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I have a few questions about eigenvectors and eigenvalues, especially in symmetric matrices. So, straight to the point:

Why, for any matrix $A_{m \times n}$, is $A^{T}A$ an symmetric matrix?

I tried to figure it out by myself, i couldn't, so i looked up in the internet and couldn't find anything. I really dont have a clue about this.

Let's talk about another problem:

Given $A$ an symmetric matrix, with, for example, all elements on its main diagonal = $k$. If $A$ is $n \times n$, then the matrix will always have $n$ independent eigenvectors? And what about when there is $n$ distinct values in its main diagonal? Is there any proof of this?

And the last one:

Given $A$ a symmetric matrix, then $dim(colA)^{\perp} = Nul A^{T}$. How to prove?

Where $colA$ is the subspace generated by $S\{ u_1,u_2,...,u_n \}$, where $u_n$ are the vectors in the columns of the matrix $A$. What i did in this last one was: for $T: W \rightarrow V, T(x)=Ax$ $$ dim W + ker T = dim V$$ in matrix form: $$ dim(colA) + dim(NulA) = n$$, where n is the number of columns of the matrix. applying this to $A^{T}$: $dim(colA^{T}) + dim(nulA^{T}) = m$, where $m$ is the number of lines of $A$. Now, we know that: $dim(colA) + dim(colA)^{\perp} = m$, substituting in the first equation, i got to: $dim(nulA^T) = dim(ColA)^{\perp}$. This proves about the dimension, but not about the space itself. How can i prove it? Thanks.

Notes:

$ColA$ = subspace generated by the columns of $A$.

$NulA$ = null space of A.

$dim V$ = dimension of $V$.

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  • $\begingroup$ For your first question, try writing out a $2\times 2$ or $3\times 3$ matrix composed of general variables $a, b, c,...$ and see what happens when you multiply $AA^T$ or $A^TA$. Can you generalize to higher dimensions? $\endgroup$
    – The Count
    Nov 26, 2016 at 17:42
  • $\begingroup$ This is what i tried, i checked it correctly. But it's not a proof. $\endgroup$ Nov 26, 2016 at 17:57
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    $\begingroup$ Of course it's not a proof. It's meant to give you insight. Can you see from doing that why it will always work? $\endgroup$
    – The Count
    Nov 26, 2016 at 21:21

1 Answer 1

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Why, for any matrix $A_{m \times n}$, is $A^{T}A$ an symmetric matrix?

For any two matrices $A, B$ with compatible dimensions, $(AB)^T = B^TA^T$ If you apply that to $A^TA$, you see that it's equal to its own transpose.

Given $A$ an symmetric matrix, with, for example, all elements on its main diagonal = $k$. If $A$ is $n \times n$, then the matrix will always have $n$ independent eigenvectors? And what about when there is $n$ distinct values in its main diagonal?

A symmetric matrix will always be diagonalizable, which, in turn, means that it will have $n$ independent eigenvectors. It doesn't matter what values are in its diagonal.

Given $A$ a symmetric matrix, then $\dim(\operatorname{col} A)^{\perp} = \operatorname{Nul} A^{T}$. How to prove?

$\dim(\operatorname{col} A)$ is the same as the rank of $A$, and the dimension of the perpendicular space is then $n$ minus said rank. By the rank-nulity theorem, this is exactly the dimension of the kernel of $A$. This has nothing to do with $A$ being symmetric.

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  • $\begingroup$ Thank you. I have a few questions about the third explanation. This means that $dim ker A = dim(colA)^{\perp}$. Does that mean that $ker A = (colA)^{\perp}$? Why? $\endgroup$ Nov 26, 2016 at 18:02
  • $\begingroup$ No. Just because two spaces have the same dimension does not mean they are equal. They are definitely connected, though. Or rather, their orthogonal complements are. The matrix $A$ gives a bijection from $\operatorname{Nul}A^\perp$ to $\operatorname{col}A$. Thus the dimensions of these two spaces are equal. $\endgroup$
    – Arthur
    Nov 26, 2016 at 18:07
  • $\begingroup$ I mean from $\ker A^\perp$ to $\operatorname{col}A$. $\endgroup$
    – Arthur
    Nov 26, 2016 at 18:14
  • $\begingroup$ So the statement kerA=(colA)⊥ is false? $\endgroup$ Nov 26, 2016 at 18:24
  • $\begingroup$ @Dovah-king Yes, it is. $\ker A$ lives in the domain of the map given by $A$, while $\operatorname{col} A$ lives in the codomain. To make it a bit clearer, if $A$ isn't square, say it's $m\times n$, then $\ker A \subseteq \Bbb R^n$ while $\operatorname{col} A$ lives in $\Bbb R^m$. $\endgroup$
    – Arthur
    Nov 26, 2016 at 18:37

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