# Is the following solution correct? Find the dimension of a vector space

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 $$P \neq 0$$.

• 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 $x_{i,j}$? – user3482749 Jan 12 at 17:11
• @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? – 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).

• 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? – Omer Jan 12 at 17:26
• @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. – egreg Jan 12 at 17:30
• 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? – Omer Jan 12 at 17:36
• @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. – egreg Jan 12 at 17:38
• @omer You're right in that last comment... – 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.