Question

Problem 3 Let L:R4 + R3 be given by L - (C)- [. (3x1 – 422 + 11x4) (1522 + 9x3 – 2124) a) [4 pts] Show that L is a linear transformation, and find the matrix representation A of L with respect to the standard bases for R4 and R3. b) [3 pts] Use part a) to find a basis for ker(L). c) [3 pts] Use part a) for find a basis for im(L).

Answer 1

Solution:

Note: Given transformation is not valid

it should be defined as

$L:R^4\to R^3$

is defined by

01 12 L = 3.01 – 4.02 + 1104 15.22 +9.73 – 21.04 -6.01 +9:22 + 4.13 - 5.04 13 14

a). Let

11 22 13 Y1 Y2 Y3 L44 ERA & a,BER 14

we have

$L\left (\alpha\begin{bmatrix} x_1\\x_2 \\x_3\\x_4 \end{bmatrix} +\beta \begin{bmatrix} y_1\\y_2 \\y_3\\y_4 \end{bmatrix} \right )=L\left ( \begin{bmatrix} \alpha x_1+\beta y_1\\ \alpha x_2+\beta y_2 \\ \alpha x_3+\beta y_3 \\ \alpha x_4+\beta y_4 \end{bmatrix} \right )$

L (1-0) - Laler 3(ar1 + Byi) – 4(ax2 + By2) +11(0.24 + By) 15(ax2 + By2) + 9(ax3 + By3) – 21(024 + Bya) -6(0.21 + Byı) + 9(0.22 + By2) + 4(0.23 + By3) – 5(ar 4 + Bya)

L Y1 Y2 Y3 3az1 +38y1 – 4a x2 – 4By2 + 110x4 + 11 Bya) 150x2 + 158y2 + 90x3 + 9By3 – 21024 – 218y4 -6a21 - 6Byı) +90.22 +93y2 + 40x3 + 4By3 – 5a14 – 5By4]

$\Rightarrow L\left (\alpha\begin{bmatrix} x_1\\x_2 \\x_3\\x_4 \end{bmatrix} +\beta \begin{bmatrix} y_1\\y_2 \\y_3\\y_4 \end{bmatrix} \right )= \begin{bmatrix} \alpha(3x_1-4x_2+11x_4)+\beta(3y_1-4y_2+11y_4)\\ \alpha(15x_2+9x_3-21x_4)+\beta(15y_2+9y_3-21y_4) \\\alpha(-6x_1+9x_2+4x_3-5x_4)+\beta(-6y_1+9y_2+4y_3-5y_4) \end{bmatrix}$

11 02 L Y1 Y2 Y3 C + 8 = CY 3.01 – 4.12 +11.04 15.02 +9.13 – 21.04 -6.21 +9.02 + 4.13 – 5.04 +8 3y1 – 4y2 + 11y4 15y2 +9y3 – 2144 -641 +9y2 + 4y3 - 5y4] 13 14 44

$\Rightarrow L\left (\alpha\begin{bmatrix} x_1\\x_2 \\x_3\\x_4 \end{bmatrix} +\beta \begin{bmatrix} y_1\\y_2 \\y_3\\y_4 \end{bmatrix} \right )=\alpha L\left (\begin{bmatrix} x_1\\x_2 \\x_3\\x_4 \end{bmatrix} \right )+\beta L\left ( \begin{bmatrix} y_1\\y_2 \\y_3\\y_4 \end{bmatrix} \right )$

Hence, is a linear transformation.

Now, Let the standard basis for $R^4$ and $R^3$ are respectively

$B_1=\left ( \begin{bmatrix} 1\\0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0\\1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0\\0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0\\0 \\ 0 \\ 1 \end{bmatrix} \right )$

and

$B_2=\left ( \begin{bmatrix} 1\\0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0\\1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0\\0 \\ 1 \end{bmatrix} \right )$

Now,

$L\left ( \begin{bmatrix} 1\\ 0 \\ 0 \\ 0 \end{bmatrix} \right )=\begin{bmatrix} 3\\ 0 \\ -6 \end{bmatrix}$

$L\left ( \begin{bmatrix} 0\\ 1 \\ 0 \\ 0 \end{bmatrix} \right )=\begin{bmatrix} -4\\ 15 \\ 9 \end{bmatrix}$

$L\left ( \begin{bmatrix} 0\\ 0 \\ 1 \\ 0 \end{bmatrix} \right )=\begin{bmatrix} 0\\ 9 \\ 4 \end{bmatrix}$

$L\left ( \begin{bmatrix} 0\\ 0 \\ 0 \\ 1 \end{bmatrix} \right )=\begin{bmatrix} 11\\ -21 \\ -5 \end{bmatrix}$

Thus, matrix representation is

$A=\begin{bmatrix} 3 & -4 &0 & 11\\ 0& 15& 9 & -21\\ -6& 9& 4 &-5 \end{bmatrix}$

Which is the required matrix representation.

b).

Now, Kernel of is given by

$ker(L)=\left \{ X \in R^4: \ \ AX=0 \right \}$

$\Rightarrow ker(L)=\left \{ \begin{bmatrix} x_1\\x_2 \\ x_3 \\ x_4 \end{bmatrix}\in R^4: \ \ \begin{bmatrix} 3 & -4 &0 & 11\\ 0& 15& 9 & -21\\ -6& 9& 4 &-5 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \\ x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \\ 0 \end{bmatrix}\right\}$

applying
R3 R3 + 2R1

01 T1 .22 C2 ker(L) ER : OOO 3 -4 0 11 15 9-21 4. 17 13 C3 1 14 14

applying $R_2 \leftrightarrow R_3$

$\Rightarrow ker(L)=\left \{ \begin{bmatrix} x_1\\x_2 \\ x_3 \\ x_4 \end{bmatrix}\in R^4: \ \ \begin{bmatrix} 3 & -4 &0 & 11\\ 0& 1& 4 &17\\ 0& 15& 9 & -21 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \\ x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \\ 0 \end{bmatrix}\right\}$

applying $R_3\to R_3-15R_2$

$\Rightarrow ker(L)=\left \{ \begin{bmatrix} x_1\\x_2 \\ x_3 \\ x_4 \end{bmatrix}\in R^4: \ \ \begin{bmatrix} 3 & -4 &0 & 11\\ 0& 1& 4 &17\\ 0& 0& -51 & -276 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \\ x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \\ 0 \end{bmatrix}\right\}$

$\Rightarrow ker(L)=\left \{ \begin{bmatrix} x_1\\x_2 \\ x_3 \\ x_4 \end{bmatrix}\in R^4: \ \3x_1-4x_2+11x_4=0,x_2+4x_3+17x_4=0,51x_3+276x_4=0 \right\}$$\Rightarrow ker(L)=\left \{ \begin{bmatrix} x_1\\x_2 \\ x_3 \\ x_4 \end{bmatrix}\in R^4: \ \3x_1=4x_2-11x_4,x_2=-4x_3-17x_4,x_3=-\frac{276}{51}x_4 \right\}$

$\Rightarrow ker(L)=\left \{ \begin{bmatrix} \frac{129}{51}x_4\\\frac{237}{51}x_4 \\ -\frac{276}{51}x_4 \\ x_4 \end{bmatrix}: x_4\in R \right\}$

$\Rightarrow ker(L)=\left \{ \begin{bmatrix} \frac{129}{51}\\\frac{237}{51} \\ -\frac{276}{51} \\ 1 \end{bmatrix}x_4: x_4\in R \right\}$

Thus, the basis for is

$basis [ker(L)]=\left \{ \begin{bmatrix} \frac{129}{51}\\\frac{237}{51} \\ -\frac{276}{51} \\ 1 \end{bmatrix} \right \}$

c). we have

$A=\begin{bmatrix} 3 & -4 &0 & 11\\ 0& 15& 9 & -21\\ -6& 9& 4 &-5 \end{bmatrix}$

$A^T=\begin{bmatrix} 3 & 0 & -6\\ -4 & 15 & 9\\ 0 & 9 & 4\\ 11 & -21 & -5 \end{bmatrix}$

applying $R_1\to \frac{1}{3}R_1$

$A^T=\begin{bmatrix} 1 & 0 & -2\\ -4 & 15 & 9\\ 0 & 9 & 4\\ 11 & -21 & -5 \end{bmatrix}$

applying $R_2\to R_2+ 4R_1,\ R_4\to R_4-11R_1$

$A^T=\begin{bmatrix} 1 & 0 & -2\\ 0 & 15 & 1\\ 0 & 9 & 4\\ 0 & -21 & 17 \end{bmatrix}$

applying $R_3\to R_3-\frac{9}{15}R_2, \ R_4\to R_4+\frac{21}{15}R_2$

$A^T=\begin{bmatrix} 1 & 0 & -2\\ 0 & 15 & 1\\ 0 & 0 & \frac{17}{5}\\ 0 & 0 & \frac{92}{5} \end{bmatrix}$

applying $R_4\to R_4-\frac{92}{17}R_3$

$A^T=\begin{bmatrix} 1 & 0 & -2\\ 0 & 15 & 1\\ 0 & 0 & \frac{17}{5}\\ 0 & 0 & 0 \end{bmatrix}$

Thus, the basis for is given by

$basis[Im(L)]=\left \{ \begin{bmatrix} 1\\0 \\ -2 \end{bmatrix},\begin{bmatrix} 0\\ 15 \\ 1 \end{bmatrix},\begin{bmatrix} 0\\0 \\ \frac{17}{5} \end{bmatrix} \right \}$

which is the required basis.

This complete the solution.