Answer to Question #256062 in Linear Algebra for Three

Question

1. Suppose T: R³ -> R³ is linear and has an upper-triangular matrix with respect to the basis (1, 0, 0), (1,2,1), (1,2,2). Then the orthonormal basis of R³ with respect to which T has an upper-triangular matrix is...
2. For u = (1,2,2) and v= (1, -2, -1), the value of ||u-v|| is...
3. Suppose that u,v E V, where V is a real vector space such that ||u|| = 4 and ||v|| = 3 Then < u + v, u - v > is...
4. In R³, let U = Span ((1, 0, 0), (0, 1/sqrt2 , 1/sqrt2 )) The u E U such that || u - ( 2,4,6)|| is as small as possible is...
5. Let T: R³ -> R³ defined as T(x,y,z) = (2x, x+y, x-z). Then adjoint operator T* (u,v,w) is...

(i) (2u+v+w, u, -w)

(ii) (2u,v+w, u-w)

(iii) (u,v, -w)

Answer 1

Section 1

Using Gram- Schmidt principle

$e_1=(1,0,0)$

$f_2=(1,2,1)-\frac{(1,2,1).(1,0,0)}{(1,0,0).(1,0,0)}(1,0,0)$

=$(0,2,1)$

$e_2=\frac{f_2}{||f_2||}=\frac{(0,2,1)}{√5}$

$e_2=(\frac{0}{√5},\frac{2}{√5},\frac{1}{√5})$

$f_3=(1,2,2)-[(1,2,2).(1,2,2)](1,0,0)$

$-[(1,2,2).(0,2,1)](0,2,1)$

$=(0,-\frac{2}{5},\frac{4}{5})$

$\frac{(0,\frac{-2}{5},\frac{4}{5})}{||f_3||}$

=$\frac{(0,\frac{-2}{5},\frac{4}{5})}{\frac{2√5}{5}}$

e₃ =$(0,\frac{-1}{√5},\frac{2}{√5})$

Part 2

u-v=(1-1, 2- -2, 2- -1)=(0,4,3)

||u-v||= $\sqrt{\smash[b]{4^2+3^2}}=5$

Part 3

<u+v, u-v> = ||u||² - 2 ||u|| ||v|| +||v||²

= 4² - 2(4)(3) +3²

= 1

Part 4

let $u$ = $x\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ $+ y\begin{pmatrix} 0 \\ \frac{1}{√2} \\ \frac {1}{√2} \end{pmatrix}$

$=\begin{pmatrix} x \\ \frac {y}{√2}\\ \frac {y}{√2} \end{pmatrix}$

$||(x,\frac {y}{√2},\frac {y}{√2}) -(2,4,6)||$

$is \> minimum$

$(x-2)^2+(\frac{y}{√2}-4)^2+(\frac {y}{√2}-6)^2$ $is \> minimum$

$x^2-4x+y^2-\frac{20y}{√2}+ \>56$ $\>is \> minimum$

$\frac{d}{dx}(x^2-4x)=0$

$\implies 2x-4=0\quad x=2$

$\frac {d}{dy}(y^2-\frac{20y}{√2})=0$

$2y-\frac {20}{√2}=0\implies\>y=\frac{10}{√2}$

Substituting the value of $x \>and \>y \>in \>u$

$u=\begin{bmatrix} 2 \\ 5 \\ 5 \end{bmatrix}$

Part 5

Transforming matrix $T$ ;

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

$\begin{pmatrix} 0 \\ 0\\ -1 \end{pmatrix}$

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

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

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

$\begin{pmatrix} 2u+v+w \\ v \\ -w \end{pmatrix}$

Therefore the adjoint operator $T^*(u,v,w)$

$=(2u+v+w,u,-w)$