Let x ∈ R3 be nonzero and let A be the matrix whose columns are x,...

Let x ∈ R³ be nonzero and let A be the matrix whose columns are x, 2x, 3x in this order. Show that x is an eigenvector of A and find a basis for the null space of A.

Expert Solution

Let x = (a,b,c)^T, where a,b,c ∈ R and none of a,b,c equals 0 so that x is not a zero vector. Then A =

a	2a	3a
b	2b	3b
c	2c	3c

The eigenvalues of A are solutions to the equation det(A-ʎI₃)= 0 or, ʎ³-aʎ²-2bʎ²-3cʎ² =0 or, ʎ²[ʎ-(a+2b+3c)]= 0. Thus, the eigenvalues of A are ʎ₁=0, ʎ₂=0 and ʎ₃= a+2b+3c. Further, the eigenvector of A associated with its eigenvalue a+2b+3c is solution to the equation [A-(a+2b+3c)I₃]X = 0. To solve this equation, we have to reduce the matrix A-( a+2b+3c)I₃ to its RREF which is

1	0	-a/c
0	1	-b/c
0	0	0

Now, if X = (x,y,z)^T, then the equation [A-( a+2b+3c)I₃]X = 0 is equivalent to x-a/c = 0 or, x = a/c and y –b/c = 0 or, y = b/c. Then X = (a/c,b/c, z) = 1/c(a,b,cz)^T, where z an arbitrary real number. Then, assuming z= 1, it is apparent that (a,b,c)^T = x is an eigenvector of A.

The null space of A is the set of solutions to the equation AX = 0. The RREF of A is

1	2	3
0	0	0
0	0	0

Now, if X =(x,y,z)^T, then the equation AX = 0 is equivalent to x+2y+3z = 0 or, x = -2y-3z. Then X = (-2y-3z,y,z)^T = y(-2,1,0)^T+z(-3,0,1)^T. Thus, {(-2,1,0)^T,(-3,0,1)^T} is a basis for the null space of A.

Note:

I have used the Wolfram calculator for computing the RREF of the above 2 matrices.

milcah answered 1 year ago

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