What are the image and kernel of matrices A and B?

[toothpaste666]

Homework Statement

I know this stuff isn't complicated but the definitions of my book are very formal and confusing. I have to find the image and the kernel of these two matrices:

$A= \begin{bmatrix}1&2&3\end{bmatrix} B= \begin{bmatrix}2&3\\6&9\end{bmatrix}$

The Attempt at a Solution

my book says the image is the span of the column vectors so for A would that be:

$im(A) = x_1[1] + x_2[2] + x_3[3]$ ?

and since 3 = 1 + 2 I think that makes 3 redundant so it would be x1[1] + x2[2]?
and the kernel is when the system = 0 so when
x1 = -2x2 - 3x3
if x2 = t and x3 = r then
$ker(A) = [-2t-3r] ??$

for B
$im(B) = x_1 \begin{bmatrix}2\\6\end{bmatrix} + x_2 \begin{bmatrix}3\\9\end{bmatrix}$
$ker(B) = \begin{bmatrix}2&3&|0\\6&9&|0\end{bmatrix} = \begin{bmatrix}1&3/2&|0\\6&9&|0\end{bmatrix} = \begin{bmatrix}1&3/2&|0\\0&0&|0\end{bmatrix}$

so x1 = -3/2 x2
if x2 = t then
$ker(B) = [-3/2t]$??

am i doing these right? I would appreciate if someone can help explain these concepts to me without using the formal abstract definitions

 

[toothpaste666]

man this is messy. Is there at tutorial for making matrices in LatEx?

 

[Mark44]

man this is messy. Is there at tutorial for making matrices in LatEx?

https://www.physicsforums.com/help/latexhelp/ -- towards the end.
Here's how I do them.
\begin{bmatrix} 2 & 3 \\ 6 & 9 \end{bmatrix}

Remove the spaces and this is how the above is rendered:
$$\begin{bmatrix} 2 & 3 \\ 6 & 9 \end{bmatrix}$$

 

[BiGyElLoWhAt]

here, this:
\left [ \begin{array}{c}
1 & 2 \\
3 & 4\\
\end{array}\right ]
renders as this:
$\left [ \begin{array}{c} 1 & 2 \\ 3 & 4\\ \end{array}\right ]$
You can replace the \left [ with \left ( or \left | (for determinants) or I think...
$\left \{ \right \}$
yup, you can use \left \{ as well, you need that \ in front of the { (as well as for the right) because the symbals { } are special for latex. This tells it that it's just another character.
Apparently the matrix components render even when you don't have Latex enabled. It's 1 (ampersand) 2 (ampersand) and double slash (\\) to end the line.

Also, 1 + 2 does in fact equal 3. However, 1*x + 2*y does not in fact equal 3z, so the 3 is not redundant.

 

[BiGyElLoWhAt]

Ahhh, this mark guy, always stealing my thunder.

 

[Mark44]

Homework Statement

I know this stuff isn't complicated but the definitions of my book are very formal and confusing. I have to find the image and the kernel of these two matrices:
A = {1, 2 , 3}
[itex] \begin{bmatrix} 2 & 3 \\ 6 & 9 \end{bmatrix} [\itex}

The Attempt at a Solution

my book says the image is the span of the column vectors so for A would that be:
im(A) = x1[1] + x2[2] + x3[3] ?

Is A a row matrix? If so, the columns span a one-dimensional space.

and since 3 = 1 + 2 I think that makes 3 redundant so it would be x1[1] + x2[2]?

2 is a multiple of 1, and 3 is a multiple of 1 as well.

and the kernel is when the system = 0 so when
x1 = -2x2 - 3x3
if x2 = t and x3 = r then
ker(A) = [-2t-3r] ??

Your matrix, if I understand what you wrote, can be considered to be a mapping from R³ to R¹. It turns out that ker(A) is two-dimensional, so you need two vectors for a basis.

for B
im(B) = x1 {{2} + x2 {{3},
{6}} {9}}
ker(B) =
{{2 , 3 | 0}, /2 {{1, 3/2 | 0}, {{1, 3/2 | 0},
{6, 9 | 0}} = {6, 9 | 0 }} -6(I) = {0, 0 | 0}}
so x1 = -3/2 x2
if x2 = t then
x1 = [-3/2t] ??

am i doing these right? I would appreciate if someone can help explain these concepts to me without using the formal abstract definitions

 

[toothpaste666]

ok thank you guys i cleaned it up a bit. how did i do?

 

[Mark44]

ok thank you guys i cleaned it up a bit. how did i do?

Lots better.

For B, note that the two vectors you show are linearly dependent. The second vector is a scalar multiple of the first (or vice-versa).

 

[toothpaste666]

are my image and kernel for A correct? B the the vector
$\begin{bmatrix}3\\9\end{bmatrix}$ is a mutliple of $\begin{bmatrix}2\\6\end{bmatrix}$
if you multiply it by k = 2/3 correct? so do I only include
$\begin{bmatrix}2\\6\end{bmatrix}$ ??
i only vaguely understand why I am doing the things I am doing

 

[toothpaste666]

oops i mean k = 3/2

 

[HallsofIvy]

If A is a linear transformation from vector space U to vector space V, then the "image" of A is the set of vectors, y, in V such that Ax= y for some vector x in U and the "kernel" is the set of all vectors, u, in U such that Au= 0.

It is still not clear to me what "A" is here. You talk about two matrices, label one of them "A" but do not label the other so it is hard to understand. I think you mean that "A" is the row matrix, $\begin{bmatrix}1 & 2 & 3\end{bmatrix}$ which takes vectors in R³ to vectors in R¹. If the vector in R³ is u= <x , y, z> then Au= x+ 2y+ 3z. For u to be in the kernel of A, we would have to have x+ 2y+ 3z= 0. What vectors, <x, y, z>, have that property? The image of A is the set of all "vectors" in R¹ (i.e. single numbers), a, such that a= x+ 2y+ 3z for some numbers, x, y, and z.

The kernel of the two by two matrix is <x, y> such that $\begin{bmatrix}2 & 3 \\ 6 & 9\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}2x+ 3y \\ 6x+ 9y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$. So we must have 2x+ 3y= 0 and 6x+ 9y= 0. What x and y satisfy that?

The image is the subspace of all vectors <a, b> such that $\begin{bmatrix}2 & 3 \\ 6 & 9 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}2x+ 3y \\ 6x+ 9y\end{bmatrix}= \begin{bmatrix}a \\ b\end{bmatrix}$. What can you say about a and b if 2x+ 3y= a and 6x+ 9y= b for some x and y?

 

[BiGyElLoWhAt]

When in doubt, Wikipedia.

 

[toothpaste666]

A is the row matrix [1, 2, 3] and the other matrix is the 2x2 matrix that I labeled B. wikipedia isn't helping because the formal definitions are just confusing me. let me know If I understand correctly so far
the im(B) is
$= im(B) = \begin{bmatrix}y_1\\y_2\end{bmatrix} = \begin{bmatrix}2&3\\6&9\end{bmatrix}$
so
$\begin{bmatrix}2&3&|y_1\\6&9&|y_2\end{bmatrix} = \begin{bmatrix}1&3/2&|y_1/2\\6&9&|y_2\end{bmatrix} = \begin{bmatrix}1&3/2&|y_1/2\\0&0&|y_2-6y_1\end{bmatrix}$

 

[toothpaste666]

so y1 = 2x1 + 3x2
and y2 = 6y1 = 12x1 + 18x2
if x1 = t and x2 = r then
$= im(B) = \begin{bmatrix}2t+3r\\12t+18r\end{bmatrix}$

sorry I am pretty confused =[

 

[toothpaste666]

is [-3/2t] the correct kernel of B?

 

[BiGyElLoWhAt]

What's the highest dimension of a column vector that can be formed by multiplying a column vector with your matrix?
You have
$\left [ \begin{array}{c} 2 & 3 \\ 6 & 9 \\ \end{array} \right ] \left [ \begin{array}{c} x_1\\ ...\\ x_n\\ \end{array} \right]$
What's the largest n allowed? This gives you the dimension that you should be looking for.
Once you find that, your goal is to figure out what vectors can be formed by multiplying B*x. This is the image. As several peole have pointed out, you don't have linlearly independant covectors in your matrix B. (covectors=row vector) To handle this, you can do something similar to what you were trying to do in solving for the kernal of A.

In this context, you're trying to take vectors on a vector space X and map them to a vector space Y. With this particular transform (B) what kind of vectors can you get in Y if you only take vectors from X and map them to Y using the transform B?

 

[HallsofIvy]

is [-3/2t] the correct kernel of B?

I don't know what you mean by "[-3/2t]". I said before that the kernel of B was the subspace of vectors <x, y> such that 2x+ 3y= 0 and 6x+ 9y= 0. Of course the second equation is just 3 times the first (if they were not dependent, (0, 0) would be the only solution). The set of all <x, y> such that 2x+ 3y= 0 form a subspace as the kernel must be. In fact, <x, y>= <3, -2> is a solution and so is any multiple.

 

[Mark44]

so y1 = 2x1 + 3x2
and y2 = 6y1 = 12x1 + 18x2

I don't follow what you're doing to get y₂.
The two equations are
$2x_1 + 3x_2 = y_1$
$6x_1 + 9x_2 = y_2$
Solve this system for $x_1$ and $x_2$, preferably using row reduction in an augmented matrix. The basic idea is to determine whether there is a solution <x₁, x₂> for any given <y₁, y₂>. As it turns out, there is not, and this fact leads you to a basis for the image of B.

if x1 = t and x2 = r then
$= im(B) = \begin{bmatrix}2t+3r\\12t+18r\end{bmatrix}$

No.

sorry I am pretty confused =[

 

[toothpaste666]

when I got 0 0 y2-6y1 in the bottom row I was going to say it was inconsistent but then i thought it is consistent
$y_2= 6y_1$ and since
$y_1 = 2x_1 + 3x_2$
i just multiplied it by 6 to get y2. if it is inconsistent does that always mean it is linearly dependent? but now that it is inconsistent I am not sure what to do to get the image. there are 2 columns of B so that means there is 2 rows of the x vector and that the y vector will be 2x1 because there is 2 rows of B and 1 column of the x vector. the first row of y (2x1) I can use 2t + 3r, but what do i put for the second row?

 

[toothpaste666]

would the image of B besimply
$im(B) = \begin{bmatrix}x_1\\3x_1\end{bmatrix}$ because any column vector is of that form for example plugging in 2 into x_1 for the first column in B and 3 into X_1 for the second column in B?

 

[Mark44]

would the image of B besimply
$im(B) = \begin{bmatrix}x_1\\3x_1\end{bmatrix}$ because any column vector is of that form for example plugging in 2 into x_1 for the first column in B and 3 into X_1 for the second column in B?

You're on the right track. Im(B) is all vector multiples of \begin{bmatrix} 1 \\ 3 \end{bmatrix}, a one dimensional subspace of $\mathbb{R}^2$.

 

[toothpaste666]

so the number of dimensions is the number of columns? what would be different is the system was consistent? would it be two dimensional?

 

[toothpaste666]

and for the kernel of B we have
$2x_1 = -3x_2$
which we need to equal 0 so
$ker(B) = k\begin{bmatrix}3\\2\end{bmatrix}$

 

[HallsofIvy]

]

and for the kernel of B we have
$2x_1 = -3x_2$
which we need to equal 0 so
$ker(B) = k\begin{bmatrix}3\\2\end{bmatrix}$

You were told this back in response #11. Why are you still asking?

 

[toothpaste666]

just trying to make sure I understand correctly

 

[toothpaste666]

so for the vectors that span the kernel of A
$x_1 = -2x_2-3x_3$
$ker(A) = \begin{bmatrix}-2x_2-3x_3\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}-2x_2\\x_2\\0\end{bmatrix}+ \begin{bmatrix}-3x_3\\0\\x_3\end{bmatrix} = x_2\begin{bmatrix}-2\\1\\0\end{bmatrix} + x3\begin{bmatrix}-3\\0\\1\end{bmatrix}$

 

[Mark44]

so for the vectors that span the kernel of A
$x_1 = -2x_2-3x_3$
$ker(A) = \begin{bmatrix}-2x_2-3x_3\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}-2x_2\\x_2\\0\end{bmatrix}+ \begin{bmatrix}-3x_3\\0\\x_3\end{bmatrix} = x_2\begin{bmatrix}-2\\1\\0\end{bmatrix} + x3\begin{bmatrix}-3\\0\\1\end{bmatrix}$

Looks OK. Alternatively, you could say that $Ker(A) = span \{\begin{bmatrix}-2\\1\\0\end{bmatrix}, \begin{bmatrix}-3\\0\\1\end{bmatrix} \}.$
Also, these two vectors are a basis for Ker(A).

 

[toothpaste666]

thank you all for your help and patience. is it true that since the kernel of A is not the zero vector then A is linearly dependent? if i were asked to find a non trivial relation between the column vectors of A would it be enough to show the span of the kernel?

 

[Mark44]

thank you all for your help and patience. is it true that since the kernel of A is not the zero vector then A is linearly dependent?

You wouldn't say that A is linearly dependent, but rather, the columns of A (considered as one-dimensional vectors) are linearly dependent.

if i were asked to find a non trivial relation between the column vectors of A would it be enough to show the span of the kernel?

No. Just find a set of nonzero constants so that c₁[1] + c₂[2] + c₃[3] = 0. Here my notation for [1] etc. is intended to mean these are (1-D) vectors. You can do this by picking values for, say, c₂ and c₃, and solving for the remaining constant.

 

[toothpaste666]

so a nontrivial relation would be
for the 2x2 matrix B would be
$5\begin{bmatrix}2\\6\end{bmatrix} + \frac{-10}{3}\begin{bmatrix}3\\9\end{bmatrix} = 0$

 

[toothpaste666]

I tried an example with three x's and made up the value for 2 and solved for the last but i ended up with 2 different values for that x one for each equation, but they didnt work for both

 

[Mark44]

so a nontrivial relation would be
for the 2x2 matrix B would be
$5\begin{bmatrix}2\\6\end{bmatrix} + \frac{-10}{3}\begin{bmatrix}3\\9\end{bmatrix} = 0$

Sure, that works. A simpler example is
$1\begin{bmatrix}2\\6\end{bmatrix} + \frac{-2}{3}\begin{bmatrix}3\\9\end{bmatrix} = 0$

 

[Mark44]

I tried an example with three x's and made up the value for 2 and solved for the last but i ended up with 2 different values for that x one for each equation, but they didnt work for both

Which matrix are you referring to here?

 

[toothpaste666]

I wanted to find a nontrivial relation between the column vectors
$\begin{bmatrix}1\\2\end{bmatrix}, \begin{bmatrix}2\\3\end{bmatrix}, \begin{bmatrix}3\\4\end{bmatrix}$
so i picked
$x_1 \begin{bmatrix}1\\2\end{bmatrix} + 2 \begin{bmatrix}2\\3\end{bmatrix} + 3 \begin{bmatrix}3\\4\end{bmatrix}$
$= \begin{bmatrix}x_1\\2x_1\end{bmatrix} + \begin{bmatrix}4\\6\end{bmatrix} + \begin{bmatrix}9\\12\end{bmatrix}$
$= \begin{bmatrix}x_1 + 4 + 9\\2x_1 + 6 + 12\end{bmatrix}$
$= \begin{bmatrix}x_1 + 13\\2x_1 + 18\end{bmatrix}$
so for the top row x1 = -13
and for the bottom row x1 = -9 but if i plug either of this into x1 in the first step neither of them work

 

[Mark44]

I wanted to find a nontrivial relation between the column vectors
$\begin{bmatrix}1\\2\end{bmatrix}, \begin{bmatrix}2\\3\end{bmatrix}, \begin{bmatrix}3\\4\end{bmatrix}$

This is a different problem from the two you posted at the start of this thread (so you should have started a new thread...)
The advice I gave about picking two of the constants applied only to matrix A. It doesn't apply to this matrix.

What you're doing here is to solve this matrix equation for the constants:
$$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4\end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3\end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}$$
To do this, row reduce the matrix on the left. Once the matrix is reduced, you can read off the constants.

so i picked
$x_1 \begin{bmatrix}1\\2\end{bmatrix} + 2 \begin{bmatrix}2\\3\end{bmatrix} + 3 \begin{bmatrix}3\\4\end{bmatrix}$
$= \begin{bmatrix}x_1\\2x_1\end{bmatrix} + \begin{bmatrix}4\\6\end{bmatrix} + \begin{bmatrix}9\\12\end{bmatrix}$
$= \begin{bmatrix}x_1 + 4 + 9\\2x_1 + 6 + 12\end{bmatrix}$
$= \begin{bmatrix}x_1 + 13\\2x_1 + 18\end{bmatrix}$
so for the top row x1 = -13
and for the bottom row x1 = -9 but if i plug either of this into x1 in the first step neither of them work