# Relationship between # dimensions in image and kernel of linear transformation called A and # dimensions in basis of image and basis of kernel of A

Will the dimension of the image of a linear transformation always equal the dimension of the basis of the image of a linear transformation?

Will the dimension of the kernel of a linear transformation always equal the dimension of the basis of the kernel of the aforementioned linear transformation?

The following will be an implication of the rank-nullity theorem: Let $A$ be the matrix that defines the linear transformation.

• "dim" means dimension

• "im" means image of

• Let $m$ = the number of columns of a matrix that defines a linear
transformation.

• Let dim(A) = the number of non linearly dependent columns or
nonredundant columns of A.

• Let dim(ker $A$) be $m$-dim(im$A$).

So far, I have always assumed that

dim(ker $A$)=dim(basis of kernel of $A$) and dim(im $A$)=dim(image of basis of $A$).

Is my assumption correct?

• The dimension if a basis is meaningless. – Bernard Feb 27 '16 at 20:07
• why is dimension of a basis meaningless? – katie Feb 27 '16 at 20:11
• No, you are missunderstanding the terminology, the dimension of a vector space $V$ is the cardinality of any basis $B$ of $V$ (the number or elements of $B$ if it is finite), the basis of a vector space is not itself a vector space, so you can't talk about its dimension but about its cardinality – la flaca Feb 27 '16 at 20:13
• I've never heard of it. However I've head of its cardinality. – Bernard Feb 27 '16 at 20:13