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?

  • 2
    $\begingroup$ 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

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 $A^TA$.

  • $\begingroup$ So what you mean is $R^n$ means the dimensions when matrix as a linear transformation above? $\endgroup$
    – JAEMTO
    Mar 4 at 7:14
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.