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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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    Thank you, I must do that Gerry and Arturo.2012-07-08

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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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    @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