Thread: Linear operator if I know its image and kernel

1. Linear operator if I know its image and kernel

Can somebody suggest how to solve these problems?

1.
U = l(a1, a2, a3)
$a_{1}$= (1, 0, 3,−1),$a_{2}$ = (1,−1, 1, 1),$a_{3}$ = (1, 1, 5,−3)
W is the solution space of:
$3x_{1}$ + $2x_{2}$$x_{3}$ = 0
$x_{1}$$x_{2}$ + $x_{4}$ = 0.
Find $\varphi$ such that:
U = Im($\varphi )$ и W = Ker($\varphi ).$

2.
L={($a_{1},a_{1}$,...)|$a_{n}$=$a_{n-1}$+$a_{n-2}$}
Find basis which elements are geometrical progression.

3.
This the matrix A:
3 -1 1
4 -1 2
2 -1 2
Find such T that the matrix D = $T^{-1}AT$ is diagonal.Find D.

2. Originally Posted by tetris
Can somebody suggest how to solve these problems?

1.
U = l(a1, a2, a3)
$a_{1}$= (1, 0, 3,−1),$a_{2}$ = (1,−1, 1, 1),$a_{3}$ = (1, 1, 5,−3)
W is the solution space of:
$3x_{1}$ + $2x_{2}$$x_{3}$ = 0
$x_{1}$$x_{2}$ + $x_{4}$ = 0.
Find $\varphi$ such that:
U = Im($\varphi )$ и W = Ker($\varphi ).$
Do you understand that there are an infinite number of possible solutions? You are only asked to find one of them.

First, find a basis for W. W is the set of all $\displaystyle (x_1, x_2, x_3, x_4)$ such that $\displaystyle 3x_1+ 2x_2- x_3= 0$ and $\displaystyle x_1- x_2+ x_4= 0$ From those, we have that $\displaystyle x_3= 3x_1+ x_2$ and $\displaystyle x_4= -x_1+ x_2$. That is, $\displaystyle (x_1, x_2, x_3, x_4)= (x_1, x_2, 3x_1+ x_2, -x_1+ x_2)$$\displaystyle = (x_1, 0, 3x_1, -x_1)+ (0, x_2, x_2, x+x_2)= x_1(1, 0, 3, -1)+ x_2(0, 1, 1, 1)$ so W is a two dimensional subspace of $\displaystyle R^4$ and {(1, 0, 3, -1), (0, 1, 1, 1)} is a basis for W.

We are told that U is spanned by (1, 0, 3, -1), (1, -1, 1, 1), and (1, 1, 5, -3). If such a linear tranformation exists, those cannot be independent. This linear transformation is on $\displaystyle R^4$ so the dimensions of it kernel and image must add to 4. Since W is two dimensional, U must also be 2 dimensional.

Of course, we check for independence by looking at a(1, 0, 3, -1)+ b(1, -1, 1, 1)+ c(1, 1, 5, -3)= (0, 0, 0, 0). That gives four equations: a+ b+ c= 0, -b+ c= 0, 3a+ b+ 5c= 0, and -a+ b- 3c= 0. Adding the first two equations, a+ 2c= 0. Adding the second and third equations also eliminates b: 3a+ 6c= 0. Dividing that last equation by 3 gives a+ 2c= 0 so those are really the same equation. Putting a= -2c into a+ b+ c= 0 gives -2c+ b+ c= b- c= 0 or b= c. That is, for any value of c, a= -2c and b= c satisfy that equation. In particular, taking c= 1, -2(1, 0, 3, -1)+ (1, -1, 1, 1)+ (1, 1, 5, -3)= (0, 0, 0, 0). We can take any two of them, say {(1, 0, 3, 1), (1, -1, 1, 1)} as basis for U.

Such a linear transformation must be from $\displaystyle R^4$ to $\displaystyle R^4$ and so can be written as a 4 by 4 matrix. The we know a basis for the kernel, U. Find two more vectors in $\displaystyle R^4$ that are independent of those two as well as independent of each other. In fact, it is easy to see that (1, 0, 0, 0) and (0, 1, 0, 0) are independent of (1, 0, 3, -1) and (0, 1, 1, 1) so {(1, 0, 0, 0), (0, 1, 0, 0), (1, 0, 3, -1), (0, 1, 1, 1) is a basis for $\displaystyle R^4$

Write A as $\displaystyle \begin{bmatrix}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p\end{bmatrix}$.

Now we want
$\displaystyle \begin{bmatrix}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}1 \\ 0 \\ 3 \\ 1\end{bmatrix}$

$\displaystyle \begin{bmatrix}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}1 \\ -1 \\ 1 \\ 1\end{bmatrix}$

$\displaystyle \begin{bmatrix}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 3 \\ -1\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0\end{bmatrix}$

and

$\displaystyle \begin{bmatrix}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 1 \\ 1\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0\end{bmatrix}$

That gives 16 equations to solve for the 16 entries of A, but they simplify a lot. For example, the first equation you get is just a= 1 and the second equation is e= 0.

2.
L={($a_{1},a_{1}$,...)|$a_{n}$=$a_{n-1}$+$a_{n-2}$}
Find basis which elements are geometrical progression.

3.
This the matrix A:
3 -1 1
4 -1 2
2 -1 2
Find such T that the matrix D = $T^{-1}AT$ is diagonal.Find D.

3. Thank you very much for the excellent explanation. My problem with that task was that i haven't figure out how to build the first and the second equations(to take random vectors). If you can help with second and the third task i will appreciated it.