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While I'm studying on Linear Algbera, I was stucked in:

Prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$

I understood what above means, and I am now curious about

Yes, I understood : N($A^TA$)=N($A$)

Then, how can I convince that rank($A^TA$)=rank($A$)?

In the above question, they say it is by rank-nullity theorem

But in my book, it seems like there is a no rank-nullity theroem, but dimension theorem:

Let $V,W$ be vector spaces, and let $T:V->W$ be linear. If $V$ is finite-dimensional, then

nullity($T$) +rank($T$) = dim($V$)

I thought like:

nullity($A^TA$) + rank($A^TA$) =dim($A^TA$) and

nullity($A$) +rank($A$) = dim($A$) So both of the rank would be the same.

But I think the dimensions of $A^TA$ and $A$ are different.

$A^TA$ becomes ${n \times n}$ matrix and $A$ is ${m \times n}$ matrix so the dimension is different.

Where I was wrong? Or I am misunderstood?

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    $\begingroup$ What's $\operatorname{dim}A$? $\endgroup$
    – user239203
    Mar 4 at 6:34
  • $\begingroup$ @Gae.S. I meant dimension $\endgroup$
    – JAEMTO
    Mar 4 at 6:35
  • $\begingroup$ The rank-nullity theorem states that the rank plus nullity equals the number of columns. $\endgroup$
    – angryavian
    Mar 4 at 6:35
  • $\begingroup$ If the "dimension" of an m×n matrix is defined to be n, then indeed m×n and n×n have same dimension and everything works $\endgroup$ Mar 4 at 6:42
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Rank is the maximum number of linearly independent columns of $A$, and dim($A$) is the number of columns of $A$. Same goes for $A^TA$. Since they both have $n$ columns, the proof you linked is correct.

If you look at $A$ and $A^TA$ as linear transformations, then we have $A: \mathbb R^n \rightarrow \mathbb R^m$, and $A^TA: \mathbb R^n \rightarrow \mathbb R^n$. So the right hand side of the dimension theorem is equal to $n$ for both $A$ and $A^TA$.

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  • $\begingroup$ So what you mean is $R^n$ means the dimensions when matrix as a linear transformation above? $\endgroup$
    – JAEMTO
    Mar 4 at 7:14
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    $\begingroup$ Yes, because $A$ takes vectors in $\mathbb R^n$ as input. It acts on an $n$-dimensional vector ($A^TA$ acts on the same vector, but the image of $A$ and $A^TA$ are in different spaces). $\endgroup$
    – PkT
    Mar 4 at 7:19
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    $\begingroup$ I now understand clearly. My book said matrix as a linear transformation, I was confused. Thanks a lot. $\endgroup$
    – JAEMTO
    Mar 4 at 7:23

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