Find a basis for the kernel of the matrix

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SUMMARY

The discussion focuses on finding a basis for the kernel of the matrix \begin{bmatrix}1 & 2 & 0 & 3 & 5 \\ 0 & 0 & 1 & 4 & 6\end{bmatrix}. The kernel is defined as the set of vectors that the matrix maps to zero, leading to the equations a + 2b + 3d + 5e = 0 and c + 4d + 6e = 0. The solution reveals that the kernel can be expressed in terms of three vectors: \begin{bmatrix}-2\\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}-3\\ 0 \\ -4 \\ 1 \\ 0\end{bmatrix}, and \begin{bmatrix}-5 \\ 0 \\ -6 \\ 0 \\ 1\end{bmatrix}. These vectors are linearly independent and form a basis for the kernel of the linear transformation.

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yeland404
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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
 
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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!":biggrin:

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?
 
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HallsofIvy said:
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!":biggrin:

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?
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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