Question

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...?

Answer 1

1)

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

$2-i=a^2-b^2+2$ abi

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

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

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

$\therefore\>a^2=2.118$

$a=^+_-1.4553$

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

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

2)

$A=\begin{pmatrix} 5 & -3 \\ -6 & 2 \end{pmatrix}$

Characteristic polynomial

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

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

$=t^2-7t-8$

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

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

For $(A-8)x=0$

$A-8=\begin{pmatrix} -3 & -3 \\ -6& -6 \end{pmatrix}$ $\to^{R_2-2R_1}_{then\>\frac{-1}{3}R_1}$ $\begin{pmatrix} 1 & 1\\ 0& 0 \end{pmatrix}$

$x_1=-x_2$

Eigen vector $=\begin{pmatrix} -1 \\ 1 \end{pmatrix}$

For $(A+1)x=0$

$A+1=\begin{pmatrix} 6 & -3\\ -6& 3 \end{pmatrix}\to^{R_2+R_1}_{Then\>\frac{1}{3}R_1}\begin{pmatrix} 2 & -1 \\ 0 & 0 \end{pmatrix}$

$2x_1=x_2$

Eigenvector = $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$

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

3)

The zero of $\R$ is defined as

$0+0+a(0)(0)=0$

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

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

Let $x_1y_1\in\R^2$ and $x_2y_2\in\R^2$

$(x_1+y_1),(x_2+y_2):(x_1+y_1)+(x_2+y_2)+a(x_1+y_1)(x_2+y_2)=0$

$(x_1y_1)+(x_2y_2)=x_1+y_1+ax_1y_1+x_2+y_2+ax_2y_2$

$=(x_1+y)+(y_1+y_2)+a(x_1y_1+x_2y_2)$

$S$ is not a subspace of $\R^2$ unless $a=0$ $b=0$

Let $\alpha\in\R$

$\alpha(x_1y)=\alpha(x+y+axy)$

$=\alpha\>x+xy+a\alpha\>xy$

$\alpha(x)\>x(y)=\alpha\>x+\alpha\>y+a\alpha^2\>xy$

$S$ is not a subspace of $\R^2$ Unless $a=0,b=0$

4)

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

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

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

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

Span W

$=$ { $\begin{bmatrix} 0 \\ 0 \\ 0\\ 1\\ 1 \end{bmatrix}$ $,\begin{bmatrix} 0 \\ 0 \\ 0\\ 0\\ 1 \end{bmatrix}$ }

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

Such that $F^5$ $=u\bigoplus\>w$

5)

The subset $S$ of the vector space $V$ is not a basis if

(i) S is linearly dependent

(ii) S is not a spanning set

Row echelon form of S is $\begin{pmatrix} 2&0&0 \\ 0&\frac{7}{2}&0 \\ 0&0&\frac{-5t}{7}\frac{+15}{7} \end{pmatrix}$

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

$\therefore\>t=3$

The vectors are not basis of $\R^3$ if $t\ne3$