Answer to Question #256058 in Linear Algebra for Three

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

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 = ...

Answer 1

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 v₂ are 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.

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

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

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

$\lambda_2=8$

for $\lambda_1$:

$6x-3y=0$

$y=2x$

$x_1=\begin{pmatrix} 1 \\ 2 \end{pmatrix}$

for $\lambda_2$:

$-3x-3y=0$

$x=-y$

$x_2=\begin{pmatrix} 1 \\ -1 \end{pmatrix}$

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