Answer to Question #256476 in Linear Algebra for Johnny

Question

1. Let S be the subspace of R⁵ defined by S = { (x₁, x₂, x₃, x₄, x₅) E R⁵ : x₁ = x_2, x₃ = 2x₄ + x₅ }. Then the dimension of S is
2. Let T: R³ -> 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) T₁ : R³ -> R² by T₁ (u, v, w) = ( u - v + 2w, 5v - w).

(ii) T₂ : P (R) -> R by T₂ (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

Answer 1

$\begin{pmatrix} x_2\\ x_2\\2x_4+x_5\\x_4\\x_5 \end{pmatrix}=x_2\begin{pmatrix} 1\\1 \\0\\0 \end{pmatrix}+x_4\begin{pmatrix} 0\\0 \\2\\1\\0 \end{pmatrix}+x_5\begin{pmatrix} 0\\0\\1\\0\\1 \end{pmatrix}$

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

rref of$S=\begin{pmatrix} 1&0&0 \\ 0&1&0\\0&0&1\\0&0&0\\0&0&0 \end{pmatrix}$

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

Dimension of S=3

PART 2

$\begin{pmatrix} x+y\\x-y\\ x+2z \end{pmatrix}=x\begin{pmatrix} 1\\ 1\\1\\ \end{pmatrix}+y\begin{pmatrix} 1\\-1\\0\\ \end{pmatrix}+z\begin{pmatrix} 0 \\0 \\2 \end{pmatrix}$

Matrix $T=\begin{pmatrix} 1&1&0\\ 1&-1&0\\1&0&2 \end{pmatrix}$

rref of $T=\begin{pmatrix} 1&0&0\\ 0&1&0\\0&0&1 \end{pmatrix}$

The vectors are linearly independent.

The basis of the range $T$ is

$[\begin{pmatrix} 1\\ 1\\1 \end{pmatrix}\>,\>\begin{pmatrix} 1 \\-1 \\0 \end{pmatrix}\>,\>\begin{pmatrix} 0 \\0\\ 2 \end{pmatrix}]$

3, Part1

Every linear transformation is associated with a matrix

$\begin{pmatrix} u-v+2w\\ 5v-w \end{pmatrix}=u\begin{pmatrix} 1 \\ 0 \end{pmatrix}+v\begin{pmatrix} -1\\ 5 \end{pmatrix}+w\begin{pmatrix} 2\\ -1 \end{pmatrix}$

$T_1=\begin{pmatrix} 1&-1&2\\ 0&5&-1 \end{pmatrix}$

$\implies$ T₁ is linear

3) Part2

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

$\int_a^b2p(x_1+x_2)\>dx$

$=\int_a^b2p(x_1)\>dx + \int_a^b2p(x_2)\>dx$

Therefore T₂ is linear

3). Part 3

$T_3P(u_1+u_2)=(u_1+u_2)P(u_1+u_2)+(u_1+u_2)$

$T_3P(u_1)=u_1P(u_1)+u_1$

$T_3P(u_2)=u_2P(u_2)+u_2$

$\therefore \>T_3P(u_1+u_2) \mathrlap{\,/}{=}T_3P(u_1)$

$+T_3P(u_2)$

Therefore T₃ is not linear