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Let $T:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be a linear transformation be such that $\langle Tx,x\rangle =0$ for all $x\in \mathbb{R}^n$.

Then,

  1. $\mathrm{trace}(T)=0$

  2. $\det(T)=0$

  3. all eigenvalues of $T$ are real

  4. $T=0$

Well, if $x=\sum_{1}^{n}a_ie_i$ then the conditions implies that $\langle\sum_{1}^{n}a_iT(e_i),\sum_{1}^{n}a_ie_i\rangle=0$ but how to proceed next? please help.I mean which are correct and which are false?thank you for help.

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    $\LaTeX$ notes: do not use `<` and `>` for inner product, use `\langle` and `\rangle`. And don't use `\ni` for "such that". That symbol isn't the (bastardized) version of the "such that" symbol that one sometimes sees; it means "has `` as an element" That is, $X\ni a$ means "$a$ is an element of $X$".2012-07-08
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    Patience, you are flooding the site with questions. Why not slow down a bit and digest some of the answers?2012-07-08
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    Parts of it are true, but parts are not. Consider a rotation of $90^{\circ}$ on $\mathbb{R}^2$. Then $\langle Tx,x\rangle = 0$ for all $x$, but the characteristic polynomial of $T$ has nonreal roots.2012-07-08
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    Thank you, I must do that Gerry and Arturo.2012-07-08

1 Answers 1

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Suppose that $\lambda$ is a (real) eigenvalue of $T$, and that $x\in\mathbb{R}^n$ is an eigenvector. Then $Tx = \lambda x$, and $x\neq\mathbf{0}$. Therefore, $$0 = \langle Tx,x\rangle = \langle \lambda x,x\rangle = \lambda \langle x,x\rangle.$$ Since $\langle x,x\rangle\neq 0$, it follows that $\lambda = 0$.

So every real eigenvalue must equal $0$. However, as the rotation of $90^{\circ}$ on $\mathbb{R}^2$ shows, you can have that $T\neq 0$, that some roots of the characteristic polynomial are not $0$, and that the determinant is nonzero.

I'll let you try to figure out whether the statement about the trace is true or not for yourself (the rotation does not provide you with a counterexample).

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    So, as trace is just sum of eigen values, and all real eigen values are $0$ so it must have $trace=0$2012-07-08
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    @Patience: That argument is invalid, because the trace is the sum of the **roots of the characteristic polynomial**. If the polynomial does not split over $\mathbb{R}$, then you have to consider the complex roots (which some people call "eigenvalues", though I don't). In the rotation I give, it just so happens that the complex roots are $i$ and $-i$, so their sum is $0$. But how do you know you cannot have a transformation with those properties whose complex roots are, say, $1+i$ and $1-i$?2012-07-08
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    @Patience the trace is the sum of all _complex_ eigenvalues counted with their mulitplicities as roots of the characteristic polynomial. The fact that the trace is $0$ stems from the fact that in a orthonormal basis $(e_1,\dots,e_n)$, the diagonal coefficients of the matrix representation of $T$ are given by $\langle Te_i,e_i\rangle$, which sum to $0$.2012-07-08