Find a basis for the kernel of the matrix

[yeland404]

Homework Statement

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

Homework Equations

how the vectors are linearly independent and span the kernel

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

 

[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?

 

[yeland404]

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?

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?

 

[Mark44]

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.