Let $P$ be an invertible matrix with n rows and n columns. Let $L$ be the following vector space: The elements in $L$ are matrices $X$ with n rows and n columns, such that $tr(PX)=0$. Find $dimL$. Here's what I did, please tell me if this is correct: Let $(p_{i,j})$ be the elements of P. $P$ is invertible and thus $P \neq 0$. That means that there exists $i,j$ such that $p_{i,j} \neq 0$. Let $X$ be a matrices with n rows and columns. Name the elements of X, X=$(x_{i,j})$. Now, $tr(PX)=0$ is equivilent to:
$$\sum_{k=1}^{n}\sum_{m=1}^{n}(p_{k,m} \cdot x_{m,k})=0$$

When we open this sum, every element of $P$ and of $X$ appears once and only once.
Without loss of generality $p_{1,1} \neq 0$ (We mentioned that at least one element of $P$ is not zero).
Then the following equation is equivilent to:
$$x_{1,1} = */p_{1,1}$$ where $*$ is some sum of the $x_{i,j}$'s and $p_{i,j}$'s, but $x_{1,1}$ is not in that sum!
Thus, we can choose the elements of $X$ except for $x_{1,1}$, and then take $x_{1,1}=*/p_{1,1}$.
So $L$ is isomorphic to $R^{n^2-1}$ and thus $dimL=n^2-1$. Correct? I am doubtful because I almost didn't use the fact that $P$ is invertible, the saame proof would work if I wast told $P \neq 0$.

  • $\begingroup$ Indeed, your proof is not valid, for precisely the reason that you mention, but even were that not the case, your result would not follow: you've shown tha tthe dimension of $L$ is at most $n^2 - 1$. Why can there not be further dependencies between the $x_{i,j}$? $\endgroup$ – user3482749 Jan 12 at 17:11
  • $\begingroup$ @user3482749 But I mentioned that L is isomorphic to a vector space with dimension n^2-1, we can define $T: R^{n^2-1} \rightarrow L$: for every $v=(v_1,...,v_{n^2-1})$, start to fill the matrix X like that: skip x(1,1) and then go row by row and fill the elements of X to be the elements in v (x(1,2)=v1 x(1,3)=v2 and so on, go row by row) and than take x(1,1)=*/p(1,1).we get an element in L, this is obviously one to one, and is onto because if we have X=(x_(i,j)) in L, then it holds that x(1,1)=*/p(1,1). we take v=(x_(1,2), x(1,3), ..., x(n,n)) [go row by row and skip x(1,1)] Why is this wrong? $\endgroup$ – Omer Jan 12 at 17:23

You have some good ideas, but the computation is flawed.

Better see this as the composition of two linear maps: \begin{align} &f\colon M_n(\mathbb{R})\to M_n(\mathbb{R}) && f(X)=PX \\[4px] &g\colon M_n(\mathbb{R})\to \mathbb{R} && g(X)=\operatorname{Tr}(X) \end{align} where $M_n(\mathbb{R})$ is the space of $n\times n$ matrices.

What does the assumption that $P$ is invertible say about $f$? What's the dimension of the kernel of $g$? Use the rank-nullity theorem for this. Finally, what's the kernel of $g\circ f$?

An extension of the result is obvious. Using the rank-nullity theorem on $g\circ f$, we see that if it is surjective, then its kernel has dimension $n^2-1$. In order to show that this holds with only assuming $P\ne0$, you need to prove that $g\circ f$ is surjective, that is, there exists $X$ with $\operatorname{Tr}(PX)\ne0$ (not obvious, but true).

  • $\begingroup$ thanks for the comment, I'll try that later, but I want to know where am I wrong. please take a look at my comment to user3482749, I added a few more details to show that L is isomorphic to R^(n^2-1) (I defined in that comment an onto and one to one linear map), isn't it correct? $\endgroup$ – Omer Jan 12 at 17:26
  • $\begingroup$ @Omer The problem is the “handwaving” you do to conclude. The intuition is correct, but the argument is not sufficient: as you say, you need to use that $P$ is invertible, which you don't. $\endgroup$ – egreg Jan 12 at 17:30
  • $\begingroup$ P is invertible thus f is onto. f is onto and thus dimIm(gf)=dimIm(g)=dimR=1. thus, dimker(gf)=n^2-1 (by the rank nullity theorem). correct? $\endgroup$ – Omer Jan 12 at 17:36
  • $\begingroup$ @Omer Almost: $f$ is an isomorphism, so the dimension of $\ker(g\circ f)$ equals the dimension of $\ker g$, which is $n^2-1$ by rank-nullity. $\endgroup$ – egreg Jan 12 at 17:38
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    $\begingroup$ @omer You're right in that last comment... $\endgroup$ – DonAntonio Jan 12 at 17:42

The trace function is a linear functional from $\;M_n(F)=\;$ the vector space of all matrices of order $\;x\times n\;$ over a field $\;F\;$ to $\;F\;$, or simple a linear transformation between both vector spaces.

By the dimensions theorem, if a linear functional is not the zero functional, then it automatically is surjective and furthermore its kernel has dimension one less than the dimension of the domain of definition. That $\;P\;$ invertible helps us to deduce $\;f(x):=tr. (PX)\;$ is not the zero functional, thus $\;\dim\ker f= \dim M_n(F)-1=n^2-1\;$ and we're done.

The above is also true for any non-zero $\;P\;$ ... and thus the assumption that $\;P\;$ is invertible is way too much, in fact.


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