Verify rank nullity theorem for the linear
transformation T : R^3 → R^3
defined by :
T(x, y, z) = (x + 2y – z, y + z, x + y – 2z)
Answers
Answer:
Let \alpha=(x_1,y_1,z_1)\isin R^3,\beta=(x_2,y_2,z_2)\isin R^3α=(x
1
,y
1
,z
1
)∈R
3
,β=(x
2
,y
2
,z
2
)∈R
3
Then T(\alpha)=(2x_1+y_1-z_1,x_1+z_1)T(α)=(2x
1
+y
1
−z
1
,x
1
+z
1
)
T(\beta)=(2x_2+y_2-z_2,x_2+z_2)T(β)=(2x
2
+y
2
−z
2
,x
2
+z
2
)
\alpha+\beta=(x_1+x_2,y_1+y_2,z_1+z_2)α+β=(x
1
+x
2
,y
1
+y
2
,z
1
+z
2
)
T(\alpha+\beta)=(2(x_1+x_2)+(y_1+y_2)-(z_1+z_2),(x_1+x_2)+(z_1+z_2))T(α+β)=(2(x
1
+x
2
)+(y
1
+y
2
)−(z
1
+z
2
),(x
1
+x
2
)+(z
1
+z
2
))
=((2x_1+y_1-z_1)+(2x_2+y_2-z_2),(x_1+z_1)+(x_2+z_2))=((2x
1
+y
1
−z
1
)+(2x
2
+y
2
−z
2
),(x
1
+z
1
)+(x
2
+z
2
))
=T(\alpha)+T(\beta)=T(α)+T(β)
Also let c\isin R.c∈R. Then c\alpha=(cx_1,cy_1,cz_1)cα=(cx
1
,cy
1
,cz
1
)
T(c\alpha)=(2cx_1+cy_1-cz_1,cx_1+cz_1)T(cα)=(2cx
1
+cy
1
−cz
1
,cx
1
+cz
1
)
=c(2x_1+y_1-z_1,x_1+z_1)=c(2x
1
+y
1
−z
1
,x
1
+z
1
)
=cT(\alpha)=cT(α)
Thus T(\alpha+\beta)=T(\alpha)+T(\beta)T(α+β)=T(α)+T(β) for all \alpha,\beta\isin R^3α,β∈R
3
and T(c\alpha)=cT(\alpha)T(cα)=cT(α) for all c\isin Rc∈R and \alpha\isin R^3.α∈R
3
.
Hence TT is a linear transformation.
KKerer T=T= {(x,y,z)\isin R^3:T(x,y,z)=(0,0)(x,y,z)∈R
3
:T(x,y,z)=(0,0)}
Let (x_1,y_1,z_1)\isin(x
1
,y
1
,z
1
)∈ KerKer TT
Then 2x_1+y_1-z_1=0,x_1+z_1=0.2x
1
+y
1
−z
1
=0,x
1
+z
1
=0.
From second equation we get x_1=-z_1x
1
=−z
1
Let x_1=kx
1
=k (say) , then z_1=-kz
1
=−k
Now from first equation we get , y_1=-3ky
1
=−3k
Therefore (x_1,y_1,z_1)=k(1,-3,-1)=c(-1,3,1)(x
1
,y
1
,z
1
)=k(1,−3,−1)=c(−1,3,1), c\isin Rc∈R
Let \alpha=(-1,3,1)α=(−1,3,1). Then KerKer T=T=linear span of \alpha=Lα=L{\alphaα } and dimdim KerKer T=1.T=1.
ImTImT is the linear span of the vectors T(\alpha_1),T(\alpha_2),T(\alpha_3)T(α
1
),T(α
2
),T(α
3
) where {\alpha_1,\alpha_2,\alpha_3α
1
,α
2
,α
3
} is any basis of R^3R
3
.
{\epsilon_1=(1,0,0),\epsilon_2=(0,1,0),\epsilon_3=(0,0,1)ϵ
1
=(1,0,0),ϵ
2
=(0,1,0),ϵ
3
=(0,0,1)} is a basis of R^3R
3
.
Now T(\epsilon_1)=(2,1),T(\epsilon_2)=(1,0),T(\epsilon_3)=(-1,1).T(ϵ
1
)=(2,1),T(ϵ
2
)=(1,0),T(ϵ
3
)=(−1,1).
ImT=LImT=L{(2,1),(1,0),(-1,1)(2,1),(1,0),(−1,1) }
These vectors are linearly dependent in R^2.R
2
.
But the subset {(2,1),(1,0)(2,1),(1,0)} is linearly independent in R^2R
2
.
\therefore ImT=L∴ImT=L{(2,1),(1,0)(2,1),(1,0) }
Therefore dimdim ImT=2.ImT=2.dim KerT+KerT+ dimdim ImTImT =dim=dim R^3R
3
Here, dimdim KerT+dimKerT+dim ImT=1+2=3ImT=1+2=3.
dimdim R^3=3R
3
=3
Therefore TT satisfies the Rank-Nullity theorem.