Answer to Question #256605 in Linear Algebra for Anuj

Question #256605
  1. Let S be the subspace of R5 defined by S = { (x1, x2, x3, x4, x5) E R5 : x1 = x2, x3 = 2x+ x}. Then the dimension of S is
  2. Let T: R-> R3 be defined as T (x,y,z) = (x+y, x-y, x+2z). Then the basis of range T is...
  3. Which of the following transformations are linear:

(i) T: R3 -> Rby T1 (u, v, w) = ( u - v + 2w, 5v - w).

(ii) T: P (R) -> R by T(P) = (integral sign from b to a) 2p(x)dx for a,b E R with a<= b

(iii) T3 : P(R) -> P(R) by T3 (P(U) = UP (U) + U


4.. Suppose T : R2 -> M22 is a linear defined by T (u,v) = [u v]

[u 2u]

then Ker (T) is...

5.. Suppose T: R6 -> R4 is a linear map such that null T = U where U is a 2-dimentional subspace of R6 . Then dim range T is...


6.. For a given 2x2 matrix A = [ 5 -3 ]

[ -6 2]

the matrix P that is diagonalizes A can be written as P = ...



1
Expert's answer
2021-10-27T15:05:11-0400

Answer:-

1.

The dimension of a nonzero subspace S, is the number of vectors in any basis for S.

Consider the vectors:

"v_1=(1,1,0,0,0),v_2=(0,0,3,1,1)"

vectors v1 and vare in S, and are linearly independent. So, the dimension of S is 2.


2.

"T(x,y,z)=x(1,1,1)+y(1,-1,0)+z(0,0,2)"

basis: "(1,1,1),(1,-1,0),(0,0,2)"

3.

i)

let "a(u_1,v_1,w_1),b(u_2,v_2,w_2)" , then:

"T_1(a+b)=(u_1+u_2-(v_1+v_2)+2(w_1+w_2),5(v_1+v_2)-(w_1+w_2))="

"=(u_1-v_1+2w_1,5v_1-w_1)+(u_2-v_2+2w_2,5v_2-w_2)=T_1(a)+T_1(b)"

"T_1(ca)=T_1(cu_1,cv_1,cw_1)=(cu_1-cv_1+2cw_1,5cv_1-cw_1)="

"=c(u_1-v_1+2w_1,5v_1-w_1)=cT_1(a)"


So, T1 is linear transformation.


ii)

"\\displaystyle{\\int^{b_1+b_2}_{a_1+a_2}}2p(x)dx\\neq \\displaystyle{\\int^{b_1}_{a_1}}2p(x)dx+\\displaystyle{\\int^{b_2}_{a_2}}2p(x)dx"


T2 is not linear transformation.


iii)

"(u_1+u_2)p(u_1+u_2)+(u_1+u_2)\\neq u_1p(u_1)+u_1+u_2p(u_2)+u_2"

T2 is not linear transformation.


4.

u+v=0

u+2u=0

Ker(T)=(0,0)


5.

by the rank-nullity theorem:

"6=dim(Ker(T))+dim(Im(T))"

So, dim range T= 6-dim(Ker(T))=6-2=4


6.

"\\begin{vmatrix}\n 5-\\lambda & -3 \\\\\n -6 & 2-\\lambda\n\\end{vmatrix}=0"


"10-7\\lambda+\\lambda^2-18=0"

"\\lambda_1=\\frac{7-\\sqrt{49+32}}{2}=-1"

"\\lambda_2=8"


for "\\lambda_1":

"6x-3y=0"

"y=2x"

"x_1=\\begin{pmatrix}\n 1 \\\\\n 2 \n\\end{pmatrix}"


for "\\lambda_2":

"-3x-3y=0"

"x=-y"

"x_2=\\begin{pmatrix}\n 1 \\\\\n -1 \n\\end{pmatrix}"

"P=\\begin{pmatrix}\n 1& 1 \\\\\n 2& -1\n\\end{pmatrix}"



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