Question #285334

15. Suppose T : R^2\to R^3 is linear defined by T(x, y) =(x + 3y, x - y, x). Then

(i) 1

(ii) 2

(iii) 3

(iv) None

16. Suppose T : R^3\to R^3 is linear and has an upper-triangular matrix with respect to the basis (1,0,0),(1,1,1),(1,1,2). Then, the orthonormal basis of R^3 with respect to which T has an upper-triangular matrix is...

(i) (1, 0, 0), (0, 1/(√2), 1/(√2)), (0, - 1/(√2), 1/(√2))

(ii) (1, 0, 0), (0, 1, 0), (0, 1/(√2), - 1/(√2)

(iii) (1, 0, 0), (0, - 1, 1), (0, 1, 1)

(iv) None


17. Which of the following defines an inner product

(i) <(x base 1, x base 2), y base 1, y base 2)>2x base 1 y base 1 +x base 2 y base 2 in R^2

(ii) <(x base 1, x base 2), y base 1, y base 2)>x base 1 y base 1 +2x base 2 y base 2 - 1 in R^2

(iii) <a base 1 + b base 1 x +c base 1 x^2, a base 2 +b base 2 x + c base 2 x^2 > = a base 1 b base 1 +a base 2 b base 2 +c base 1 c base 2 in P base 2




1
Expert's answer
2022-01-09T14:37:07-0500

15)


T(x,y)=(x+3y,xy,x)T(x,y)=(x+3y,x-y,x)


T(x,y)=(x+3yxyx)T(x,y)=\begin{pmatrix} x+3y \\ x-y\\ x \end{pmatrix}



=x(111)+y(310)=x\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}+y\begin{pmatrix} 3 \\ -1\\ 0 \end{pmatrix}



T=(131110)T=\begin{pmatrix} 1 & 3\\ 1 & -1\\ 1&0 \end{pmatrix}


rref of T=(100100)T=\begin{pmatrix} 1& 0 \\ 0 & 1\\ 0&0 \end{pmatrix}


Rank of T is 2

Correct option is (ii)(ii)


16)


Using Gram-Schmidt process


Let V1=(1,0,0)V_1=(1,0,0)

V2=(1,1,1)[(1,1,1).(1,0,0)](1,0,0).(1,0,0).(1,0,0)V_2=(1,1,1)-\frac{[(1,1,1).(1,0,0)]}{(1,0,0).(1,0,0)}.(1,0,0)


=(1,1,1)(1,0,0)=(1,1,1)-(1,0,0)

=(0,1,1)=(0,1,1)


V3=(1,1,2)[(1,1,2).(1,0,0)](1,0,0).(1,0,0).(1,0,0)[(1,1,2).(0,1,1)](0,1,1).(0,1,1)(0,1,1)V_3=(1,1,2)-\frac{[(1,1,2).(1,0,0)]}{(1,0,0).(1,0,0)}.(1,0,0)-\frac{[(1,1,2).(0,1,1)]}{(0,1,1).(0,1,1)}(0,1,1)




=(1,1,2)(1,0,0)32(0,1,1)=(1,1,2)-(1,0,0)-\frac{3}{2}(0,1,1)

°(0,12,12)°(0,-\frac{1}{2},\frac{1}{2})



e1=(1,0,0)e_1=(1,0,0)

e2=12(0,1,1)e_2=\frac{1}{√2}(0,1,1)


e3=12(0,1,1)e_3=\frac{1}{√2}(0,-1,1)


Correct option is (i)



17)


For a finite-dimension polynomial vector

Space:

<p,q>=p0q0+p1q1+.....+pnqn<p,q>=p_0q_0+p_1q_1+.....+p_nq_n


Option (iii) is the correct one









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