# Image and kernel of a matrix

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
Mentor
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
Mentor
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.
toothpaste666 said:
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.

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

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
Mentor
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
Mentor
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?

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
Mentor
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.
toothpaste666 said:
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

oops sorry! that makes sense. is the reason that only applies to A because it is a square matrix?

Mark44
Mentor
oops sorry! that makes sense. is the reason that only applies to A because it is a square matrix?
A is not a square matrix: it is 3 X 1. B is a square matrix. The vectors in post #34 are unrelated to the A and B matrices.

Please start a new thread.