Answer to Question #283963 in Linear Algebra for Njabsy

Question #283963

1. the square root of (2-i) is?


2. for a given ,matrix A=[(5,-6),(-3,2)] the matrix P that diagonalizes A is?


3. for a,b is an element of R, let S be a subset of R^2 defined as S={(x,y) is an element of R^3:x+y+axy=b}. Then S is a subspace of R^2 if...?


4. suppose U={(x,y,x+y,z,2y+z) is an element of F^5:x,y,z is an element of F}, then a subspace W of F^5 such that F^5=U denote W is...?


5. the vectors (1,-1,2),(2,3,1),(3,2,t) are not basis of R^3 if...?



1
Expert's answer
2022-01-17T16:54:07-0500

1)


Let (2i)=a+bi\sqrt{(2-i)}=a+bi where a,bRa,b\in\R



2i=a2b2+22-i=a^2-b^2+2 abi


a2b2=2a^2-b^2=2 and 2ab=12ab=-1      b=12a\implies\>b=\frac{-1}{2a}


a2(12a)2=2    4a48a21=0a^2-(\frac{-1}{2a})^2=2\implies4a^4-8a^2-1=0



a2=8+64+168a^2=\frac{8^+_-\sqrt{64+16}}{8} But aRa\in\R


a2=2.118\therefore\>a^2=2.118


a=+1.4553a=^+_-1.4553


b=12a=+0.3436b=\frac{-1}{2a}=^+_-0.3436


2i=+(1.45530.3436i)\sqrt{2-i}=^+_-(1.4553-0.3436i)



2)


A=(5362)A=\begin{pmatrix} 5 & -3 \\ -6 & 2 \end{pmatrix}


Characteristic polynomial

P(t)=det(AtI)=(5t362t)P(t)=det(A-tI)=\begin{pmatrix} 5-t&-3 \\ -6&2-t \end{pmatrix}


=(5t)(2t)18=(5-t)(2-t)-18

=t27t8=t^2-7t-8


Solving t27t8=0,t=8or1t^2-7t-8=0\>,t=8 \>or-1

Eigenvalues are λ=8     λ=1\lambda=8\>\>\>\>\>\lambda=-1


For (A8)x=0(A-8)x=0

A8=(3366)A-8=\begin{pmatrix} -3 & -3 \\ -6& -6 \end{pmatrix} then13R1R22R1\to^{R_2-2R_1}_{then\>\frac{-1}{3}R_1} (1100)\begin{pmatrix} 1 & 1\\ 0& 0 \end{pmatrix}



x1=x2x_1=-x_2

Eigen vector =(11)=\begin{pmatrix} -1 \\ 1 \end{pmatrix}


For (A+1)x=0(A+1)x=0

A+1=(6363)Then13R1R2+R1(2100)A+1=\begin{pmatrix} 6 & -3\\ -6& 3 \end{pmatrix}\to^{R_2+R_1}_{Then\>\frac{1}{3}R_1}\begin{pmatrix} 2 & -1 \\ 0 & 0 \end{pmatrix}



2x1=x22x_1=x_2

Eigenvector = (12)\begin{pmatrix} 1 \\ 2 \end{pmatrix}


P=(1112)P=\begin{pmatrix} -1 & 1\\ 1 & 2 \end{pmatrix}









3)

The zero of R\R is defined as

0+0+a(0)(0)=00+0+a(0)(0)=0


This shows that SS is a subspace of R2\R^2 if and only b=0b=0


Also SS is a subspace of R2\R^2 if it is closed under addition and scalar multiplication


Let x1y1R2x_1y_1\in\R^2 and x2y2R2x_2y_2\in\R^2

(x1+y1),(x2+y2):(x1+y1)+(x2+y2)+a(x1+y1)(x2+y2)=0(x_1+y_1),(x_2+y_2):(x_1+y_1)+(x_2+y_2)+a(x_1+y_1)(x_2+y_2)=0



(x1y1)+(x2y2)=x1+y1+ax1y1+x2+y2+ax2y2(x_1y_1)+(x_2y_2)=x_1+y_1+ax_1y_1+x_2+y_2+ax_2y_2


=(x1+y)+(y1+y2)+a(x1y1+x2y2)=(x_1+y)+(y_1+y_2)+a(x_1y_1+x_2y_2)



SS is not a subspace of R2\R^2 unless a=0a=0 b=0b=0


Let αR\alpha\in\R

α(x1y)=α(x+y+axy)\alpha(x_1y)=\alpha(x+y+axy)

=αx+xy+aαxy=\alpha\>x+xy+a\alpha\>xy


α(x)x(y)=αx+αy+aα2xy\alpha(x)\>x(y)=\alpha\>x+\alpha\>y+a\alpha^2\>xy


SS is not a subspace of R2\R^2 Unless a=0,b=0a=0,b=0



4)


U=x(10100)+y(01102)+z(00011)U=x\begin{pmatrix} 1 \\ 0\\ 1\\ 0\\ 0 \end{pmatrix}+y\begin{pmatrix} 0 \\ 1 \\ 1\\ 0\\ 2 \end{pmatrix}+z\begin{pmatrix} 0 \\ 0 \\ 0\\ 1\\ 1 \end{pmatrix}



UU is a subspace spanned by linearly independent set S={(1,0,1,0,0)+(0,1,1,0,2)+(0,0,0,1,1)}S=\begin{Bmatrix} ( 1,0,1,0,0)+ (0,1,1,0,2)+(0,0,0,1,1) \end{Bmatrix}


By adjoining elements of F5F^5 that will not affect linear independence of

(0,0,0,1,1)(0,0,0,1,1) and (0,0,0,0,1)(0,0,0,0,1)

Span W


== { [00011]\begin{bmatrix} 0 \\ 0 \\ 0\\ 1\\ 1 \end{bmatrix} ,[00001],\begin{bmatrix} 0 \\ 0 \\ 0\\ 0\\ 1 \end{bmatrix} }



Or W={(0,0,0,x,x+y)}W=\begin{Bmatrix} (0,0,0,x,x+y)\\ \end{Bmatrix}


Such that F5F^5 =uw=u\bigoplus\>w



5)

The subset SS of the vector space VV is not a basis if

(i) S is linearly dependent

(ii) S is not a spanning set



Row echelon form of S is (2000720005t7+157)\begin{pmatrix} 2&0&0 \\ 0&\frac{7}{2}&0 \\ 0&0&\frac{-5t}{7}\frac{+15}{7} \end{pmatrix}


    5t7+157=0\implies\frac{-5t}{7}+\frac{15}{7}=0

t=3\therefore\>t=3


The vectors are not basis of R3\R^3 if t3t\ne3




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment