and have not been verified, so use at your own risk. It is strongly recommended that you attempt to answer all the questions first before checking the solutions below (if they exist).

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1Thursday, January 31

Questions

1.1Question 1

  1. By rank-nullity
  2. No, not necessarily

1.2Question 2

  1. $$\begin{pmatrix} 3 & 1 \\ 0 & 0 \end{pmatrix}$$
  2. $v-3w$

1.3Question 3

  1. Yes
  2. No. Put them in a matrix, row-reduce

1.4Question 4

  1. Yes
  2. I think so.
  3. I don't think so. What if $T=0$? Then $S$ could be anything.

2Tuesday, February 19

2.1Question 1

Let $T$ be a linear operator on $\mathbb R^3$ defined by

$$T(x_1, x_2, x_3) = (3x_1 + x_3, -2x_1 + x_2, -x_1+2x_2 +4x_3)$$

(1) What is the matrix of $T$ in the standard ordered basis for $\mathbb R^3$?
(2) What is the matrix of $T$ in the ordered basis $\{\alpha_1, \alpha_2, \alpha_3\}$ where $\alpha_1 = (1, 0, 1)$, $\alpha_2 = (-1, 2, 1)$ and $\alpha_3 = (2, 1, 1)$?
(3) Provide a rule for $T^{-1}$ like the one which defines $T$.

2.1.1Answers

(1) $$\begin{pmatrix} 3 & 0 & 1 \\ -2 & 1 & 0 \\ -1 & 2 & 4 \end{pmatrix}$$

(Just take the coefficients for each dimension of the vector and make them into the rows of the matrix).

(2) First, we determine the effect of $T$ on each vector in the basis:

$$T\alpha_1 = (4, -2, 3) \quad T\alpha_2 = (-2, 4, 9) \quad T\alpha_3 = (7, -3, 4)$$

So the matrix is just

$$\begin{pmatrix} 4 & -2 & 3 \\ -2 & 4 & 9 \\ 7 & -3 & 4 \end{pmatrix}^T$$

(3) The inverse of the matrix in (1) is

$$\frac{1}{9}\begin{pmatrix} 4 & 2 & -1 \\ 8 & 13 & -2 \\ -3 & -6 & 3 \end{pmatrix}$$

(Determined computationally. Check it by hand if you wish.) Thus the rule is

$$T(x_1, x_2, x_3) = (\frac{4}{9}x_1 + \frac{2}{9}x_2 - \frac{1}{9}x_3, \frac{8}{9}x_1 + \frac{13}{9}x_2 -\frac{2}{9}x_3, -\frac{3}{9}x_1 - \frac{6}{9}x_2 + \frac{3}{9}x_3)$$

To prove that it works, we take the output of $T$ given $(x_1, x_2, x_3)$, and hope that $T^{-1}$ applied to this will give us $(x_1, x_2, x_3)$ again:

$$\begin{align} T(3x_1+x_3, -2x_1+x_2, -x_1+2x_2+4x_3) = & \bigg ( \frac{4}{9}(3x_1 + x_3) + \frac{2}{9}(-2x_1+x_2) - \frac{1}{9}(-x_1 +2x_2+4x_3), \\ & \frac{8}{9}(3x_1+x_3) + \frac{13}{9}(-2x_1+x_2) - \frac{2}{9}(-x_1+2x_2+4x_3), \\ & -\frac{3}{9}(3x_1+x_3) - \frac{6}{9}(-2x_1+x_2) + \frac{3}{9}(-x_1+2x_2+4x_3) \bigg ) \\ = & \bigg ( \left(\frac{12}{9} -\frac{4}{9} +\frac{1}{9} \right )x_1 + \left ( \frac{2}{9}-\frac{2}{9} \right )x_2 + \left ( \frac{4}{9} -\frac{4}{9} \right )x_3, \\ & \left ( \frac{24}{9} - \frac{26}{9} + \frac{2}{9} \right )x_1 + \left ( \frac{13}{9} - \frac{4}{9} \right )x_2 + \left ( \frac{8}{9} - \frac{8}{9} \right )x_3 \\ & \left ( -\frac{9}{9} +\frac{12}{9} -\frac{3}{9}\right)x_1 + \left ( -\frac{6}{9}+\frac{6}{9} \right )x_2 + \left ( -\frac{3}{9} + \frac{12}{9} \right )x_3 \bigg ) \\ = & (x_1, x_2, x_3) \; \blacksquare \end{align}$$

2.2Question 2

Consider the transformation $T$ which is represented by the matrix

$$A = \begin{pmatrix} 2 & 6 & -6 \\ 0 & 5 & -2 \\ 0 & 0 & 4 \end{pmatrix}$$

(1) Find its eigenvalues.
(2) Is this matrix diagonalisable? If so, diagonalise it. (Recall that a matrix $A$ is called diagonalisable if there exists a diagonal matrix $D$ and an invertible matrix $P$ such that $A = PDP^{-1}$.

2.2.1Answer

(1) 2, 5, and 4. Since the matrix is upper triangular, the eigenvalues are just the values along the diagonal (Proposition 5.18),

(2) Yes, since there are 3 distinct eigenvalues. Let's find the eigenvectors for each.

For $\lambda_1 = 2$:

$$T-\lambda_1 I = A - 2I = \begin{pmatrix} 0 & 6 & -6 \\ 0 & 3 & -2 \\ 0 & 0 & 2 \end{pmatrix} \mapsto \begin{pmatrix} 0 & 1 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad \therefore v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$

For $\lambda_2 = 5$:

$$T-\lambda_2 I = A-5I = \begin{pmatrix} -3 & 6 & -6 \\ 0 & 0 & -2 \\ 0 & 0 & -1 \end{pmatrix} \mapsto \begin{pmatrix} 1 & -2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \quad \therefore v_2 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$$

For $\lambda_3 = 4$:

$$T-\lambda_3 I = A-4I = \begin{pmatrix} -2 & 6 & -6 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{pmatrix} \mapsto \begin{pmatrix} 1 & 0 & -3 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{pmatrix} \quad \therefore v_3 = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$$

Then we make these the column vectors of $P$, and calculate $P^{-1}$:

$$P = \begin{pmatrix}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \quad P^{-1} = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}$$

$D$, of course, is just a diagonal matrix with the eigenvalues of $A$ along its diagonal (in the same order as the corresponding eigenvectors):

$$D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 4 \end{pmatrix}$$

Source: MATH 223 HTSEFP. I missed the lectures where we talked about diagonalising for this course.

2.2.2Accuracy and discussion

(1) and (2): Wolfram|Alpha confirms.

2.3Question 3

Let $\mathcal M_{2\times 2}(\mathbb R)$ be the vector space of all $2 \times 2$ matrices with real entries. Let

$$B = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$$

and let $T$ be the linear map on $\mathcal M_{2\times 2}(\mathbb R)$ defined by $T(A) = AB - BA$.

(1) Find the bases for the null space of $T$ and the range of $T$.
(2) Verify the rank-nullity theorem. (Note that the dimension of the range of $T$ is called the rank of $T$ and the dimension of the nullspace of $T$ is called the nullity of $T$.)

2.3.1Answer

(1) Just apply $T$ to an arbitrary matrix $A \in V$:

$$T(A) = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix}a & 2b \\ c & 2d \end{pmatrix} - \begin{pmatrix} a & b \\ 2c & 2d \end{pmatrix} = \begin{pmatrix} 0 & b \\ -c & 0 \end{pmatrix}$$

So a suitable basis for the range is

$$\left \{ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right \}$$

For the nullspace, it is clear that we need $b = 0$ and $c = 0$. $a$ and $d$ can be anything. So a suitable basis for the nullspace is:

$$\left \{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} \right \}$$

(2) Yep. The rank is 2 and the nullity is 2. The dimension of the vector space is 4. $2+2=4 \;\checkmark$

2.4Question 4

Let $T \in \mathcal L(\mathbb R^2)$ which is represented in the standard basis by the matrix $\displaystyle B =\begin{pmatrix}1 & 0 \\ 0 & 2 \end{pmatrix}$. Let $W$ be the subspace of $\mathbb R^2$ spanned by $e_1 = (1, 0)$.

(1) Prove that $W$ is invariant under $W$.
(2) Prove that there is no $W'$ invariant under $T$ such that $\mathbb R^2 = W \oplus W'$. (Note that in this form, the question cannot be answered, due to a typo somewhere.)

2.4.1Answer

(1) Let $w = (a, 0) \in W$ (where $a \in \mathbb R$). Then

$$Tw = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \cdot \begin{pmatrix} a \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ 0 \end{pmatrix} \in W \; \blacksquare$$

(2) No can do