Answer to Question #101515 in Linear Algebra for Henderson

Question

C03

1. Write the characteristic equation of A = 9 1
-2 -3

2. Use the function to find the image of v= ( 1 4 9).
T(v1, v2, v3)=(v2-v1, v2 is not equal to v2, 2v3)

3a. Determine whether or not T R^2 arrow R is a linear transformation.
T(a,b)= a^2.

3b. Determine whether or not T R^3 arrow R is a linear transformation.
T(a,b)= a^3

4. Let A = 6 16 Show that -2 is an eigenvector with the corresponding eigenvalue of -2.
-1 -4 1

Answer 1

1

The characteristic matrix has the form

$A-\lambda E=\begin{pmatrix} 9-\lambda & 1 \\ -2 & -3-\lambda \end{pmatrix}\\$

The characteristic equation has the form

$\begin{vmatrix} 9-\lambda & 1 \\ -2 & -3-\lambda \end{vmatrix}=0\\ (9-\lambda)(-3-\lambda)-(-2)=0\\ -27-9\lambda+3\lambda+\lambda^{2}+2=0\\ \lambda^{2}-6\lambda-25=0$

2

The image of v=(1,4,9)

$T(v_{1},v_{2},v_{3})=(v_{2}-v_{1},v_{2}\not=v_{2},2v_{3})$

$T(v)=T(1;4;9)=(4-1,x,2\cdot9)=(3,x,18)$

where $x\not=4$

3a

$T(a,b)=a^{2}$

If T is a linear transformation then

$\forall x=(x_{1},x_{2})\in R^{2}, y=(y_1,y_2)\in R^{2}$

1)$T(x+y)=T(x)+T(y)$

2)$T(\alpha x)=\alpha T(x)$

1)$T(x+y)=T(x_1+y_1,x_2+y_2)=(x_1+y_1)^{2}=x_1^2+2x_1y_1+y_1^2$

$T(x)=T(x_1,x_2)=x_1^2, T(y)=T(y_1,y_2)=y_1^2$

$T(x)+T(y)=x_1^2+y_1^2$

$T(x+y)\not=T(x)+T(y)$

T is not a linear transformation

3b

$T(a,b,c)=a^3$

If T is a linear transformation then

$\forall x=(x_1,x_2,x_3)\in R^3, y=(y_1,y_2,y_3)\in R^3$

$1)T(x+y)=T(x)+T(y)\\ 2) T(\alpha x)=\alpha T(x)$

$1) T(x+y)=T(x_1+y_1,x_2+y_2,x_3+y_3)=\\ =(x_1+y_1)^3=x_1^3+3x_1^2y_1+3x_1y_1^2+y_1^3\\ T(x)=T(x_1,x_2,x_3)=x_1^3$

$T(y)=T(y_1,y_2,y_3)=y_1^3\\ T(x)+T(y)=x_1^3+y_1^3$

$T(x+y)\not =T(x)+T(y)$

T is not a linear transformation

4

The characteristic equation has the form

$|A-\lambda E|=\begin{vmatrix} 6-\lambda & 16 \\ -1 & -4-\lambda \end{vmatrix}=0\\ (6-\lambda)(-4-\lambda)+16=0\\ -24-6\lambda+4\lambda+\lambda^2+16=0\\ \lambda^2-2\lambda-8=0\\ D=4+4\cdot8=36\\ \lambda_1=\frac{2-6}{2}=\frac{-4}{2}=-2\\ \lambda_2=\frac{2+6}{2}=\frac{8}{2}=4\\$

So, $\lambda_1=-2, \lambda_2=4$ .

Find a vector $x=(x_1,x_2), x\not=0$ that satisfies the condition

$(A-\lambda_1E)x=0\\ 1) \lambda_1=-2\\ \begin{pmatrix} 8 & 16 \\ -1 & -2 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$

$\left \{ \begin{matrix} 8x_1+16x_2=0 \\ -x_1-2x_{2}=0 \end{matrix}\right.$

$x_1=-2x_2, x_2\in R$

$x=(-2;1)$

$2)\lambda_2=4\\ \begin{pmatrix} 2 & 16 \\ -1 & -8 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$

$\left \{ \begin{matrix} 2x_1+16x_2=0 \\ -x_1-8x_{2}=0 \end{matrix} \right.$

$x_1=-8x_2, x_2\in R\\ x=(-8;1)$