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Sok
MATH2250S24PS5.p...
Raney
Linear Algebra
Spring 2024
Problem Set 5

Show that 
x=[[-2],[1],[1]] is an eigenvector of 
A=[[0,1,-1],[1,1,1],[1,2,0]] and find the corresponding eigenvalue.
Show that 
lambda=i=sqrt(-1) is an eigenvalue of 
A=[[0,-1],[1,0]], and find a corresponding eigenvector.
Let 
A=[[2,4],[6,0]]. Find the eigenvalues of 
A and bases for the corresponding eigenspaces.
Let 
A be an idempotent matrix (that is, 
A^(2)=A ). Show that 
lambda=0 and 
lambda=1 are the only possible eigenvalues of 
A.
Let 
T:P_(2)rarrP_(2) be the linear transformation defined by


T(p(x))=xp^(')(x)
(a) Which, if any, of the polynomials

p_(1)(x)=1,p_(2)(x)=x,p_(3)(x)=x^(2)
are in 
ker(T) ?
(b) Which, if any, of the polynomials

p_(1)(x)=1,p_(2)(x)=x,p_(3)(x)=x^(2)
are in 
ran(T) ?
6. Find the dimension of the vector space 
V={p(x)=a+bx+cx^(2)inP_(2):p(1)=0} and give a basis for 
V.
7. Define a linear transformation 
T:M_(2×2)rarrM_(2×2) by

T(A)=AB-BA
where 
B=[[1,1],[0,1]]. Find a basis for 
ker(T).
(D)
Dashboard
Calendar

[[0=6],[0=-1],[0]:}
To-do
Notifications

⊠
Inbox

Sok\newlineMATH22502250S2424PS55.p...\newlineRaney\newlineLinear Algebra\newlineSpring 20242024\newlineProblem Set 55\newline11. Show that x=[211] \mathbf{x}=\left[\begin{array}{c}-2 \\ 1 \\ 1\end{array}\right] is an eigenvector of A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] and find the corresponding eigenvalue.\newline22. Show that λ=i=1 \lambda=i=\sqrt{-1} is an eigenvalue of A=[0110] A=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] , and find a corresponding eigenvector.\newline33. Let A=[2460] A=\left[\begin{array}{ll}2 & 4 \\ 6 & 0\end{array}\right] . Find the eigenvalues of A A and bases for the corresponding eigenspaces.\newline44. Let A A be an idempotent matrix (that is, A2=A A^{2}=A ). Show that λ=0 \lambda=0 and λ=1 \lambda=1 are the only possible eigenvalues of A A .\newline55. Let A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 11 be the linear transformation defined by\newlineT(p(x))=xp(x) T(p(x))=x p^{\prime}(x) \newline(a) Which, if any, of the polynomials\newlinep1(x)=1,p2(x)=x,p3(x)=x2 p_{1}(x)=1, p_{2}(x)=x, p_{3}(x)=x^{2} \newlineare in A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 22 ?\newline(b) Which, if any, of the polynomials\newlinep1(x)=1,p2(x)=x,p3(x)=x2 p_{1}(x)=1, p_{2}(x)=x, p_{3}(x)=x^{2} \newlineare in A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 33 ?\newline66. Find the dimension of the vector space A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 44 and give a basis for A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 55.\newline77. Define a linear transformation A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 66 by\newlineT(A)=ABBA T(A)=A B-B A \newlinewhere A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 77. Find a basis for A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 22.\newline(D)\newlineDashboard\newlineCalendar\newlineA=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 99\newlineTo-do\newlineNotifications\newlineλ=i=1 \lambda=i=\sqrt{-1} 00\newlineInbox

Full solution

Q. Sok\newlineMATH22502250S2424PS55.p...\newlineRaney\newlineLinear Algebra\newlineSpring 20242024\newlineProblem Set 55\newline11. Show that x=[211] \mathbf{x}=\left[\begin{array}{c}-2 \\ 1 \\ 1\end{array}\right] is an eigenvector of A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] and find the corresponding eigenvalue.\newline22. Show that λ=i=1 \lambda=i=\sqrt{-1} is an eigenvalue of A=[0110] A=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] , and find a corresponding eigenvector.\newline33. Let A=[2460] A=\left[\begin{array}{ll}2 & 4 \\ 6 & 0\end{array}\right] . Find the eigenvalues of A A and bases for the corresponding eigenspaces.\newline44. Let A A be an idempotent matrix (that is, A2=A A^{2}=A ). Show that λ=0 \lambda=0 and λ=1 \lambda=1 are the only possible eigenvalues of A A .\newline55. Let A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 11 be the linear transformation defined by\newlineT(p(x))=xp(x) T(p(x))=x p^{\prime}(x) \newline(a) Which, if any, of the polynomials\newlinep1(x)=1,p2(x)=x,p3(x)=x2 p_{1}(x)=1, p_{2}(x)=x, p_{3}(x)=x^{2} \newlineare in A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 22 ?\newline(b) Which, if any, of the polynomials\newlinep1(x)=1,p2(x)=x,p3(x)=x2 p_{1}(x)=1, p_{2}(x)=x, p_{3}(x)=x^{2} \newlineare in A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 33 ?\newline66. Find the dimension of the vector space A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 44 and give a basis for A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 55.\newline77. Define a linear transformation A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 66 by\newlineT(A)=ABBA T(A)=A B-B A \newlinewhere A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 77. Find a basis for A=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 22.\newline(D)\newlineDashboard\newlineCalendar\newlineA=[011111120] A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0\end{array}\right] 99\newlineTo-do\newlineNotifications\newlineλ=i=1 \lambda=i=\sqrt{-1} 00\newlineInbox
  1. Check Eigenvector: To show that xx is an eigenvector of AA, we need to check if Ax=λxAx = \lambda x for some scalar λ\lambda.
    Calculation: Ax=[011 111 120][2 1 1]A\cdot x = \begin{bmatrix}0 & 1 & -1\ 1 & 1 & 1\ 1 & 2 & 0\end{bmatrix} \cdot \begin{bmatrix}-2\ 1\ 1\end{bmatrix}
    =[(02)+(11)+(11) (12)+(11)+(11) (12)+(21)+(01)]= \begin{bmatrix}(0\cdot -2)+(1\cdot 1)+(-1\cdot 1)\ (1\cdot -2)+(1\cdot 1)+(1\cdot 1)\ (1\cdot -2)+(2\cdot 1)+(0\cdot 1)\end{bmatrix}
    =[0+11 2+1+1 2+2+0]= \begin{bmatrix}0+1-1\ -2+1+1\ -2+2+0\end{bmatrix}
    =[0 0 0]= \begin{bmatrix}0\ 0\ 0\end{bmatrix}
    Since Ax=0Ax = 0, which is not a scalar multiple of xx, xx is not an eigenvector.

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