Question

Let n E N and A e Mnxn(R). Define trace(A) = x1=1 di,i (i. e. the sum of the diagonal entries) and tr : Mnxn(R) + R, A trace(A).

Answer 1

Consider the linear map tr: M. (IR) IR nxn which maps A E Maxn (R) to its trace. The tr : Max (IR) IR is onto For any XER There enrist A E MOX COR) Such that trace (A) = x for example, Consider the matrin X Az ryn Here A - [acj) nxn Where au and an ao f (ij) & (1,1)

Then trace (A) = x +0+.. ito (n-1) times 3 x That is tr (A) = 4 tr: Mnxn OR) R is Onto.

Then

Range of this map = 1R > im (tr) im (tv) - TR dim (im (tv) - dim (K) = 1 Then That is dim (im (tr)) = 1 We have dim (Maxn GR)) = n²

Then by rank -nullity theorem,

Rank(tr) + Nullity (tr) = dim(Mnxn (R)

$\Rightarrow\dim (im (tr))+\dim (ker (tr))=\dim(M_{n\times n}(\mathbb{R}))$

$\Rightarrow\dim (ker (tr))=\dim(M_{n\times n}(\mathbb{R}))-\dim (im (tr))$

$\Rightarrow\dim (ker (tr))=n^2-1$

Hence we get,

$\dim (ker (tr))=n^2-1$ and $\dim (im (tr))=1$ .

Rank-Nullity Theorem :

Let  $\displaystyle V$ and $\displaystyle W$ are two vector spaces over a same field $\mathbb{F}$ , where $\displaystyle V$ is finite dimensional. Let $\displaystyle T:\displaystyle V\to\displaystyle W$ be a linear transformation. Then

$\operatorname {Rank}(T)+\operatorname{Nullity}(T)=\dim(V)$

where $\operatorname {Rank}(T)=\dim\left(im (\displaystyle{T})\right)$ and $\operatorname {Nullity}(T)=\dim\left(ker (\displaystyle{T})\right)$ .

Note :

• For any nonzero real number a, {a} is a basis for $\mathbb{R}$ , therefore $\dim\mathbb{R}=1$
• consider the subset  $\left\{A_{ij}:1\leq i,j\leq n\right\}$ of $\displaystyle M_{n\times n}(\mathbb{R})$ , where $A_{ij}\in\displaystyle M_{n\times n}(\mathbb{R})$ such that 1 in the entry and all other entries are zero. Then $\left\{A_{ij}:1\leq i,j\leq n\right\}$ is a basis for $\displaystyle M_{n\times n}(\mathbb{R})$ and hence  $\dim\left(\displaystyle M_{n\times n}(\mathbb{R})\right)=n^2$.