Overview
Exam #4 will be administered in LECTURE on Friday, April 24 and will cover all Matrix Algebra content covered this semester (Lectures 13–26). There will be a greater emphasis on material covered after Exam 3 (sections 3.3, 3.4, 3.5, 3.6, 4.1, 4.2, 4.4, 4.5, 4.6, and 4.7 ), but you should be prepared to answer questions from any of the sections covered on Exam 3 (sections 1.1, 1.2, 1.3, 1.5, 1.6, 1.7, 1.9, and 3.2).
If you submitted a ANSS memo with accommodations for exams, you will be contacted via email with details regarding the alternate location of your exam. Only students who have emailed the instructor their accommodations memorandum by Wednesday 4/22 at 5PM will be able to use their testing accommodations on Exam 3.
You will have 70 minutes to complete the exam. Students must arrive on time and listen to instructions regarding seating requirements. Students must leave all backpacks in the front of the exam room up against the wall. Do not block any exits or obstruct any walkways. You will not be granted any additional time to complete your exam if you arrive after the exam has begun. Students will be required to sign in and collected exams will be cross-checked with the sign-in sheets. Exams from students who do not sign in will not be graded.
Questions
Q1. Span and Linear Combinations
Let v ⃗ 1 = [ 1 2 0 ] \vec{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} v 1 = 1 2 0 v ⃗ 2 = [ 0 − 1 1 ] \vec{v}_2 = \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} v 2 = 0 − 1 1 S p { v ⃗ 1 , v ⃗ 2 } Sp\{\vec{v}_1, \vec{v}_2\} Sp { v 1 , v 2 } S p { v ⃗ 1 , v ⃗ 2 } Sp\{\vec{v}_1, \vec{v}_2\} Sp { v 1 , v 2 } v ⃗ 1 \vec{v}_1 v 1 v ⃗ 2 \vec{v}_2 v 2
(a) v ⃗ = [ 1 1 − 1 ] \vec{v} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} v = 1 1 − 1
(b) w ⃗ = [ 2 3 1 ] \vec{w} = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} w = 2 3 1
Answer of This Question We check if each vector can be written as c 1 v ⃗ 1 + c 2 v ⃗ 2 c_1\vec{v}_1 + c_2\vec{v}_2 c 1 v 1 + c 2 v 2
(a) v ⃗ = [ 1 1 − 1 ] \vec{v} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} v = 1 1 − 1
Set up the system c 1 [ 1 2 0 ] + c 2 [ 0 − 1 1 ] = [ 1 1 − 1 ] c_1\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} c 1 1 2 0 + c 2 0 − 1 1 = 1 1 − 1
{ c 1 = 1 2 c 1 − c 2 = 1 c 2 = − 1 \begin{cases} c_1 = 1 \\ 2c_1 - c_2 = 1 \\ c_2 = -1 \end{cases} ⎩ ⎨ ⎧ c 1 = 1 2 c 1 − c 2 = 1 c 2 = − 1 From the first equation: c 1 = 1 c_1 = 1 c 1 = 1 c 2 = − 1 c_2 = -1 c 2 = − 1 2 ( 1 ) − ( − 1 ) = 3 ≠ 1 2(1) - (-1) = 3 \neq 1 2 ( 1 ) − ( − 1 ) = 3 = 1 Contradiction.
Answer (a): v ⃗ ∉ S p { v ⃗ 1 , v ⃗ 2 } \vec{v} \notin Sp\{\vec{v}_1, \vec{v}_2\} v ∈ / Sp { v 1 , v 2 }
(b) w ⃗ = [ 2 3 1 ] \vec{w} = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} w = 2 3 1
Set up the system c 1 [ 1 2 0 ] + c 2 [ 0 − 1 1 ] = [ 2 3 1 ] c_1\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} c 1 1 2 0 + c 2 0 − 1 1 = 2 3 1
{ c 1 = 2 2 c 1 − c 2 = 3 c 2 = 1 \begin{cases} c_1 = 2 \\ 2c_1 - c_2 = 3 \\ c_2 = 1 \end{cases} ⎩ ⎨ ⎧ c 1 = 2 2 c 1 − c 2 = 3 c 2 = 1 From the first equation: c 1 = 2 c_1 = 2 c 1 = 2 c 2 = 1 c_2 = 1 c 2 = 1 2 ( 2 ) − 1 = 3 2(2) - 1 = 3 2 ( 2 ) − 1 = 3 Consistent ✓ \checkmark ✓
Answer (b): w ⃗ ∈ S p { v ⃗ 1 , v ⃗ 2 } \vec{w} \in Sp\{\vec{v}_1, \vec{v}_2\} w ∈ Sp { v 1 , v 2 } w ⃗ = 2 v ⃗ 1 + v ⃗ 2 \vec{w} = 2\vec{v}_1 + \vec{v}_2 w = 2 v 1 + v 2
Final Answer
(a) v ⃗ ∉ S p { v ⃗ 1 , v ⃗ 2 } \vec{v} \notin Sp\{\vec{v}_1, \vec{v}_2\} v ∈ / Sp { v 1 , v 2 }
(b) w ⃗ = 2 v ⃗ 1 + v ⃗ 2 \vec{w} = 2\vec{v}_1 + \vec{v}_2 w = 2 v 1 + v 2
Q2. Null Space
Determine which of the following vectors are in the null space of the matrix A = [ 2 2 3 3 ] A = \begin{bmatrix} 2 & 2 \\ 3 & 3 \end{bmatrix} A = [ 2 3 2 3 ]
(a) a ⃗ = [ 1 − 1 ] \vec{a} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} a = [ 1 − 1 ]
(b) b ⃗ = [ 2 − 3 ] \vec{b} = \begin{bmatrix} 2 \\ -3 \end{bmatrix} b = [ 2 − 3 ]
(c) c ⃗ = [ − 2 2 ] \vec{c} = \begin{bmatrix} -2 \\ 2 \end{bmatrix} c = [ − 2 2 ]
(d) d ⃗ = [ 1 0 ] \vec{d} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} d = [ 1 0 ]
(e) e ⃗ = [ 0 0 ] \vec{e} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} e = [ 0 0 ]
Answer of This Question A vector x ⃗ \vec{x} x A A A A x ⃗ = 0 ⃗ A\vec{x} = \vec{0} A x = 0
Check each vector:
(a) A a ⃗ = [ 2 2 3 3 ] [ 1 − 1 ] = [ 2 − 2 3 − 3 ] = [ 0 0 ] A\vec{a} = \begin{bmatrix} 2 & 2 \\ 3 & 3 \end{bmatrix}\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 2-2 \\ 3-3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} A a = [ 2 3 2 3 ] [ 1 − 1 ] = [ 2 − 2 3 − 3 ] = [ 0 0 ]
(b) A b ⃗ = [ 2 2 3 3 ] [ 2 − 3 ] = [ 4 − 6 6 − 9 ] = [ − 2 − 3 ] ≠ 0 ⃗ A\vec{b} = \begin{bmatrix} 2 & 2 \\ 3 & 3 \end{bmatrix}\begin{bmatrix} 2 \\ -3 \end{bmatrix} = \begin{bmatrix} 4-6 \\ 6-9 \end{bmatrix} = \begin{bmatrix} -2 \\ -3 \end{bmatrix} \neq \vec{0} A b = [ 2 3 2 3 ] [ 2 − 3 ] = [ 4 − 6 6 − 9 ] = [ − 2 − 3 ] = 0
(c) A c ⃗ = [ 2 2 3 3 ] [ − 2 2 ] = [ − 4 + 4 − 6 + 6 ] = [ 0 0 ] A\vec{c} = \begin{bmatrix} 2 & 2 \\ 3 & 3 \end{bmatrix}\begin{bmatrix} -2 \\ 2 \end{bmatrix} = \begin{bmatrix} -4+4 \\ -6+6 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} A c = [ 2 3 2 3 ] [ − 2 2 ] = [ − 4 + 4 − 6 + 6 ] = [ 0 0 ]
(d) A d ⃗ = [ 2 2 3 3 ] [ 1 0 ] = [ 2 3 ] ≠ 0 ⃗ A\vec{d} = \begin{bmatrix} 2 & 2 \\ 3 & 3 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \neq \vec{0} A d = [ 2 3 2 3 ] [ 1 0 ] = [ 2 3 ] = 0
(e) A e ⃗ = [ 2 2 3 3 ] [ 0 0 ] = [ 0 0 ] A\vec{e} = \begin{bmatrix} 2 & 2 \\ 3 & 3 \end{bmatrix}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} A e = [ 2 3 2 3 ] [ 0 0 ] = [ 0 0 ]
Final Answer
In N ( A ) N(A) N ( A ) a ⃗ , c ⃗ , e ⃗ \vec{a}, \vec{c}, \vec{e} a , c , e N ( A ) N(A) N ( A ) b ⃗ , d ⃗ \vec{b}, \vec{d} b , d
Q3. Range (Column Space)
Determine which of the following vectors are in the range of the matrix A = [ − 1 3 2 − 6 ] A = \begin{bmatrix} -1 & 3 \\ 2 & -6 \end{bmatrix} A = [ − 1 2 3 − 6 ] A A A A A A A A A
(a) b ⃗ 1 = [ 1 − 3 ] \vec{b}_1 = \begin{bmatrix} 1 \\ -3 \end{bmatrix} b 1 = [ 1 − 3 ]
(b) b ⃗ 2 = [ 4 − 8 ] \vec{b}_2 = \begin{bmatrix} 4 \\ -8 \end{bmatrix} b 2 = [ 4 − 8 ]
Answer of This Question A vector b ⃗ \vec{b} b A A A A x ⃗ = b ⃗ A\vec{x} = \vec{b} A x = b b ⃗ \vec{b} b A A A
Let the columns of A = [ − 1 3 2 − 6 ] A = \begin{bmatrix} -1 & 3 \\ 2 & -6 \end{bmatrix} A = [ − 1 2 3 − 6 ] a ⃗ 1 = [ − 1 2 ] \vec{a}_1 = \begin{bmatrix} -1 \\ 2 \end{bmatrix} a 1 = [ − 1 2 ] a ⃗ 2 = [ 3 − 6 ] \vec{a}_2 = \begin{bmatrix} 3 \\ -6 \end{bmatrix} a 2 = [ 3 − 6 ]
Note that a ⃗ 2 = − 3 a ⃗ 1 \vec{a}_2 = -3\vec{a}_1 a 2 = − 3 a 1 c o l ( A ) = S p { a ⃗ 1 } col(A) = Sp\{\vec{a}_1\} co l ( A ) = Sp { a 1 }
(a) b ⃗ 1 = [ 1 − 3 ] \vec{b}_1 = \begin{bmatrix} 1 \\ -3 \end{bmatrix} b 1 = [ 1 − 3 ]
Solve c 1 [ − 1 2 ] + c 2 [ 3 − 6 ] = [ 1 − 3 ] c_1\begin{bmatrix} -1 \\ 2 \end{bmatrix} + c_2\begin{bmatrix} 3 \\ -6 \end{bmatrix} = \begin{bmatrix} 1 \\ -3 \end{bmatrix} c 1 [ − 1 2 ] + c 2 [ 3 − 6 ] = [ 1 − 3 ]
{ − c 1 + 3 c 2 = 1 2 c 1 − 6 c 2 = − 3 \begin{cases} -c_1 + 3c_2 = 1 \\ 2c_1 - 6c_2 = -3 \end{cases} { − c 1 + 3 c 2 = 1 2 c 1 − 6 c 2 = − 3
The second equation is − 2 -2 − 2 − 2 ( 1 ) = − 2 ≠ − 3 -2(1) = -2 \neq -3 − 2 ( 1 ) = − 2 = − 3 Inconsistent.
Answer (a): b ⃗ 1 ∉ c o l ( A ) \vec{b}_1 \notin col(A) b 1 ∈ / co l ( A )
(b) b ⃗ 2 = [ 4 − 8 ] \vec{b}_2 = \begin{bmatrix} 4 \\ -8 \end{bmatrix} b 2 = [ 4 − 8 ]
Solve c 1 [ − 1 2 ] + c 2 [ 3 − 6 ] = [ 4 − 8 ] c_1\begin{bmatrix} -1 \\ 2 \end{bmatrix} + c_2\begin{bmatrix} 3 \\ -6 \end{bmatrix} = \begin{bmatrix} 4 \\ -8 \end{bmatrix} c 1 [ − 1 2 ] + c 2 [ 3 − 6 ] = [ 4 − 8 ]
{ − c 1 + 3 c 2 = 4 2 c 1 − 6 c 2 = − 8 \begin{cases} -c_1 + 3c_2 = 4 \\ 2c_1 - 6c_2 = -8 \end{cases} { − c 1 + 3 c 2 = 4 2 c 1 − 6 c 2 = − 8
The second equation is − 2 -2 − 2 − 2 ( 4 ) = − 8 -2(4) = -8 − 2 ( 4 ) = − 8 Consistent.
Choosing c 2 = 0 c_2 = 0 c 2 = 0 c 1 = − 4 c_1 = -4 c 1 = − 4 b ⃗ 2 = − 4 a ⃗ 1 + 0 a ⃗ 2 \vec{b}_2 = -4\vec{a}_1 + 0\vec{a}_2 b 2 = − 4 a 1 + 0 a 2
Answer (b): b ⃗ 2 ∈ c o l ( A ) \vec{b}_2 \in col(A) b 2 ∈ co l ( A ) b ⃗ 2 = − 4 a ⃗ 1 + 0 a ⃗ 2 \vec{b}_2 = -4\vec{a}_1 + 0\vec{a}_2 b 2 = − 4 a 1 + 0 a 2
Final Answer
(a) b ⃗ 1 ∉ c o l ( A ) \vec{b}_1 \notin col(A) b 1 ∈ / co l ( A )
(b) b ⃗ 2 ∈ c o l ( A ) \vec{b}_2 \in col(A) b 2 ∈ co l ( A ) b ⃗ 2 = − 4 a ⃗ 1 + 0 a ⃗ 2 \vec{b}_2 = -4\vec{a}_1 + 0\vec{a}_2 b 2 = − 4 a 1 + 0 a 2
Q4. Subspaces of a Matrix
Let A = [ 1 0 1 1 2 0 0 1 1 0 − 1 0 ] A = \begin{bmatrix} 1 & 0 & 1 & 1 \\ 2 & 0 & 0 & 1 \\ 1 & 0 & -1 & 0 \end{bmatrix} A = 1 2 1 0 0 0 1 0 − 1 1 1 0
(a) Write N ( A ) N(A) N ( A )
(b) Write c o l ( A ) col(A) co l ( A )
(c) Write r o w ( A ) row(A) ro w ( A )
Answer of This Question First, row reduce A = [ 1 0 1 1 2 0 0 1 1 0 − 1 0 ] A = \begin{bmatrix} 1 & 0 & 1 & 1 \\ 2 & 0 & 0 & 1 \\ 1 & 0 & -1 & 0 \end{bmatrix} A = 1 2 1 0 0 0 1 0 − 1 1 1 0
[ 1 0 1 1 2 0 0 1 1 0 − 1 0 ] → R 2 → R 2 − 2 R 1 , R 3 → R 3 − R 1 [ 1 0 1 1 0 0 − 2 − 1 0 0 − 2 − 1 ] \left[\begin{array}{cccc} 1 & 0 & 1 & 1 \\ 2 & 0 & 0 & 1 \\ 1 & 0 & -1 & 0 \end{array}\right] \xrightarrow{R_2 \to R_2-2R_1,\; R_3 \to R_3-R_1} \left[\begin{array}{cccc} 1 & 0 & 1 & 1 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & -2 & -1 \end{array}\right] 1 2 1 0 0 0 1 0 − 1 1 1 0 R 2 → R 2 − 2 R 1 , R 3 → R 3 − R 1 1 0 0 0 0 0 1 − 2 − 2 1 − 1 − 1 → R 3 → R 3 − R 2 [ 1 0 1 1 0 0 − 2 − 1 0 0 0 0 ] → R 2 → − 1 2 R 2 [ 1 0 1 1 0 0 1 1 2 0 0 0 0 ] \xrightarrow{R_3 \to R_3-R_2} \left[\begin{array}{cccc} 1 & 0 & 1 & 1 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{R_2 \to -\frac{1}{2}R_2} \left[\begin{array}{cccc} 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 & 0 \end{array}\right] R 3 → R 3 − R 2 1 0 0 0 0 0 1 − 2 0 1 − 1 0 R 2 → − 2 1 R 2 1 0 0 0 0 0 1 1 0 1 2 1 0 → R 1 → R 1 − R 2 [ 1 0 0 1 2 0 0 1 1 2 0 0 0 0 ] \xrightarrow{R_1 \to R_1-R_2} \left[\begin{array}{cccc} 1 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 & 0 \end{array}\right] R 1 → R 1 − R 2 1 0 0 0 0 0 0 1 0 2 1 2 1 0 This is the RREF. Pivot columns: 1, 3. Free variables: x 2 , x 4 x_2, x_4 x 2 , x 4
(a) Null space N ( A ) N(A) N ( A )
From RREF: x 1 + 1 2 x 4 = 0 x_1 + \frac{1}{2}x_4 = 0 x 1 + 2 1 x 4 = 0 x 3 + 1 2 x 4 = 0 x_3 + \frac{1}{2}x_4 = 0 x 3 + 2 1 x 4 = 0
{ x 1 = − 1 2 x 4 x 2 = x 2 (free) x 3 = − 1 2 x 4 x 4 = x 4 (free) \begin{cases} x_1 = -\frac{1}{2}x_4 \\ x_2 = x_2 \text{ (free)} \\ x_3 = -\frac{1}{2}x_4 \\ x_4 = x_4 \text{ (free)} \end{cases} ⎩ ⎨ ⎧ x 1 = − 2 1 x 4 x 2 = x 2 (free) x 3 = − 2 1 x 4 x 4 = x 4 (free) In vector form:
x ⃗ = x 2 [ 0 1 0 0 ] + x 4 [ − 1 2 0 − 1 2 1 ] \vec{x} = x_2\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4\begin{bmatrix} -\frac{1}{2} \\ 0 \\ -\frac{1}{2} \\ 1 \end{bmatrix} x = x 2 0 1 0 0 + x 4 − 2 1 0 − 2 1 1 Answer (a): N ( A ) = S p { [ 0 1 0 0 ] , [ − 1 2 0 − 1 2 1 ] } N(A) = Sp\left\{ \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -\frac{1}{2} \\ 0 \\ -\frac{1}{2} \\ 1 \end{bmatrix} \right\} N ( A ) = Sp ⎩ ⎨ ⎧ 0 1 0 0 , − 2 1 0 − 2 1 1 ⎭ ⎬ ⎫
(b) Column space c o l ( A ) col(A) co l ( A )
Pivot columns are 1 and 3, so c o l ( A ) col(A) co l ( A ) A A A
Answer (b): c o l ( A ) = S p { [ 1 2 1 ] , [ 1 0 − 1 ] } col(A) = Sp\left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \right\} co l ( A ) = Sp ⎩ ⎨ ⎧ 1 2 1 , 1 0 − 1 ⎭ ⎬ ⎫
(c) Row space r o w ( A ) row(A) ro w ( A )
The nonzero rows of the RREF form a basis:
Answer (c): r o w ( A ) = S p { [ 1 0 0 1 2 ] , [ 0 0 1 1 2 ] } row(A) = Sp\left\{ \begin{bmatrix} 1 & 0 & 0 & \frac{1}{2} \end{bmatrix}, \begin{bmatrix} 0 & 0 & 1 & \frac{1}{2} \end{bmatrix} \right\} ro w ( A ) = Sp { [ 1 0 0 2 1 ] , [ 0 0 1 2 1 ] }
Q5. Basis and Dimension for Column, Row, and Null Spaces
Let A = [ 1 2 2 − 5 6 − 1 − 2 − 1 1 − 1 4 8 5 − 8 9 3 6 1 5 − 6 ] A = \begin{bmatrix} 1 & 2 & 2 & -5 & 6 \\ -1 & -2 & -1 & 1 & -1 \\ 4 & 8 & 5 & -8 & 9 \\ 3 & 6 & 1 & 5 & -6 \end{bmatrix} A = 1 − 1 4 3 2 − 2 8 6 2 − 1 5 1 − 5 1 − 8 5 6 − 1 9 − 6 A A A
Answer of This Question A = [ 1 2 2 − 5 6 − 1 − 2 − 1 1 − 1 4 8 5 − 8 9 3 6 1 5 − 6 ] A = \begin{bmatrix} 1 & 2 & 2 & -5 & 6 \\ -1 & -2 & -1 & 1 & -1 \\ 4 & 8 & 5 & -8 & 9 \\ 3 & 6 & 1 & 5 & -6 \end{bmatrix} A = 1 − 1 4 3 2 − 2 8 6 2 − 1 5 1 − 5 1 − 8 5 6 − 1 9 − 6 Row reduction:
R 2 → R 2 + R 1 R_2 \to R_2 + R_1 R 2 → R 2 + R 1 [ 0 , 0 , 1 , − 4 , 5 ] [0, 0, 1, -4, 5] [ 0 , 0 , 1 , − 4 , 5 ] R 3 → R 3 − 4 R 1 R_3 \to R_3 - 4R_1 R 3 → R 3 − 4 R 1 [ 0 , 0 , − 3 , 12 , − 15 ] [0, 0, -3, 12, -15] [ 0 , 0 , − 3 , 12 , − 15 ] R 4 → R 4 − 3 R 1 R_4 \to R_4 - 3R_1 R 4 → R 4 − 3 R 1 [ 0 , 0 , − 5 , 20 , − 24 ] [0, 0, -5, 20, -24] [ 0 , 0 , − 5 , 20 , − 24 ]
[ 1 2 2 − 5 6 0 0 1 − 4 5 0 0 − 3 12 − 15 0 0 − 5 20 − 24 ] \left[\begin{array}{ccccc} 1 & 2 & 2 & -5 & 6 \\ 0 & 0 & 1 & -4 & 5 \\ 0 & 0 & -3 & 12 & -15 \\ 0 & 0 & -5 & 20 & -24 \end{array}\right] 1 0 0 0 2 0 0 0 2 1 − 3 − 5 − 5 − 4 12 20 6 5 − 15 − 24 R 3 → R 3 + 3 R 2 R_3 \to R_3 + 3R_2 R 3 → R 3 + 3 R 2 [ 0 , 0 , 0 , 0 , 0 ] [0, 0, 0, 0, 0] [ 0 , 0 , 0 , 0 , 0 ] R 4 → R 4 + 5 R 2 R_4 \to R_4 + 5R_2 R 4 → R 4 + 5 R 2 [ 0 , 0 , 0 , 0 , 1 ] [0, 0, 0, 0, 1] [ 0 , 0 , 0 , 0 , 1 ]
Swap R 3 ↔ R 4 R_3 \leftrightarrow R_4 R 3 ↔ R 4
[ 1 2 2 − 5 6 0 0 1 − 4 5 0 0 0 0 1 0 0 0 0 0 ] \left[\begin{array}{ccccc} 1 & 2 & 2 & -5 & 6 \\ 0 & 0 & 1 & -4 & 5 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] 1 0 0 0 2 0 0 0 2 1 0 0 − 5 − 4 0 0 6 5 1 0 To RREF:
R 2 → R 2 − 5 R 3 R_2 \to R_2 - 5R_3 R 2 → R 2 − 5 R 3 [ 0 , 0 , 1 , − 4 , 0 ] [0, 0, 1, -4, 0] [ 0 , 0 , 1 , − 4 , 0 ] R 1 → R 1 − 6 R 3 R_1 \to R_1 - 6R_3 R 1 → R 1 − 6 R 3 [ 1 , 2 , 2 , − 5 , 0 ] [1, 2, 2, -5, 0] [ 1 , 2 , 2 , − 5 , 0 ] R 1 → R 1 − 2 R 2 R_1 \to R_1 - 2R_2 R 1 → R 1 − 2 R 2 [ 1 , 2 , 0 , 3 , 0 ] [1, 2, 0, 3, 0] [ 1 , 2 , 0 , 3 , 0 ]
RREF = [ 1 2 0 3 0 0 0 1 − 4 0 0 0 0 0 1 0 0 0 0 0 ] \text{RREF} = \left[\begin{array}{ccccc} 1 & 2 & 0 & 3 & 0 \\ 0 & 0 & 1 & -4 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] RREF = 1 0 0 0 2 0 0 0 0 1 0 0 3 − 4 0 0 0 0 1 0 Pivot columns: 1, 3, 5. Free variables: x 2 , x 4 x_2, x_4 x 2 , x 4
Column space: Columns 1, 3, 5 of the original A A A
basis = { [ 1 − 1 4 3 ] , [ 2 − 1 5 1 ] , [ 6 − 1 9 − 6 ] } , dim = 3 \text{basis} = \left\{ \begin{bmatrix} 1 \\ -1 \\ 4 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ -1 \\ 5 \\ 1 \end{bmatrix}, \begin{bmatrix} 6 \\ -1 \\ 9 \\ -6 \end{bmatrix} \right\}, \quad \dim = 3 basis = ⎩ ⎨ ⎧ 1 − 1 4 3 , 2 − 1 5 1 , 6 − 1 9 − 6 ⎭ ⎬ ⎫ , dim = 3 Row space: Nonzero rows of RREF (written as row vectors):
basis = { [ 1 2 0 3 0 ] , [ 0 0 1 − 4 0 ] , [ 0 0 0 0 1 ] } , dim = 3 \text{basis} = \left\{ \begin{bmatrix} 1 & 2 & 0 & 3 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 1 & -4 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \end{bmatrix} \right\}, \quad \dim = 3 basis = { [ 1 2 0 3 0 ] , [ 0 0 1 − 4 0 ] , [ 0 0 0 0 1 ] } , dim = 3 Null space:
{ x 1 + 2 x 2 + 3 x 4 = 0 x 3 − 4 x 4 = 0 x 5 = 0 ⟹ { x 1 = − 2 x 2 − 3 x 4 x 3 = 4 x 4 x 5 = 0 \begin{cases} x_1 + 2x_2 + 3x_4 = 0 \\ x_3 - 4x_4 = 0 \\ x_5 = 0 \end{cases} \implies \begin{cases} x_1 = -2x_2 - 3x_4 \\ x_3 = 4x_4 \\ x_5 = 0 \end{cases} ⎩ ⎨ ⎧ x 1 + 2 x 2 + 3 x 4 = 0 x 3 − 4 x 4 = 0 x 5 = 0 ⟹ ⎩ ⎨ ⎧ x 1 = − 2 x 2 − 3 x 4 x 3 = 4 x 4 x 5 = 0 x ⃗ = x 2 [ − 2 1 0 0 0 ] + x 4 [ − 3 0 4 1 0 ] \vec{x} = x_2\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4\begin{bmatrix} -3 \\ 0 \\ 4 \\ 1 \\ 0 \end{bmatrix} x = x 2 − 2 1 0 0 0 + x 4 − 3 0 4 1 0 basis = { [ − 2 1 0 0 0 ] , [ − 3 0 4 1 0 ] } , dim = 2 \text{basis} = \left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 4 \\ 1 \\ 0 \end{bmatrix} \right\}, \quad \dim = 2 basis = ⎩ ⎨ ⎧ − 2 1 0 0 0 , − 3 0 4 1 0 ⎭ ⎬ ⎫ , dim = 2 Final Answer
Column space basis: { [ 1 − 1 4 3 ] , [ 2 − 1 5 1 ] , [ 6 − 1 9 − 6 ] } \left\{ \begin{bmatrix} 1 \\ -1 \\ 4 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ -1 \\ 5 \\ 1 \end{bmatrix}, \begin{bmatrix} 6 \\ -1 \\ 9 \\ -6 \end{bmatrix} \right\} ⎩ ⎨ ⎧ 1 − 1 4 3 , 2 − 1 5 1 , 6 − 1 9 − 6 ⎭ ⎬ ⎫ dim = 3 \dim = 3 dim = 3
Row space basis: { [ 1 2 0 3 0 ] , [ 0 0 1 − 4 0 ] , [ 0 0 0 0 1 ] } \left\{ \begin{bmatrix} 1 & 2 & 0 & 3 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 1 & -4 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \end{bmatrix} \right\} { [ 1 2 0 3 0 ] , [ 0 0 1 − 4 0 ] , [ 0 0 0 0 1 ] } dim = 3 \dim = 3 dim = 3
Null space basis: { [ − 2 1 0 0 0 ] , [ − 3 0 4 1 0 ] } \left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 4 \\ 1 \\ 0 \end{bmatrix} \right\} ⎩ ⎨ ⎧ − 2 1 0 0 0 , − 3 0 4 1 0 ⎭ ⎬ ⎫ dim = 2 \dim = 2 dim = 2
Q6. Basis and Dimension of a Subspace
Let W = { x ⃗ = [ x 1 x 2 x 3 ] ∈ R 3 : x 1 − x 2 + x 3 = 0 } W = \left\{ \vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in \mathbb{R}^3 : x_1 - x_2 + x_3 = 0 \right\} W = ⎩ ⎨ ⎧ x = x 1 x 2 x 3 ∈ R 3 : x 1 − x 2 + x 3 = 0 ⎭ ⎬ ⎫
(a) Find a basis for W W W
(b) What is the dimension of W W W
Answer of This Question W = { x ⃗ ∈ R 3 : x 1 − x 2 + x 3 = 0 } W = \{ \vec{x} \in \mathbb{R}^3 : x_1 - x_2 + x_3 = 0 \} W = { x ∈ R 3 : x 1 − x 2 + x 3 = 0 }
We can write x 1 = x 2 − x 3 x_1 = x_2 - x_3 x 1 = x 2 − x 3 x 2 x_2 x 2 x 3 x_3 x 3
(a) Any vector in W W W
x ⃗ = [ x 2 − x 3 x 2 x 3 ] = x 2 [ 1 1 0 ] + x 3 [ − 1 0 1 ] \vec{x} = \begin{bmatrix} x_2-x_3 \\ x_2 \\ x_3 \end{bmatrix} = x_2\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} x = x 2 − x 3 x 2 x 3 = x 2 1 1 0 + x 3 − 1 0 1 These two vectors are linearly independent (one is not a scalar multiple of the other).
Answer (a): A basis for W W W { [ 1 1 0 ] , [ − 1 0 1 ] } \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \right\} ⎩ ⎨ ⎧ 1 1 0 , − 1 0 1 ⎭ ⎬ ⎫
(b) The dimension is the number of vectors in the basis.
Answer (b): dim ( W ) = 2 \dim(W) = 2 dim ( W ) = 2
Final Answer
(a) Basis: { [ 1 1 0 ] , [ − 1 0 1 ] } \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \right\} ⎩ ⎨ ⎧ 1 1 0 , − 1 0 1 ⎭ ⎬ ⎫
(b) dim ( W ) = 2 \dim(W) = 2 dim ( W ) = 2
Q7. Gram-Schmidt Process and Coordinates
Let W = S p { v ⃗ 1 , v ⃗ 2 , v ⃗ 3 } W = Sp\{\vec{v}_1, \vec{v}_2, \vec{v}_3\} W = Sp { v 1 , v 2 , v 3 } v ⃗ 1 = [ 2 2 1 ] \vec{v}_1 = \begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix} v 1 = 2 2 1 v ⃗ 2 = [ − 2 1 2 ] \vec{v}_2 = \begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix} v 2 = − 2 1 2 v ⃗ 3 = [ 18 0 0 ] \vec{v}_3 = \begin{bmatrix} 18 \\ 0 \\ 0 \end{bmatrix} v 3 = 18 0 0
(a) Use Gram-Schmidt to construct an orthonormal basis { u ⃗ 1 , u ⃗ 2 , u ⃗ 3 } \{\vec{u}_1, \vec{u}_2, \vec{u}_3\} { u 1 , u 2 , u 3 } W W W
(b) Find the coordinates of x ⃗ = [ 1 1 1 ] \vec{x} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} x = 1 1 1 u ⃗ 1 , u ⃗ 2 , u ⃗ 3 \vec{u}_1, \vec{u}_2, \vec{u}_3 u 1 , u 2 , u 3 a 1 , a 2 , a 3 a_1, a_2, a_3 a 1 , a 2 , a 3 x ⃗ = a 1 u ⃗ 1 + a 2 u ⃗ 2 + a 3 u ⃗ 3 \vec{x} = a_1\vec{u}_1 + a_2\vec{u}_2 + a_3\vec{u}_3 x = a 1 u 1 + a 2 u 2 + a 3 u 3
Answer of This Question v ⃗ 1 = [ 2 2 1 ] , v ⃗ 2 = [ − 2 1 2 ] , v ⃗ 3 = [ 18 0 0 ] \vec{v}_1 = \begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}, \;\vec{v}_2 = \begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix}, \;\vec{v}_3 = \begin{bmatrix} 18 \\ 0 \\ 0 \end{bmatrix} v 1 = 2 2 1 , v 2 = − 2 1 2 , v 3 = 18 0 0
(a) Gram-Schmidt orthogonalization:
Step 1: w ⃗ 1 = v ⃗ 1 = [ 2 2 1 ] , ∥ w ⃗ 1 ∥ = 4 + 4 + 1 = 3 \vec{w}_1 = \vec{v}_1 = \begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}, \quad \|\vec{w}_1\| = \sqrt{4+4+1} = 3 w 1 = v 1 = 2 2 1 , ∥ w 1 ∥ = 4 + 4 + 1 = 3
Step 2:
w ⃗ 2 = v ⃗ 2 − v ⃗ 2 ⋅ w ⃗ 1 ∥ w ⃗ 1 ∥ 2 w ⃗ 1 \vec{w}_2 = \vec{v}_2 - \frac{\vec{v}_2 \cdot \vec{w}_1}{\|\vec{w}_1\|^2}\vec{w}_1 w 2 = v 2 − ∥ w 1 ∥ 2 v 2 ⋅ w 1 w 1
v ⃗ 2 ⋅ w ⃗ 1 = ( − 2 ) ( 2 ) + ( 1 ) ( 2 ) + ( 2 ) ( 1 ) = − 4 + 2 + 2 = 0 \vec{v}_2 \cdot \vec{w}_1 = (-2)(2) + (1)(2) + (2)(1) = -4 + 2 + 2 = 0 v 2 ⋅ w 1 = ( − 2 ) ( 2 ) + ( 1 ) ( 2 ) + ( 2 ) ( 1 ) = − 4 + 2 + 2 = 0
w ⃗ 2 = v ⃗ 2 = [ − 2 1 2 ] , ∥ w ⃗ 2 ∥ = 4 + 1 + 4 = 3 \vec{w}_2 = \vec{v}_2 = \begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix}, \quad \|\vec{w}_2\| = \sqrt{4+1+4} = 3 w 2 = v 2 = − 2 1 2 , ∥ w 2 ∥ = 4 + 1 + 4 = 3 Step 3:
w ⃗ 3 = v ⃗ 3 − v ⃗ 3 ⋅ w ⃗ 1 ∥ w ⃗ 1 ∥ 2 w ⃗ 1 − v ⃗ 3 ⋅ w ⃗ 2 ∥ w ⃗ 2 ∥ 2 w ⃗ 2 \vec{w}_3 = \vec{v}_3 - \frac{\vec{v}_3 \cdot \vec{w}_1}{\|\vec{w}_1\|^2}\vec{w}_1 - \frac{\vec{v}_3 \cdot \vec{w}_2}{\|\vec{w}_2\|^2}\vec{w}_2 w 3 = v 3 − ∥ w 1 ∥ 2 v 3 ⋅ w 1 w 1 − ∥ w 2 ∥ 2 v 3 ⋅ w 2 w 2
v ⃗ 3 ⋅ w ⃗ 1 = 18 ( 2 ) = 36 \vec{v}_3 \cdot \vec{w}_1 = 18(2) = 36 v 3 ⋅ w 1 = 18 ( 2 ) = 36 v ⃗ 3 ⋅ w ⃗ 2 = 18 ( − 2 ) = − 36 \vec{v}_3 \cdot \vec{w}_2 = 18(-2) = -36 v 3 ⋅ w 2 = 18 ( − 2 ) = − 36
w ⃗ 3 = [ 18 0 0 ] − 36 9 [ 2 2 1 ] − − 36 9 [ − 2 1 2 ] = [ 18 0 0 ] − 4 [ 2 2 1 ] + 4 [ − 2 1 2 ] \vec{w}_3 = \begin{bmatrix} 18 \\ 0 \\ 0 \end{bmatrix} - \frac{36}{9}\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix} - \frac{-36}{9}\begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 18 \\ 0 \\ 0 \end{bmatrix} - 4\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix} + 4\begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix} w 3 = 18 0 0 − 9 36 2 2 1 − 9 − 36 − 2 1 2 = 18 0 0 − 4 2 2 1 + 4 − 2 1 2
= [ 18 − 8 − 8 0 − 8 + 4 0 − 4 + 8 ] = [ 2 − 4 4 ] , ∥ w ⃗ 3 ∥ = 4 + 16 + 16 = 6 = \begin{bmatrix} 18-8-8 \\ 0-8+4 \\ 0-4+8 \end{bmatrix} = \begin{bmatrix} 2 \\ -4 \\ 4 \end{bmatrix}, \quad \|\vec{w}_3\| = \sqrt{4+16+16} = 6 = 18 − 8 − 8 0 − 8 + 4 0 − 4 + 8 = 2 − 4 4 , ∥ w 3 ∥ = 4 + 16 + 16 = 6 Normalize to get orthonormal basis:
u ⃗ 1 = 1 3 [ 2 2 1 ] , u ⃗ 2 = 1 3 [ − 2 1 2 ] , u ⃗ 3 = 1 6 [ 2 − 4 4 ] \vec{u}_1 = \frac{1}{3}\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}, \quad \vec{u}_2 = \frac{1}{3}\begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix}, \quad \vec{u}_3 = \frac{1}{6}\begin{bmatrix} 2 \\ -4 \\ 4 \end{bmatrix} u 1 = 3 1 2 2 1 , u 2 = 3 1 − 2 1 2 , u 3 = 6 1 2 − 4 4 (b) Coordinates of x ⃗ = [ 1 1 1 ] \vec{x} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} x = 1 1 1
Since { u ⃗ 1 , u ⃗ 2 , u ⃗ 3 } \{\vec{u}_1, \vec{u}_2, \vec{u}_3\} { u 1 , u 2 , u 3 }
a 1 = x ⃗ ⋅ u ⃗ 1 = 1 3 ( 2 + 2 + 1 ) = 5 3 a_1 = \vec{x} \cdot \vec{u}_1 = \frac{1}{3}(2+2+1) = \frac{5}{3} a 1 = x ⋅ u 1 = 3 1 ( 2 + 2 + 1 ) = 3 5
a 2 = x ⃗ ⋅ u ⃗ 2 = 1 3 ( − 2 + 1 + 2 ) = 1 3 a_2 = \vec{x} \cdot \vec{u}_2 = \frac{1}{3}(-2+1+2) = \frac{1}{3} a 2 = x ⋅ u 2 = 3 1 ( − 2 + 1 + 2 ) = 3 1
a 3 = x ⃗ ⋅ u ⃗ 3 = 1 6 ( 2 − 4 + 4 ) = 1 3 a_3 = \vec{x} \cdot \vec{u}_3 = \frac{1}{6}(2-4+4) = \frac{1}{3} a 3 = x ⋅ u 3 = 6 1 ( 2 − 4 + 4 ) = 3 1 Verification: 5 3 u ⃗ 1 + 1 3 u ⃗ 2 + 1 3 u ⃗ 3 = 5 9 [ 2 2 1 ] + 1 9 [ − 2 1 2 ] + 1 18 [ 2 − 4 4 ] = 1 18 [ 18 18 18 ] = [ 1 1 1 ] \frac{5}{3}\vec{u}_1 + \frac{1}{3}\vec{u}_2 + \frac{1}{3}\vec{u}_3 = \frac{5}{9}\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix} + \frac{1}{9}\begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix} + \frac{1}{18}\begin{bmatrix} 2 \\ -4 \\ 4 \end{bmatrix} = \frac{1}{18}\begin{bmatrix} 18 \\ 18 \\ 18 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} 3 5 u 1 + 3 1 u 2 + 3 1 u 3 = 9 5 2 2 1 + 9 1 − 2 1 2 + 18 1 2 − 4 4 = 18 1 18 18 18 = 1 1 1 ✓ \checkmark ✓
Final Answer
(a) u ⃗ 1 = 1 3 [ 2 2 1 ] , u ⃗ 2 = 1 3 [ − 2 1 2 ] , u ⃗ 3 = 1 6 [ 2 − 4 4 ] \vec{u}_1 = \frac{1}{3}\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}, \;\vec{u}_2 = \frac{1}{3}\begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix}, \;\vec{u}_3 = \frac{1}{6}\begin{bmatrix} 2 \\ -4 \\ 4 \end{bmatrix} u 1 = 3 1 2 2 1 , u 2 = 3 1 − 2 1 2 , u 3 = 6 1 2 − 4 4
(b) a 1 = 5 3 , a 2 = 1 3 , a 3 = 1 3 a_1 = \frac{5}{3}, \;a_2 = \frac{1}{3}, \;a_3 = \frac{1}{3} a 1 = 3 5 , a 2 = 3 1 , a 3 = 3 1
Q8. Eigenvalues, Multiplicities, and Defectiveness
For the following matrices, find the eigenvalues and eigenvectors of A A A A A A
(a) A = [ 3 0 0 3 ] A = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} A = [ 3 0 0 3 ]
(b) A = [ 3 1 0 3 ] A = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} A = [ 3 0 1 3 ]
Answer of This Question (a) A = [ 3 0 0 3 ] A = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} A = [ 3 0 0 3 ]
This is 3 I 3I 3 I det ( A − λ I ) = ( 3 − λ ) 2 = 0 \det(A - \lambda I) = (3-\lambda)^2 = 0 det ( A − λ I ) = ( 3 − λ ) 2 = 0
Eigenvalue: λ = 3 \lambda = 3 λ = 3 Algebraic multiplicity: 2
Eigenvectors: ( A − 3 I ) x ⃗ = 0 ⃗ ⟹ [ 0 0 0 0 ] x ⃗ = 0 ⃗ (A - 3I)\vec{x} = \vec{0} \implies \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\vec{x} = \vec{0} ( A − 3 I ) x = 0 ⟹ [ 0 0 0 0 ] x = 0 R 2 \mathbb{R}^2 R 2
Basis for eigenspace: { [ 1 0 ] , [ 0 1 ] } \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\} { [ 1 0 ] , [ 0 1 ] }
Geometric multiplicity: 2Defective? No (algebraic = geometric for all eigenvalues)
(b) A = [ 3 1 0 3 ] A = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} A = [ 3 0 1 3 ]
Characteristic equation: det ( A − λ I ) = ( 3 − λ ) 2 = 0 \det(A - \lambda I) = (3-\lambda)^2 = 0 det ( A − λ I ) = ( 3 − λ ) 2 = 0
Eigenvalue: λ = 3 \lambda = 3 λ = 3 Algebraic multiplicity: 2
Eigenvectors: ( A − 3 I ) x ⃗ = 0 ⃗ ⟹ [ 0 1 0 0 ] [ x 1 x 2 ] = [ 0 0 ] ⟹ x 2 = 0 , x 1 (A - 3I)\vec{x} = \vec{0} \implies \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies x_2 = 0, x_1 ( A − 3 I ) x = 0 ⟹ [ 0 0 1 0 ] [ x 1 x 2 ] = [ 0 0 ] ⟹ x 2 = 0 , x 1
Basis for eigenspace: { [ 1 0 ] } \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\} { [ 1 0 ] }
Geometric multiplicity: 1Defective? Yes (algebraic multiplicity 2 > geometric multiplicity 1)
Final Answer
(a) λ = 3 \lambda = 3 λ = 3
(b) λ = 3 \lambda = 3 λ = 3
Q9. Eigenvalues and Eigenspaces
A = [ 1 − 1 1 1 ] A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} A = [ 1 1 − 1 1 ] A A A
Answer of This Question A = [ 1 − 1 1 1 ] A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} A = [ 1 1 − 1 1 ] Characteristic equation:
det ( A − λ I ) = det [ 1 − λ − 1 1 1 − λ ] = ( 1 − λ ) 2 + 1 = 0 \det(A - \lambda I) = \det\begin{bmatrix} 1-\lambda & -1 \\ 1 & 1-\lambda \end{bmatrix} = (1-\lambda)^2 + 1 = 0 det ( A − λ I ) = det [ 1 − λ 1 − 1 1 − λ ] = ( 1 − λ ) 2 + 1 = 0
( 1 − λ ) 2 = − 1 ⟹ 1 − λ = ± i (1-\lambda)^2 = -1 \implies 1-\lambda = \pm i ( 1 − λ ) 2 = − 1 ⟹ 1 − λ = ± i Eigenvalues:
λ 1 = 1 + i , λ 2 = 1 − i \lambda_1 = 1 + i, \quad \lambda_2 = 1 - i λ 1 = 1 + i , λ 2 = 1 − i Eigenvectors for λ 1 = 1 + i \lambda_1 = 1 + i λ 1 = 1 + i
( A − ( 1 + i ) I ) x ⃗ = [ − i − 1 1 − i ] [ x 1 x 2 ] = [ 0 0 ] (A - (1+i)I)\vec{x} = \begin{bmatrix} -i & -1 \\ 1 & -i \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} ( A − ( 1 + i ) I ) x = [ − i 1 − 1 − i ] [ x 1 x 2 ] = [ 0 0 ] From the first row: − i x 1 − x 2 = 0 ⟹ x 2 = − i x 1 -ix_1 - x_2 = 0 \implies x_2 = -ix_1 − i x 1 − x 2 = 0 ⟹ x 2 = − i x 1
Let x 1 = 1 ⟹ x 2 = − i x_1 = 1 \implies x_2 = -i x 1 = 1 ⟹ x 2 = − i [ 1 − i ] \begin{bmatrix} 1 \\ -i \end{bmatrix} [ 1 − i ]
Eigenvectors for λ 2 = 1 − i \lambda_2 = 1 - i λ 2 = 1 − i
( A − ( 1 − i ) I ) x ⃗ = [ i − 1 1 i ] [ x 1 x 2 ] = [ 0 0 ] (A - (1-i)I)\vec{x} = \begin{bmatrix} i & -1 \\ 1 & i \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} ( A − ( 1 − i ) I ) x = [ i 1 − 1 i ] [ x 1 x 2 ] = [ 0 0 ] From the first row: i x 1 − x 2 = 0 ⟹ x 2 = i x 1 ix_1 - x_2 = 0 \implies x_2 = ix_1 i x 1 − x 2 = 0 ⟹ x 2 = i x 1
Let x 1 = 1 ⟹ x 2 = i x_1 = 1 \implies x_2 = i x 1 = 1 ⟹ x 2 = i [ 1 i ] \begin{bmatrix} 1 \\ i \end{bmatrix} [ 1 i ]
Final Answer
λ 1 = 1 + i \lambda_1 = 1+i λ 1 = 1 + i { [ 1 − i ] } \left\{ \begin{bmatrix} 1 \\ -i \end{bmatrix} \right\} { [ 1 − i ] } λ 2 = 1 − i \lambda_2 = 1-i λ 2 = 1 − i { [ 1 i ] } \left\{ \begin{bmatrix} 1 \\ i \end{bmatrix} \right\} { [ 1 i ] }
Q10. Matrix Diagonalization
Let A = [ 4 − 3 2 − 1 ] A = \begin{bmatrix} 4 & -3 \\ 2 & -1 \end{bmatrix} A = [ 4 2 − 3 − 1 ] A 6 A^6 A 6
Answer of This Question A = [ 4 − 3 2 − 1 ] A = \begin{bmatrix} 4 & -3 \\ 2 & -1 \end{bmatrix} A = [ 4 2 − 3 − 1 ] Step 1: Find eigenvalues
det ( A − λ I ) = det [ 4 − λ − 3 2 − 1 − λ ] = ( 4 − λ ) ( − 1 − λ ) + 6 = λ 2 − 3 λ + 2 = ( λ − 1 ) ( λ − 2 ) = 0 \det(A - \lambda I) = \det\begin{bmatrix} 4-\lambda & -3 \\ 2 & -1-\lambda \end{bmatrix} = (4-\lambda)(-1-\lambda) + 6 = \lambda^2 - 3\lambda + 2 = (\lambda-1)(\lambda-2) = 0 det ( A − λ I ) = det [ 4 − λ 2 − 3 − 1 − λ ] = ( 4 − λ ) ( − 1 − λ ) + 6 = λ 2 − 3 λ + 2 = ( λ − 1 ) ( λ − 2 ) = 0 Eigenvalues: λ 1 = 1 \lambda_1 = 1 λ 1 = 1 λ 2 = 2 \lambda_2 = 2 λ 2 = 2
Step 2: Find eigenvectors
For λ 1 = 1 \lambda_1 = 1 λ 1 = 1
( A − I ) x ⃗ = [ 3 − 3 2 − 2 ] [ x 1 x 2 ] = [ 0 0 ] ⟹ x 1 = x 2 (A - I)\vec{x} = \begin{bmatrix} 3 & -3 \\ 2 & -2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies x_1 = x_2 ( A − I ) x = [ 3 2 − 3 − 2 ] [ x 1 x 2 ] = [ 0 0 ] ⟹ x 1 = x 2
Eigenvector: p ⃗ 1 = [ 1 1 ] \vec{p}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} p 1 = [ 1 1 ]
For λ 2 = 2 \lambda_2 = 2 λ 2 = 2
( A − 2 I ) x ⃗ = [ 2 − 3 2 − 3 ] [ x 1 x 2 ] = [ 0 0 ] ⟹ 2 x 1 = 3 x 2 ⟹ x 1 = 3 2 x 2 (A - 2I)\vec{x} = \begin{bmatrix} 2 & -3 \\ 2 & -3 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies 2x_1 = 3x_2 \implies x_1 = \frac{3}{2}x_2 ( A − 2 I ) x = [ 2 2 − 3 − 3 ] [ x 1 x 2 ] = [ 0 0 ] ⟹ 2 x 1 = 3 x 2 ⟹ x 1 = 2 3 x 2
Eigenvector: p ⃗ 2 = [ 3 2 ] \vec{p}_2 = \begin{bmatrix} 3 \\ 2 \end{bmatrix} p 2 = [ 3 2 ]
Step 3: Diagonalization
P = [ 1 3 1 2 ] , D = [ 1 0 0 2 ] P = \begin{bmatrix} 1 & 3 \\ 1 & 2 \end{bmatrix}, \quad D = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} P = [ 1 1 3 2 ] , D = [ 1 0 0 2 ] Check: A = P D P − 1 A = PDP^{-1} A = P D P − 1 P − 1 = 1 2 − 3 [ 2 − 3 − 1 1 ] = [ − 2 3 1 − 1 ] P^{-1} = \frac{1}{2-3}\begin{bmatrix} 2 & -3 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix} P − 1 = 2 − 3 1 [ 2 − 1 − 3 1 ] = [ − 2 1 3 − 1 ]
Step 4: Compute A 6 A^6 A 6
A 6 = P D 6 P − 1 = [ 1 3 1 2 ] [ 1 0 0 64 ] [ − 2 3 1 − 1 ] A^6 = PD^6P^{-1} = \begin{bmatrix} 1 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 64 \end{bmatrix}\begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix} A 6 = P D 6 P − 1 = [ 1 1 3 2 ] [ 1 0 0 64 ] [ − 2 1 3 − 1 ] = [ 1 192 1 128 ] [ − 2 3 1 − 1 ] = [ − 2 + 192 3 − 192 − 2 + 128 3 − 128 ] = [ 190 − 189 126 − 125 ] = \begin{bmatrix} 1 & 192 \\ 1 & 128 \end{bmatrix}\begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} -2+192 & 3-192 \\ -2+128 & 3-128 \end{bmatrix} = \begin{bmatrix} 190 & -189 \\ 126 & -125 \end{bmatrix} = [ 1 1 192 128 ] [ − 2 1 3 − 1 ] = [ − 2 + 192 − 2 + 128 3 − 192 3 − 128 ] = [ 190 126 − 189 − 125 ] Final Answer
A = P D P − 1 A = PDP^{-1} A = P D P − 1 P = [ 1 3 1 2 ] P = \begin{bmatrix} 1 & 3 \\ 1 & 2 \end{bmatrix} P = [ 1 1 3 2 ] D = [ 1 0 0 2 ] D = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} D = [ 1 0 0 2 ] P − 1 = [ − 2 3 1 − 1 ] P^{-1} = \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix} P − 1 = [ − 2 1 3 − 1 ] A 6 = [ 190 − 189 126 − 125 ] A^6 = \begin{bmatrix} 190 & -189 \\ 126 & -125 \end{bmatrix} A 6 = [ 190 126 − 189 − 125 ]
Q11. Diagonalization of Symmetric Matrix
Let A = [ 2 0 − 1 0 3 0 − 1 0 2 ] A = \begin{bmatrix} 2 & 0 & -1 \\ 0 & 3 & 0 \\ -1 & 0 & 2 \end{bmatrix} A = 2 0 − 1 0 3 0 − 1 0 2 A A A A A A
Answer of This Question A = [ 2 0 − 1 0 3 0 − 1 0 2 ] A = \begin{bmatrix} 2 & 0 & -1 \\ 0 & 3 & 0 \\ -1 & 0 & 2 \end{bmatrix} A = 2 0 − 1 0 3 0 − 1 0 2 Since A A A
Step 1: Find eigenvalues
det ( A − λ I ) = det [ 2 − λ 0 − 1 0 3 − λ 0 − 1 0 2 − λ ] \det(A - \lambda I) = \det\begin{bmatrix} 2-\lambda & 0 & -1 \\ 0 & 3-\lambda & 0 \\ -1 & 0 & 2-\lambda \end{bmatrix} det ( A − λ I ) = det 2 − λ 0 − 1 0 3 − λ 0 − 1 0 2 − λ Expanding along the second row:
( 3 − λ ) ⋅ det [ 2 − λ − 1 − 1 2 − λ ] = ( 3 − λ ) ( ( 2 − λ ) 2 − 1 ) (3-\lambda) \cdot \det\begin{bmatrix} 2-\lambda & -1 \\ -1 & 2-\lambda \end{bmatrix} = (3-\lambda)\left((2-\lambda)^2 - 1\right) ( 3 − λ ) ⋅ det [ 2 − λ − 1 − 1 2 − λ ] = ( 3 − λ ) ( ( 2 − λ ) 2 − 1 )
= ( 3 − λ ) ( λ 2 − 4 λ + 3 ) = ( 3 − λ ) ( λ − 1 ) ( λ − 3 ) = − ( λ − 1 ) ( λ − 3 ) 2 = (3-\lambda)(\lambda^2 - 4\lambda + 3) = (3-\lambda)(\lambda-1)(\lambda-3) = -(\lambda-1)(\lambda-3)^2 = ( 3 − λ ) ( λ 2 − 4 λ + 3 ) = ( 3 − λ ) ( λ − 1 ) ( λ − 3 ) = − ( λ − 1 ) ( λ − 3 ) 2 Eigenvalues: λ 1 = 1 \lambda_1 = 1 λ 1 = 1 λ 2 = 3 \lambda_2 = 3 λ 2 = 3
Step 2: Find eigenvectors
For λ 1 = 1 \lambda_1 = 1 λ 1 = 1
( A − I ) x ⃗ = [ 1 0 − 1 0 2 0 − 1 0 1 ] [ x 1 x 2 x 3 ] = [ 0 0 0 ] (A - I)\vec{x} = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} ( A − I ) x = 1 0 − 1 0 2 0 − 1 0 1 x 1 x 2 x 3 = 0 0 0
x 1 − x 3 = 0 ⟹ x 1 = x 3 x_1 - x_3 = 0 \implies x_1 = x_3 x 1 − x 3 = 0 ⟹ x 1 = x 3 2 x 2 = 0 ⟹ x 2 = 0 2x_2 = 0 \implies x_2 = 0 2 x 2 = 0 ⟹ x 2 = 0 p ⃗ 1 = [ 1 0 1 ] \vec{p}_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} p 1 = 1 0 1
For λ 2 = 3 \lambda_2 = 3 λ 2 = 3
( A − 3 I ) x ⃗ = [ − 1 0 − 1 0 0 0 − 1 0 − 1 ] [ x 1 x 2 x 3 ] = [ 0 0 0 ] (A - 3I)\vec{x} = \begin{bmatrix} -1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & -1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} ( A − 3 I ) x = − 1 0 − 1 0 0 0 − 1 0 − 1 x 1 x 2 x 3 = 0 0 0
x 1 + x 3 = 0 ⟹ x 3 = − x 1 x_1 + x_3 = 0 \implies x_3 = -x_1 x 1 + x 3 = 0 ⟹ x 3 = − x 1 x 2 x_2 x 2 p ⃗ 2 = [ 0 1 0 ] \vec{p}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} p 2 = 0 1 0 p ⃗ 3 = [ 1 0 − 1 ] \vec{p}_3 = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} p 3 = 1 0 − 1
Step 3: Diagonalization
P = [ 1 0 1 0 1 0 1 0 − 1 ] , D = [ 1 0 0 0 3 0 0 0 3 ] P = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & -1 \end{bmatrix}, \quad D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} P = 1 0 1 0 1 0 1 0 − 1 , D = 1 0 0 0 3 0 0 0 3 Since A A A
u ⃗ 1 = 1 2 [ 1 0 1 ] , u ⃗ 2 = [ 0 1 0 ] , u ⃗ 3 = 1 2 [ 1 0 − 1 ] \vec{u}_1 = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \;\vec{u}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \;\vec{u}_3 = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} u 1 = 2 1 1 0 1 , u 2 = 0 1 0 , u 3 = 2 1 1 0 − 1 Q = [ 1 2 0 1 2 0 1 0 1 2 0 − 1 2 ] , D = [ 1 0 0 0 3 0 0 0 3 ] Q = \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \end{bmatrix}, \quad D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} Q = 2 1 0 2 1 0 1 0 2 1 0 − 2 1 , D = 1 0 0 0 3 0 0 0 3 And A = Q D Q T A = QDQ^T A = Q D Q T
Final Answer
Yes, A A A
Eigenvalues: λ = 1 \lambda = 1 λ = 1 λ = 3 \lambda = 3 λ = 3
A = P D P − 1 A = PDP^{-1} A = P D P − 1 P = [ 1 0 1 0 1 0 1 0 − 1 ] P = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & -1 \end{bmatrix} P = 1 0 1 0 1 0 1 0 − 1 D = [ 1 0 0 0 3 0 0 0 3 ] D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} D = 1 0 0 0 3 0 0 0 3 Orthogonal diagonalization: A = Q D Q T A = QDQ^T A = Q D Q T Q Q Q
James included in Electrical Engineering and MATH-2010