Matrix corresponding to linear transformation is invertible iff it is onto?

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SUMMARY

A square matrix A is invertible if and only if it is onto, meaning the image of A spans the entire codomain of the linear transformation. This relationship is established through the rank-nullity theorem, which states that A being onto implies that the dimension of the image of A equals n, leading to a zero-dimensional kernel. Consequently, A is one-to-one, confirming its invertibility.

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Aziza
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Let A be a nxn matrix corresponding to a linear transformation.
Is it true that A is invertible iff A is onto? (ie, the image of A is the entire codomain of the transformation)
In other words, is it sufficient to show that A is onto so as to show that A is invertible?
That was what my professor said but I am having trouble understanding this..could someone please prove this or direct me to a proof?
 
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Aziza said:
Let A be a nxn matrix corresponding to a linear transformation.
Is it true that A is invertible iff A is onto? (ie, the image of A is the entire codomain of the transformation)

Yes

In other words, is it sufficient to show that A is onto so as to show that A is invertible?
That was what my professor said but I am having trouble understanding this..could someone please prove this or direct me to a proof?


$$A\,\,\text{is onto}\,\,\Longleftrightarrow \dim(Im A)=n\Longleftrightarrow \dim(\ker A)=0\Longleftrightarrow A\,\,\text{ is }\,\,1-1$$

DonAntonio
 

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