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$$ A= \begin{pmatrix} 1 & -2 & 1 & 3 & 0\\ 2 & -4 & 4 & 6 & 4\\ -2 & 4 & -1 & -6 & 2 \\ 1 &-2 & -3 & 3 & -8 \end{pmatrix} $$

(I was given this matrix and i am assuming this is an augmented matrix, the variables we are using are a,b,c,d and e that correspond to the columns)

So obviously the first step that i took was to find the kernel of A

I did row reduction and i got the matrix below

$$ A= \begin{pmatrix} 1 & -2 & 0 & 3 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 2 \\ 0 &0 & 0 & 0 & 0 \end{pmatrix} $$

if i wrote down a system of linear equations that correspond to the matrix above, i would get the equations below

a - 2b + 3d = 0
c + 2e = 0

if we let b = t, d = m and e = n(t, m and n are any real numbers) then we get...

a = 2t - 3m
b = t
c = -2n
d = m
e = n

so if we collect the arbitrary terms in the equations above, we get the kernel, which is written below

kernel of A = { (2t, t, 0, 0, 0) , (-3m, 0, 0, m, 0) , (0, 0, -2n, 0 ,n): t, m and n are real numbers }

is this the kernel of A? did i make a mistake somewhere. I would appreciate it if someone checked my work.

Soon_to_be_code_master
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  • The domain of $A$ is $\mathbb R^5$, but you’ve got elements of $\mathbb R^4$ in your kernel, so that’s clearly wrong. See this answer for a guide to reading a kernel basis from the RREF of a matrix. – amd Nov 04 '17 at 22:58
  • Finding the kernel of a $4 \times 5$ matrix is a different problem than solving a $4 \times 4$ system of equations with a $4 \times 1$ target vector. –  Nov 04 '17 at 23:46
  • Your formula for the kernel of the $4 \times 4$ matrix is incorrect; you presumably meant to write $(2t, t, 0 ,0) + (-3m, 0, 0, m)$. –  Nov 04 '17 at 23:46
  • Finally, you don't have to find the kernel before you find a basis for the kernel; it is quite possible (and common!) to do it the other way around. In fact, you've done something extremely similar in the process of obtaining your formula for the kernel of the $4 \times 4$ matrix: you identified $(2,1,0,0)$ and $(-3,0,0,1)$ as a basis for its kernel, and used that fact to help you write down the kernel! –  Nov 04 '17 at 23:48
  • I have made the appropriate corrections in my original post, please double check everything just in case i made an error........ so i am assuming that the basis of the kernel are the following vectors (2,1,0,0,0), (-3,0,0,1,0),(0,0,-2,0,1), is this correct? – Soon_to_be_code_master Nov 05 '17 at 00:10

1 Answers1

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You have the solution $$\begin{pmatrix}a\\ b\\ c\\ d\\ e\end{pmatrix}=\begin{pmatrix} 2b-3d\\ b\\ c -2e\\ d\\ e\end{pmatrix}=b\begin{pmatrix}2\\ 1\\ 0\\ 0\\ 0\end{pmatrix}+d\begin{pmatrix}-3\\ 0\\ 0\\ 1\\ 0\end{pmatrix}+e\begin{pmatrix}0\\ 0\\ -2\\ 0\\ 1\end{pmatrix}$$

So the vectors $$\begin{pmatrix}2\\ 1\\ 0\\ 0\\ 0\end{pmatrix},\begin{pmatrix}-3\\ 0\\ 0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ -2\\ 0\\ 1\end{pmatrix}$$ are a basis for the kernel.

amWhy
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Rene Schipperus
  • 39,526
  • @Marine Galantin: If suspect an error, particularly in an Accepted Answer, the better approach is to broach a correction in a Comment to the author, even if the suspicion is more of a certainty (as here, where $c$ has disappeared from the right hand side). Fortunately you have plenty enough reputation to leave such a Comment, and if the author does not respond after a few days, then it would be more justified to make a substantive edit. – hardmath Jan 12 '19 at 18:29