Answer to Question #208902 in Linear Algebra for Kabir

Question

a) Determine whether or not the following are subspaces?

i. 𝑾 = {(𝒂,𝒃, 𝒄) ∈ ℝ𝟑

|𝒂 + 𝒃 + 𝒄 = 𝟎} of ℝ𝟑

ii. The symmetric matrices of 𝑴𝒏𝒏 (the vector space of 𝒏 × 𝒏

matrices)

iii. All polynomials of degree 2.

b) 

i. For which real values of 𝝀 do the following vectors form a 

linearly dependent set in ℝ𝟑

?

𝒗𝟏 = (𝝀, −

𝟏

𝟐

, −

𝟏

𝟐

) , 𝒗𝟐 = (−

𝟏

𝟐

, 𝝀, −

𝟏

𝟐

) , 𝒗𝟑 = (−

𝟏

𝟐

, −

𝟏

𝟐

, 𝝀)

ii. Find a basis and dimension of the solution space for the 

following homogenous linear equations: 

𝒙𝟏 + 𝟐𝒙𝟐 − 𝒙𝟑 + 𝟒𝒙𝟒 = 𝟎

𝟐𝒙𝟏 − 𝒙𝟐 + 𝟑𝒙𝟑 + 𝟑𝒙𝟒 = 𝟎

𝟒𝒙𝟏 + 𝒙𝟐 + 𝟑𝒙𝟑 + 𝟗𝒙𝟒 = 𝟎

𝒙𝟐 − 𝒙𝟑 + 𝒙𝟒 = 𝟎

𝟐𝒙𝟏 + 𝟑𝒙𝟐 − 𝒙𝟑 + 𝟕𝒙𝟒 = 𝟎

Answer 1

a)

i.

$W=\{(a,b,c)\in R^3|a+b+c=0\} of R^3\\ x=(a,b,c), a+b+c=0\\ y=(a_1,b_1,c_1), a_1+b_1+c_1=0\\ (1)x+y=(a+a_1,b+b_1,c+c_1), \\ a+a_1+b+b_1+c+c_1=0+0=0\\ x+y\in W\\ i) x+y=y+x\\ ii) (x+y)+z=x+(y+z)\\ iii) \exists 0=(0,0,0)\in W, x+0=x\\ iiii) \forall x(a,b,c)\in W \\ \exists (-x)=(-a,-b,-c)\in W (-a+(-b)+(-c)=-(a+b+c)=0):\\ x+(-x)=0$

$(2) r\in R, x=(a,b,c)\in W, a+b+c=0\\ r\cdot x=r(a,b,c)=(ra,rb,rc)\in W\\ ra+rb+rc=r(a+b+c)=0\\ i) r,s\in R: (r+s)x=r\cdot x+s\cdot x\\ ii) r(x+y)=r\cdot x+r\cdot y\\ iii) (rs)x=r \cdot(s\cdot x)\\ iiii)1\cdot x=x$

W are subspaces

ii.

$M_{nn}=\{\begin{pmatrix} a_{11} & a_{12}&...&a_{1n} \\ a_{12} & a_{22}&...&a_{2n}\\ ...&...&...&...\\ a_{1n} & a_{2n}&...&a_{nn} \end{pmatrix}, a_{ij}\in R\}\\ A=\begin{pmatrix} a_{11} & a_{12}&...&a_{1n} \\ a_{12} & a_{22}&...&a_{2n}\\ ...&...&...&...\\ a_{1n} & a_{2n}&...&a_{nn} \end{pmatrix}\in M_{nn}\\ B=\begin{pmatrix} b_{11} & b_{12}&...&b_{1n} \\ b_{12} & b_{22}&...&b_{2n}\\ ...&...&...&...\\ b_{1n} & b_{2n}&...&b_{nn} \end{pmatrix}\in M_{nn}$

$A+B=\begin{pmatrix} a_{11} & a_{12}&...&a_{1n} \\ a_{12} & a_{22}&...&a_{2n}\\ ...&...&...&...\\ a_{1n} & a_{2n}&...&a_{nn} \end{pmatrix}+\\+\begin{pmatrix} b_{11} & b_{12}&...&b_{1n} \\ b_{12} & b_{22}&...&b_{2n}\\ ...&...&...&...\\ b_{1n} & b_{2n}&...&b_{nn} \end{pmatrix}=\\ =\begin{pmatrix} a_{11}+b_{11} & a_{12}+b_{12}&...&a_{1n}+b_{1n} \\ a_{12} + b_{12}& a_{22}+ b_{22}&...&a_{2n}+ b_{2n}\\ ...&...&...&...\\ a_{1n}+ b_{1n} & a_{2n}+ b_{2n}&...&a_{nn}+b_{nn} \end{pmatrix}\in M_{nn}$

$i) A+B=B+A\\ ii) (A+B)+C=A+(B+C)\\ iii) \exists 0=\begin{pmatrix} 0 & 0&...&0 \\ 0 & 0&...&0\\ ...&...&...&...\\ 0 & 0&...&0 \end{pmatrix}\in W_{nn}, A+0=A\\$

$iiii) \forall A=\begin{pmatrix} a_{11} & a_{12}&...&a_{1n} \\ a_{12} & a_{22}&...&a_{2n}\\ ...&...&...&...\\ a_{1n} & a_{2n}&...&a_{nn} \end{pmatrix}\in M_{nn},\\ \exists (-A)=\begin{pmatrix} -a_{11} & -a_{12}&...&-a_{1n} \\ -a_{12} & -a_{22}&...&-a_{2n}\\ ...&...&...&...\\ -a_{1n} & -a_{2n}&...&-a_{nn} \end{pmatrix} \in M_{nn}:\\ A+(-A)=0$

$(2) r\in R, A\in M_{nn}\\ r\cdot A=r\cdot \begin{pmatrix} a_{11} & a_{12}&...&a_{1n} \\ a_{12} & a_{22}&...&a_{2n}\\ ...&...&...&...\\ a_{1n} & a_{2n}&...&a_{nn} \end{pmatrix}=\\ =\begin{pmatrix} r\cdot a_{11} &r\cdot a_{12}&...&r\cdot a_{1n} \\ r\cdot a_{12} & r\cdot a_{22}&...&r\cdot a_{2n}\\ ...&...&...&...\\ r\cdot a_{1n} & r\cdot a_{2n}&...&r\cdot a_{nn} \end{pmatrix}\in M_{nn}$

$i) r,s\in R: (r+s)A=r\cdot A+s\cdot A\\ ii) r(A+B)=r\cdot A+r\cdot B\\ iii) (rs)A=r \cdot(s\cdot A)\\ iiii)1\cdot A=A$

$M_{nn}$ are subspaces

iii.

$P_2=\{ax^2+bx+c| a,b,c\in R\}\\ f=ax^2+bx+c\in P_2\\ f_1=a_1x^2+b_1x+c_1\in P_2\\ (1)f+f_1=ax^2+bx+c+a_1x^2+b_1x+c_1=\\ =(a+a_1)x^2+(b+b_1)x+(c+c_1)\in P_2$

$i) f+f_1=f_1+f\\ ii) (f+f_1)+f_2=f+(f_1+f_2)\\ iii) \exists 0=0\cdot x^2+0\cdot x+0)\in P_2, f+0=f\\ iiii) \forall f\in P_2 \\ \exists (-f)=-ax^2+(-b)x+(-c)\in P_2:\\ f+(-f)=0$

$(2) r\in R, f=ax^2+bx+c\in P_2\\ r\cdot f=r(ax^2+bx+c)=\\ =r\cdot ax^2+r\cdot bx+r\cdot c\in P_2\\ i) r,s\in R: (r+s)f=r\cdot f+s\cdot f\\ ii) r(f+f_1)=r\cdot f+r\cdot f_1\\ iii) (rs)f=r \cdot(s\cdot f)\\ iiii)1\cdot f=f$

$P_2$ are subspaces

b)

i.

$v_1=(\lambda,-12,-12)\\ v_2= (-12,\lambda,-12)\\ v_3=(-12,-12,\lambda)\\ a\cdot v_1+b\cdot v_2+c\cdot v_3=0\\ a\cdot \lambda-12b-12c=0\\ -12a+b\cdot \lambda-12c=0\\ -12a-12b+c\cdot \lambda=0$

$\Delta=\begin{vmatrix} \lambda & -12&-12 \\ -12&\lambda& -12\\ -12&-12&\lambda \end{vmatrix}=\lambda ^3-432\lambda -3456\neq0\\ \lambda ^3-432\lambda -3456=0\\ \lambda ^3+12\lambda^2-12\lambda^2-432\lambda -3456=0\\ \lambda ^2(\lambda+12)-12(\lambda^2+36\lambda +288)=0\\ \lambda ^2(\lambda+12)-12(\lambda+12)(\lambda +24)=0\\ (\lambda+12)(\lambda ^2-12\lambda-288)=0\\ \lambda=-12, \lambda=-12,\lambda=24$

For $\lambda=-12, \lambda=24$ the following vectors form a 

linearly dependent set in $R^3$.

ii.

$x_1+2x_2-x_3+4x_4=0\\ 2x_1-x_2+3x_3+3x_4=0\\ 4x_1+x_2+3x_3+9x_4=0\\ x_2-x_3+x_4=0\\ 2x_1+3x_2-x_3+7x_4=0\\ \begin{pmatrix} 1& 2&-1&4&|0 \\ 2&-1&3&3&|0\\ 4&1&3&9&|0\\ 0&1&-1&1&|0\\ 2&3&-1&7&|0 \end{pmatrix}\\ IIr+Ir(-2)\\IIIr+Ir(-4)\\IIIIr+Ir(-2)\\$

$\begin{pmatrix} 1& 2&-1&4&|0 \\ 0&-5&5&-5&|0\\ 0&-7&7&-7&|0\\ 0&1&-1&1&|0\\ 0&-3&3&-6&|0 \end{pmatrix}\\ III+IIIIr\cdot 5\\IIIr+IIIIr\cdot 7\\IIIIr+IIIIr \cdot 3$

$\begin{pmatrix} 1& 2&-1&4&|0 \\ 0&1&-1&1&|0\\ 0&0&0&0&|0\\ 0&0&0&0&|0 \end{pmatrix}\\$

$x_1+2x_2-x_3+4x_4=0\\ x_2-x_3+x_4=0\\ x_1=-x_3-2x_4\\ x_2=x_3-x_4$

Basis

$x_1=(-1,1,1,0)\\ x_2=(-2,-1,0,1)$

dim =2