Answer to Question #285334 in Linear Algebra for Sabelo Xulu

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,x-y,x)"


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



"=x\\begin{pmatrix}\n 1 \\\\\n 1\\\\\n1\n\\end{pmatrix}+y\\begin{pmatrix}\n 3 \\\\\n -1\\\\\n0\n\\end{pmatrix}"



"T=\\begin{pmatrix}\n 1 & 3\\\\\n 1 & -1\\\\\n1&0\n\\end{pmatrix}"


rref of "T=\\begin{pmatrix}\n 1& 0 \\\\\n 0 & 1\\\\\n0&0\n\\end{pmatrix}"


Rank of T is 2

Correct option is "(ii)"


16)


Using Gram-Schmidt process


Let "V_1=(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)"

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

"\u00b0(0,-\\frac{1}{2},\\frac{1}{2})"



"e_1=(1,0,0)"

"e_2=\\frac{1}{\u221a2}(0,1,1)"


"e_3=\\frac{1}{\u221a2}(0,-1,1)"


Correct option is (i)



17)


For a finite-dimension polynomial vector

Space:

"<p,q>=p_0q_0+p_1q_1+.....+p_nq_n"


Option (iii) is the correct one









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