3. Let A be a matrix of a linear transformation p : R³ → R³ with respect to the natural basis. Find a basis of the image and the kernel of p. -() 1 2 3 4 5 6 A = 7 89

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Chapter7: Eigenvalues And Eigenvectors
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Problem 3CM: Let T:RnRm be the linear transformation defined by T(v)=Av, where A=[30100302]. Find the dimensions...
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3. Let A be a matrix of a linear transformation y : R3 → R³ with respect to the natural
basis. Find a basis of the image and the kernel of p.
1 2 3
A =
4 5 6
7 8 9
||
Transcribed Image Text:3. Let A be a matrix of a linear transformation y : R3 → R³ with respect to the natural basis. Find a basis of the image and the kernel of p. 1 2 3 A = 4 5 6 7 8 9 ||
Expert Solution
Step 1

Consider the given matrix as A=123456789 which represents the linear transformation between R3 and R3.

For the kernel of the linear transformation set AX=0, where X is the column vector of the 3 cross 1 order.

 Substitute the matrix in the equation AX=0.

AX=0123456789xyz=0x+2y+3z=04x+5y+6z=07x+8y+9z=0

 

Step 2

Convert the matrix into row reduced echelon form by the set of row operations as below:

A=123456789   ~1230-3-60-6-12  R2-4R1+R2R3-7R1+R3   ~123012000          R2-134R36R6+R3   ~10-1012000      R1-2R2+R1

The above row reduced echelon form gives x-z=0, y+2z=0 or x=z, y=-2z.

Thus, the basis for the kernel φ is obtain below:

xyz=z-2zz       =z1-21Basis of kerφ=1-21

 

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