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What's kernel in linear algebra?
Jan 29, 2024
What's kernel in linear algebra?
Solution by Steps
step 1
To understand the kernel in linear algebra, we first consider the concept of null space
step 2
The null space of a matrix is the set of all vectors that, when multiplied by the matrix, result in the zero vector
step 3
The kernel of a linear transformation (or matrix) is equivalent to the null space of the matrix representing that linear transformation
step 4
For the matrix (1amp;2amp;33amp;2amp;12amp;1amp;3)\begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{pmatrix}, we would find the set of vectors that satisfy (1amp;2amp;33amp;2amp;12amp;1amp;3)(xyz)=(000)\begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
step 5
If the only solution to this equation is the trivial solution (0,0,0)(0, 0, 0), then the kernel is just the set containing the zero vector
step 6
In contrast, if the matrix were the identity matrix (1amp;0amp;00amp;1amp;00amp;0amp;1)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, the kernel would also be trivial, containing only the zero vector, since any non-zero vector would not be mapped to the zero vector
Answer
The kernel in linear algebra is the null space of a matrix, which is the set of all vectors that are mapped to the zero vector by the matrix.
Key Concept
Kernel in Linear Algebra
Explanation
The kernel is a fundamental concept in linear algebra that represents the set of vectors that are mapped to the zero vector by a linear transformation or matrix. It is synonymous with the null space of the matrix.
give me some examples
Sure! I'd be happy to provide you with some examples. Could you please specify the topic or concept you'd like examples for? That way, I can tailor the examples to your needs and help you better.
give me some examples of finding kernel
Solution by Steps
step 1
To find the kernel of a linear transformation given by a matrix, we need to solve the homogeneous system of linear equations Ax=0 Ax = 0 , where A A is the matrix and x x is the vector in the kernel
step 2
The matrix given is A=[1amp;2amp;34amp;5amp;67amp;8amp;9] A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} . We need to find the vector x=[x1x2x3] x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} such that Ax=0 Ax = 0
step 3
Setting up the augmented matrix for the system Ax=0 Ax = 0 and performing row reduction to find the reduced row echelon form (RREF)
step 4
The RREF of matrix A A will reveal the free variables, which correspond to the basis vectors of the kernel
step 5
Solving the system after row reduction will give us the kernel of the linear transformation
Answer
The kernel of the linear transformation given by the matrix A A is the set of all vectors x x that satisfy Ax=0 Ax = 0 .
Key Concept
Kernel of a Linear Transformation
Explanation
The kernel of a linear transformation is the set of all vectors that are mapped to the zero vector by the transformation. It is found by solving the homogeneous system of linear equations associated with the transformation.
please show me the entire solviing rocess of this example
Solution by Steps
step 1
Differentiate the volume of a sphere with respect to time: ddt(43πr3)=−2π \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) = -2\pi
step 2
Apply the chain rule: ddt(43πr3)=4πr2drdt \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) = 4\pi r^2 \frac{dr}{dt}
step 3
Substitute r=5 r = 5 into the equation: 4π(5)2drdt=−2π 4\pi (5)^2 \frac{dr}{dt} = -2\pi
step 4
Solve for drdt \frac{dr}{dt} : 100πdrdt=−2π 100\pi \frac{dr}{dt} = -2\pi
step 5
Divide both sides by 100π 100\pi : drdt=−2π100π \frac{dr}{dt} = \frac{-2\pi}{100\pi}
step 6
Simplify the fraction: drdt=−150 \frac{dr}{dt} = -\frac{1}{50}
Answer
drdt=−150 \frac{dr}{dt} = -\frac{1}{50}
Key Concept
Differentiation with respect to time and the chain rule
Explanation
The rate of change of the radius with respect to time is found by differentiating the volume of a sphere and using the chain rule to relate drdt \frac{dr}{dt} to the given rate of change of volume.
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Solution by Steps
step 1
Differentiate the surface area of a sphere with respect to time: ddt(4πr2)=8πrdrdt \frac{d}{dt} (4\pi r^2) = 8\pi r \frac{dr}{dt}
step 2
Substitute r=5 r = 5 and drdt=150 \frac{dr}{dt} = \frac{1}{50} into the equation: ddt(4π(5)2)=8π(5)(150) \frac{d}{dt} (4\pi (5)^2) = 8\pi (5) \left(\frac{1}{50}\right)
step 3
Calculate the rate of change of the surface area: dSdt=8π(5)(150) \frac{dS}{dt} = 8\pi (5) \left(\frac{1}{50}\right)
step 4
Simplify the expression: dSdt=8π(550) \frac{dS}{dt} = 8\pi \left(\frac{5}{50}\right)
step 5
Reduce the fraction: dSdt=8π(110) \frac{dS}{dt} = 8\pi \left(\frac{1}{10}\right)
step 6
Multiply to find the rate of change of surface area: dSdt=8π10 \frac{dS}{dt} = \frac{8\pi}{10}
step 7
Simplify the fraction: dSdt=4π5 \frac{dS}{dt} = \frac{4\pi}{5}
Answer
dSdt=4π5 \frac{dS}{dt} = \frac{4\pi}{5}
Key Concept
Differentiation of surface area with respect to time
Explanation
The rate of change of the surface area with respect to time is found by differentiating the surface area formula for a sphere and substituting the given values for r r and drdt \frac{dr}{dt} .
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Solution by Steps
step 1
Calculate the value of the expression: 8π(5)(150) 8\pi (5) \left(\frac{1}{50}\right)
step 2
Multiply the constants and the fraction: 8π×5×150 8\pi \times 5 \times \frac{1}{50}
step 3
Simplify the expression: 8π×110 8\pi \times \frac{1}{10}
step 4
Multiply to find the value: 8π10 \frac{8\pi}{10}
step 5
Reduce the fraction: 4π5 \frac{4\pi}{5}
Answer
4π5 \frac{4\pi}{5}
Key Concept
Multiplication of constants and fractions
Explanation
The value of the expression is found by multiplying the constants and simplifying the resulting fraction.
Problem 3.2: (2.5 \#24. Introduction to Linear Algebra: Strang) Use GaussJordan elimination on [Uamp;I]\left[\begin{array}{ll}U & I\end{array}\right] to find the upper triangular U−1U^{-1} : UU−1=I[1amp;aamp;b0amp;1amp;c0amp;0amp;1][x1amp;x2amp;x3]=[1amp;0amp;00amp;1amp;00amp;0amp;1] U U^{-1}=I\left[\begin{array}{lll} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} x_{1} & x_{2} & x_{3} \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]
Solution by Steps
step 1
To find the inverse of the upper triangular matrix U U , we will use the Gauss-Jordan elimination method on the augmented matrix [Uamp;I]\left[\begin{array}{ll}U & I\end{array}\right]
step 2
The augmented matrix is [1amp;aamp;bamp;1amp;0amp;00amp;1amp;camp;0amp;1amp;00amp;0amp;1amp;0amp;0amp;1]\left[\begin{array}{ccc|ccc} 1 & a & b & 1 & 0 & 0 \\ 0 & 1 & c & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]
step 3
We start with the third row, which already has the form we want for the inverse matrix, so no changes are needed there
step 4
Next, we make the second column of the identity matrix match the second column of U−1 U^{-1} by subtracting a a times the third row from the second row. This gives us [1amp;aamp;bamp;1amp;0amp;00amp;1amp;0amp;0amp;1amp;−c0amp;0amp;1amp;0amp;0amp;1]\left[\begin{array}{ccc|ccc} 1 & a & b & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -c \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]
step 5
Now, we make the first column of the identity matrix match the first column of U−1 U^{-1} by subtracting b b times the third row from the first row and then subtracting a a times the second row from the first row. This gives us the final form of the inverse matrix U−1 U^{-1} as [1amp;−aamp;ac−b0amp;1amp;−c0amp;0amp;1]\left[\begin{array}{ccc} 1 & -a & ac-b \\ 0 & 1 & -c \\ 0 & 0 & 1 \end{array}\right]
Answer
U−1=[1amp;−aamp;ac−b0amp;1amp;−c0amp;0amp;1] U^{-1} = \left[\begin{array}{ccc} 1 & -a & ac-b \\ 0 & 1 & -c \\ 0 & 0 & 1 \end{array}\right]
Key Concept
Gauss-Jordan Elimination for Inverse Matrix
Explanation
The Gauss-Jordan elimination method transforms an augmented matrix into reduced row-echelon form to find the inverse of a matrix. For an upper triangular matrix, this process involves using the lower rows to eliminate the off-diagonal elements of the upper rows in the identity matrix portion of the augmented matrix.
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