Department

MathematicsCourse Code

MAT224H1Professor

Martin, BurdaThis

**preview**shows page 1. to view the full**4 pages of the document.**Wednesday19/01/2011, Lecture notes byY. Burda

1Rank-nullitytheorem

Avery simple example of kernel and image (with picture):

Let f:R2→R2begiven 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)|x∈R}=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:V→Wbe 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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