Question #256476
  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



1
Expert's answer
2021-10-29T02:52:51-0400

(x2x22x4+x5x4x5)=x2(1100)+x4(00210)+x5(00101)\begin{pmatrix} x_2\\ x_2\\2x_4+x_5\\x_4\\x_5 \end{pmatrix}=x_2\begin{pmatrix} 1\\1 \\0\\0 \end{pmatrix}+x_4\begin{pmatrix} 0\\0 \\2\\1\\0 \end{pmatrix}+x_5\begin{pmatrix} 0\\0\\1\\0\\1 \end{pmatrix}


Matrix S=(100100021010001)S=\begin{pmatrix} 1& 0&0\\ 1&0&0\\0&2&1\\0&1&0\\0&0&1 \end{pmatrix}


rref ofS=(100010001000000)S=\begin{pmatrix} 1&0&0 \\ 0&1&0\\0&0&1\\0&0&0\\0&0&0 \end{pmatrix}


The three vectors are linearly independent and they are the basis of S

Dimension of S=3


PART 2


(x+yxyx+2z)=x(111)+y(110)+z(002)\begin{pmatrix} x+y\\x-y\\ x+2z \end{pmatrix}=x\begin{pmatrix} 1\\ 1\\1\\ \end{pmatrix}+y\begin{pmatrix} 1\\-1\\0\\ \end{pmatrix}+z\begin{pmatrix} 0 \\0 \\2 \end{pmatrix}



Matrix T=(110110102)T=\begin{pmatrix} 1&1&0\\ 1&-1&0\\1&0&2 \end{pmatrix}



rref of T=(100010001)T=\begin{pmatrix} 1&0&0\\ 0&1&0\\0&0&1 \end{pmatrix}


The vectors are linearly independent.

The basis of the range TT is


[(111),(110),(002)][\begin{pmatrix} 1\\ 1\\1 \end{pmatrix}\>,\>\begin{pmatrix} 1 \\-1 \\0 \end{pmatrix}\>,\>\begin{pmatrix} 0 \\0\\ 2 \end{pmatrix}]


3, Part1


Every linear transformation is associated with a matrix


(uv+2w5vw)=u(10)+v(15)+w(21)\begin{pmatrix} u-v+2w\\ 5v-w \end{pmatrix}=u\begin{pmatrix} 1 \\ 0 \end{pmatrix}+v\begin{pmatrix} -1\\ 5 \end{pmatrix}+w\begin{pmatrix} 2\\ -1 \end{pmatrix}



T1=(112051)T_1=\begin{pmatrix} 1&-1&2\\ 0&5&-1 \end{pmatrix}


    \implies T1 is linear


3) Part2


abk2p(x)dx=kab2p(x)dx\int_a^bk2p(x)\>dx=k\int_a^b2p(x)\>dx


ab2p(x1+x2)dx\int_a^b2p(x_1+x_2)\>dx


=ab2p(x1)dx+ab2p(x2)dx=\int_a^b2p(x_1)\>dx + \int_a^b2p(x_2)\>dx


Therefore T2 is linear


3). Part 3



T3P(u1+u2)=(u1+u2)P(u1+u2)+(u1+u2)T_3P(u_1+u_2)=(u_1+u_2)P(u_1+u_2)+(u_1+u_2)


T3P(u1)=u1P(u1)+u1T_3P(u_1)=u_1P(u_1)+u_1


T3P(u2)=u2P(u2)+u2T_3P(u_2)=u_2P(u_2)+u_2


T3P(u1+u2)/=T3P(u1)\therefore \>T_3P(u_1+u_2) \mathrlap{\,/}{=}T_3P(u_1)


+T3P(u2)+T_3P(u_2)


Therefore T3 is not linear



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