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Let $n \geq 2$ and let $W$ be the subspace of $\mathbb M_n(\mathbb R)$ consisting of all matrices whose trace is zero. If $ A = (a_{ij})$ and $B = (b_{ij})$, for $1 \leq$ i,j $\leq n$ , are elements in $\mathbb M_n(\mathbb R)$, define their inner-product by $$(A,B) = \sum_{i,j=0}^n a_{ij}b_{ij}$$ Identify the subspace $W^\perp$ of elements orthogonal to the subspace W.

My work : I am actually confuse in the symbols and on the basis of what I understand, I took a $2\times 2$ then $3\times 3$ matrices then i reached to the conclusion that "if i take two $n\times n$ matrices A and B with entries $a_{11},a_{12}\ldots ,a_{1n},\ldots, a_{nn}$ $b_{11},b_{12},\ldots, b_{1n},\ldots, b_{nn}$

Then inner product is $a_{11}b_{11}+a_{12}b_{12}+....a_{nn}b_{nn}$ " (here please check if I am thinking def right ?)

But after I equate it to zero and use trace for one matrix A (since other will be in $W^⊥$ I it will not have trace necessary zero so I took a random matrix B) Then $$a_{11}+a_{22}+ \ldots +a_{nn}=0$$ Then I can have value of any one of $a_{ij}$ and put it there in that def of inner product but what do I have to do after that? All $b_{nn}$ are not necessarily non-zero and all $a_{nn}$ are also not necessarily non-zero so I am not sure if I can take them comman and compare both sides and will get something useful and in this step how the def of $W^\perp$ exactly looks like?

I'm not sure the way I'm thinking is right or not?

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  • $\begingroup$ Your definition is not right: the sum runs over all rows and all columns, not only over diagonal elements. You have $a_{11}b_{11} + a_{12}b_{12} + \ldots +a_{1n}b_{1n} + a_{21}b_{21} +a_{22}b_{22} + \ldots +a_{2n}b_{2n} + $ and so on. $\endgroup$ – Student Aug 7 '17 at 18:19
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    $\begingroup$ Thank but I wrote the same thing as you but someone edited latex and with that defn also ..but I shd have added some more terms .I will do that $\endgroup$ – Pranita Gupta Aug 7 '17 at 19:59
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Recall that the trace map is the function $\DeclareMathOperator{tr}{tr}\tr:\Bbb M_n(\Bbb R)\to\Bbb R$ given by $\tr(A)=\sum_k a_{kk}$. One easily shows that $\tr$ is linear. In this problem, we are interested in the kernel of $\tr$, which you call $W$ but I will refer to as $\mathfrak{sl}_n(\Bbb R)=\ker(\tr)$. Since $\tr$ has rank one, the rank-nullity theorem implies that $$ \dim\mathfrak{sl_n}(\Bbb R)=\dim\Bbb M_n(\Bbb R)-\dim\DeclareMathOperator{image}{image}\image(\tr)=n^2-1 $$ Now, we wish to describe $\mathfrak{sl}_n(\Bbb R)^\perp$ where we are viewing $\Bbb M_n(\Bbb R)$ as an inner-product space with inner product given by $$ \langle A, B\rangle=\sum_{i,j}a_{ij}b_{ij} $$ The dimension formula for orthogonal complements in finite-dimensional vector spaces implies that $$ \dim\mathfrak{sl}_n(\Bbb R)^\perp=\dim\Bbb M_n(\Bbb R)-\dim\mathfrak{sl}_n(\Bbb R)=n^2-(n^2-1)=1 $$ Thus $\mathfrak{sl}_n(\Bbb R)^\perp$ is a one-dimensional subspace of $\Bbb M_n(\Bbb R)$. Can you compute a basis of $\mathfrak{sl}_n(\Bbb R)^\perp$?

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We have one choice that is we can use it as : since $a_{11}+a_{22}+\ldots+a{nn}=0$ then $k(a_{11}+a_{22}+\ldots+a{nn}=0)$ so any matrix B which take only diagonal entries of a matrix A and just scale it and don't multiply with other entries of A then B should be a scalar matrix.

Other thing is if I put values of tr(A)=0 in that equation which I mention in question of inner product (and suppose all $a_{ij} \not = 0$ ) then I got $b_{ii}=b_{jj}$ for all i,j, and other entries zero so it's a scalar matrix .

"Now here I took all $a_{ij} \not = 0$ in this point i'm not sure."

Note : now I'm sure of the definition so don't worry about that. :)

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