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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?

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    $\begingroup$ The dimension if a basis is meaningless. $\endgroup$ – Bernard Feb 27 '16 at 20:07
  • $\begingroup$ why is dimension of a basis meaningless? $\endgroup$ – katie Feb 27 '16 at 20:11
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    $\begingroup$ 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 $\endgroup$ – la flaca Feb 27 '16 at 20:13
  • $\begingroup$ I've never heard of it. However I've head of its cardinality. $\endgroup$ – Bernard Feb 27 '16 at 20:13
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Loosely speaking, the dimension of the image equals the number of elements in the basis of that image. Likewise for the kernel. Hence, the answer to your question, "Is my assumption correct?" is "no" if you want to use terminology correctly.

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