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Suppose $X$ is a real $n\times n$ matrix. Suppose $m>0$ and let $\operatorname{tr}(X)$ denote the trace of $X$. If $\operatorname {tr}(X^{\top}X)=m$, can i conclude that $X$ is invertible?


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up vote 4 down vote accepted

No, take $$X = X^T = \pmatrix{1&0\\0&0}$$ then $$tr(X^TX) = 1\gt 0$$ but $X$ ist not regular.

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The fact that $\operatorname{Tr}(X^TX)$ is positive just mean that the matrix is non-zero. so any non-zero matrix which is not invertible will do the job.

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I would say that the sum of the eigenvalues is not $0$, not just that its a non-zero matrix – Belgi Dec 3 '12 at 11:25
What is $Tr(A^TA)$? – Davide Giraudo Dec 3 '12 at 11:34
@DavideGiraudo what is $A$? just kidding... – draks ... Dec 3 '12 at 11:38
Sorry I meant $X$ (any matrix $n\times n$ with real entries). – Davide Giraudo Dec 3 '12 at 11:39
My bad, I thought about the trace of the orifinal matrix for some reason – Belgi Dec 3 '12 at 11:49

No, consider $X=\pmatrix{ 1 & 0 \\ 0 & 0 } $. Then $\operatorname{tr}(X^tX)=1$ but $X$ is not invertible.

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