# Find a basis for the kernel of the matrix

1. Sep 18, 2011

### yeland404

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

find a basis of the kernel of the matrix that
1 2 0 3 5
0 0 1 4 6

2. Relevant equations

how the vectors are linearly independent and span the kernel

3. The attempt at a solution
Does it mean I need to samplify the 1 2 0 3 5
0 0 1 4 6
into 1 0
0 1
I can not understand the statement

2. Sep 18, 2011

### HallsofIvy

Do you not know the definition of kernel? If so, the first thing you should thought is "Oh, wow, look at that strange word. I had better look up its definition in my textbook!"

The kernel of a linear transformation is the space of all vectors it maps to 0. Here, that would be
$$\begin{bmatrix}1 & 2 & 0 & 3 & 5 \\ 0 & 0 & 1 & 4 & 6\end{bmatrix}\begin{bmatrix}a \\ b \\ c \\ d \\ e\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$$.

That is the same as the two equations a+ 2b+ 3d+ 5e= 0 and c+ 4d+ 6e= 0. You can use those two equations to solve for two of the variables in terms of the other three so all vectors in the kernel can be written in terms of three vectors. What are those three vectors?

Last edited by a moderator: Sep 18, 2011
3. Sep 18, 2011

### yeland404

yes，I understand the define of kernel, then it turns to be kernel（T)= $$\begin{bmatrix}-2\\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix}t+\begin{bmatrix}-3\\ 0 \\ -4 \\ 1 \\ 0\end{bmatrix}r+\begin{bmatrix}-5 \\ 0 \\ -6 \\ 0 \\ 1\end{bmatrix}s$$, while set b= t, d=r, e=s, so now there are the three vectors in kernel. But, how to deal with the Basis?

Last edited by a moderator: Sep 19, 2011
4. Sep 19, 2011

### Staff: Mentor

The three vectors are a basis for the kernel. By observation I can see that they are linearly independent, and there are three of them, which is how many you need for a basis for the kernel of your linear transformation.