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Lecture

Rank Nullity Theorem, coordinates, matricies of transformation


Department
Mathematics
Course Code
MAT224H1
Professor
Martin, Burda

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Wednesday19/01/2011, Lecture notes byY. Burda
1Rank-nullitytheorem
Avery simple example of kernel and image (with picture):
Let f:R2R2begiven byf((x
y)) =(x
0),i.e. the orthogonal projection
onto the x-axis. The kernelis bydefinition the set of all vectors (x
y)such
that their image under fis the zero vector, i.e. {(x
y)|(x
0)=(0
0)}.Thus
ker f=h(0
1)i.
The image of fis bydefinition the set of all images of vectors (x
y)under
f:Imf={f(x
y)|(x
y)R2}={(x
0)|xR}=h(1
0)i.
It turns out that the rank and the nullityof atransformation are always
related byavery simple formula, called rank-nullitytheorem.
Theorem (Rank-nullitytheorem).Let f:VWbe alinear transforma-
tion of K-vector spaces. Assume that Vis finite-dimensional. Then
dim ker f+dim Imf=dim V
Proof. The subspace ker fis finite-dimensional, since it is asubspace of a
finite-dimensional space V.Choose its basis v1,...,vk.Being abasis of some
subspace, the vectors v1,...,vkare linearly independent. Wecan complete
them to abasis of Vbyadding some new vectors u1,...,um.
For now wehavethe following:
v1,...,vkis abasis of ker fand v1,...,vk,u1,...,umis abasis of V;in
particular ker fis k-dimensional, while Vitself is k+m-dimensional.
Since Vis spanned bythe vectors v1,...,vk,u1,...,um,its image f(V)
is spanned bythe vectors f(v1),...,f(vk),f(u1),...,f(um). Notice however
that f(v1)=0, . . . , f(vk)=0since v1,...,vklie in the kernel of f.Thus
Imf=hf(u1),...,f(um)i.
If wecan provethat f(u1),...,f(um)are also linearly independent, we’ll
beable to deduce that they form abasis of Imfand in particular the dimen-
sion of Imfwould haveto bem,thus proving thetheorem.
So now weprovethat f(u1),...,f(um)are linearly independent:
Suppose that µ1f(u1)+. . .+µmf(um)=0. Then f(µ1u1+. . .+µmum)=0
since fis linear. It means that µ1u1+. . . +µmumis in the kernel of f.In
particular it can bewritten as alinear combination of basis vectors ofker f:
µ1u1+. . . +µmum=λ1v1+. . . +λkvk
1
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