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-   -   Finding Kernel and Image of Matrix transformation (http://www.physicsforums.com/showthread.php?t=541179)

 Locoism Oct16-11 11:27 PM

Finding Kernel and Image of Matrix transformation

1. The problem statement, all variables and given/known data

Matrix A =
0 1 0
0 0 1
12 8 -1

Let E1 = a(A)(A+2I)2
Let E2 = b(A)(A-3I)

For each of these, calculate the image and the kernel

2. Relevant equations

I found a(A) to be 1/25
and b(A) to be 1/25*(A-7I)
Also, if I am not mistaken, I think KernelE1 = ImageE2 and vice versa

Matrix E1 =
4 4 1
12 12 3
36 36 9

Matrix E2 =
21 -10 1
12 29 -11
-132 -76 40

3. The attempt at a solution

Um... well if v1, v2, v3 are the column vectors of E1 respectively, and w1, w2, w3 are those of E2, isn't {w1, w2, w3}α the Kernel of E1 and Image of E2 (and the other way around)???

Part two says to find a new basis such that the linear transformation corresponding to A is represented by
-2 1 0
0 -2 0
0 0 3

Where do I even begin this one?

PS: Is there a way to add matrices on this forum? It's a little messy this way.

 HallsofIvy Oct17-11 08:02 AM

Re: Finding Kernel and Image of Matrix transformation

To write matrices or other math formulas, use LaTeX. There are a number of tutorials on the internet. On this board you begin LaTeX code by [ tex ] (without the spaces) and end with [ \tex ] (without the spaces). You can do "in line" LaTeX with [ itex ] and [ /itex ]. To do a matrix use \ begin{bmatrix} and \ end{bmatrix} (without the spaces) for a matrix with braces:
$$\begin{bmatrix}4 & 4 & 1 \\ 12 & 12 & 3\\ 36 & 36 & 9\end{bmatrix}$$
use "pmatrix" to get the same thing with ( ):
$$\begin{pmatrix}4 & 4 & 1 \\ 12 & 12 & 3\\ 36 & 36 & 9\end{pmatrix}$$

Quote:
 Matrix E1 = 4 4 1 12 12 3 36 36 9
No, it's not. That is $(A+ 2I)^2$. Isn't $E_1= A(A+ 2I)^2$? (I don't know what that "a" in the formula was supposed to be.)

If I have done the calculation correctly,
$$E_1= A(A+ 2I)^2= \begin{bmatrix}12 & 12 & 3 \\ 36 & 36 & 9\\ 108 & 108 & 27\end{bmatrix}$$
the kernel of $E_1$ would be the set of all (x, y, z) such that
$$\begin{bmatrix}12 & 12 3 \\ 36 & 36 & 9\\ 108 & 108 & 27\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}12x+ 12y+ 3z \\ 36x+ 36y+ 9z \\ 108x+ 108y+ 27z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$$
which gives the three equations 12x+ 12y+ 3z= 0, 36x+ 36y+ 9z= 0, 108x+ 108y+ 27z= 0 which are all multiples of one another. In fact, they all reduce to 4x+ 4y+ z= 0 or z= -4x- 4y. That means that any vector in the kernel can be written as
$$\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ -4x- 4y\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ -4\end{bmatrix}+ \begin{bmatrix}0 \\ 1 \\ -4\end{bmatrix}$$
so the kernel is the two-dimensional subspace spanned by those two vectors.

The Image of $E_1$ is the set of all (x, y, z) such that
$$\begin{bmatrix}12 & 12 3 \\ 36 & 36 & 9\\ 108 & 108 & 27\end{bmatrix}\begin{bmatrix}a \\ b\\ c \end{bmatrix}= \begin{bmatrix}12a+ 12b+ 3c \\ 36a+ 36y+ 9z \\ 108x+ 108y+ 27z\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$$
for some (a, b, c). That gives the three equations 12a+ 12b+ 3c= x, 36b+ 36c+ 9z= y, and 108a+ 108b+ 27c= z. If we subtract 3 times the first equation from the second we get y- 3x= 0. If we subtract 3 times the second equation from the third, we get z- 3y= 0. That is, z= 3y= 3(3x)= 9x. Any vector in the Image of $E_1$ must be of the form
$$\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix} x \\ 3x \\ 9x\end{bmatrix}= x\begin{bmatrix}1 \\ 3 \\ 9\end{bmatrix}$$

However, that may have nothing to do with your problem because it is not at all clear how you intended to define "$E_1$".

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