Rank-Nullity Theorem

Let $f:E\to F$ be a linear mapping. The rank of $f$ is the dimension of the image, $rk(f) := dim(Im(f))$ The rank is an invariant of the map.

The dimension of the kernel, $dim(ker(f))$ is said nullity of a mapping.

Theorem

Under the hypothese: the sum of rank and nullity of a mapping is equal to the dimension of the vector space $(E,+,.)$: $dim(ker(f)) + dim(Im(f)) = dim(E)$ Usually $dim(Im(f))$ is the hardest to calculate and this theorem allows an easy way to compute it as $dim(E)-dim(ker(f))$

Proof

If $dim(ker(f))=0$ then $f$ is injective. If the basis spans the image, the vectors are linearly independent.

Corollaries of Rank-Nullity Theorem

Let $f:E\to E$ be an endomorphism where $(E,+,.)$ is a finite-dimensional vector space.

• If $f$ is injective then it is also surjective
• if $f$ is surjective then it is also injective

Corollary

Let $f:E\to F$ be a linear mapping with $(E,+,.)$ and $(F,+,.)$.

• Consider $dim(E)>dim(F)$. Follows that the mapping is not injective.
• Consider $dim(E)<dim(F)$. Follows that the mapping is not surjective

Eigenvalues and Eigenvectors

Let $f:E\to E$ be an endomorphism where $(E,+,.)$ is a vectors space with dimension $K$. Every vector $x$ such that $f(x)=\lambda x$ with a $\lambda$ scalar and $x\in E\div \{o_E\}$ is said eigenvector of the endomorphism $f$ related to the eigenvalue $\lambda$ Eigenvalue and eigenvector are building blocks of reality

Dan terms: $\lambda$ = eigenvalue (num of x) $x$ = eigenvector (any number) homogeneous = (0,0)

Eigenspace

Let $f:E\to E$ be an endomorphism The set $V(\lambda)\subset E$ with $\lambda\in K$ defined as $V(\lambda)=e\{o_E\}\cup\{x\in E|f(x)=\lambda x\}$ This is said eigenspace of the endomorphism $f$ related to the eigenvalue $\lambda$. The dimension of the eigenspace is said geometric multiplicity of the eigne value $\lambda$ and is indicated with $\gamma_m$

Determining Eigenvalues and Eigenvectors

Let $f:E\to E$ be an endomorphism. Let $p(\lambda)$ be the order $n$ characteristic polynomial related to the endomorphism. The number of roots of $\rho(\lambda)$ is called algebraic multiplicity.

Introduction to Diagonalization

One variable for each line in a diagonal manner. Not all the mappings/matrices can be diagonalized, Need to have enough linearly independent columns of $P$.

Let $f:E\to E$ be an endomorphism. The matrix is diagonalizable if and only if it has n linearly independent eigenvectors:

• All the eigenvalues are distinct
• The algebraic multiplicity of each eigenvalue coincides with its geometric multiplicity

Symmetric Mappings

If the mapping is characterised by a symmetric matrix, then can prove that;

• The mapping is always diagonalisable
• The transformation matrix $P$ can be built to be orthogonal ($P^{-1}=P^{T})$. Also means the new reference system given by the eigenvectors is also a convenient orthogonal system