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Good evening everybody,

I'm stuck with the following problem from an old exam in Linear Algebra. One is given two Eigenvectors with corresponding two Eigenvalues and told that the trace is negative. I figured out a third orthogonal eigenvector and tried to get to the matrix via diagonalisation, using a parameter alpha for the third eigenvector. Then I set the trace of this matrix equal to the sum of the known eigenvalues plus alpha. But my method does not work out; it does not give me a negative value for alpha. Where am I going wrong? How could I make this work?

Thank you so much.

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  • $\begingroup$ It would be easier to answer if you provide the actual matrix and the two known eigenvectors. $\endgroup$ – anderstood Dec 5 '15 at 23:01
  • $\begingroup$ Sure thing: we are given the Eigenvector (0, 0, 1) with Eigenvalue 0 and another eigenvector (2, 1, 0) with Eigenvalue 1. This is all the information given in the text. @anderstood $\endgroup$ – Mitch Baker Dec 5 '15 at 23:04
  • $\begingroup$ If there was a matrix, it would be super easy, but the lack thereof makes it non-trivial, at least for me... $\endgroup$ – Mitch Baker Dec 5 '15 at 23:21
  • $\begingroup$ See how far you get without the trace hint by writing the matrix in variables and multiplying it by the known eigenvectors (including the one you found, with eigenvalue $\alpha$). $\endgroup$ – TokenToucan Dec 5 '15 at 23:22
  • $\begingroup$ @CuddlyCuttlefish Thanks for the hint; unfortunately I arrive at an alpha which does not work out for the other eigenvectors. $\endgroup$ – Mitch Baker Dec 6 '15 at 0:00
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You know that $M=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\in\mathcal{M}_3(\mathbb{R})$ is of the form $\begin{pmatrix}a&b&c\\b&e&f\\c&f&i\end{pmatrix},$ that $M\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$ which implies $c=f=i=0,$ that $M\begin{pmatrix}2\\1\\0\end{pmatrix}=\begin{pmatrix}2\\1\\0\end{pmatrix}$ which implies $2a+b=2$ and $2b+e=1.$ You also know that $a+e\leq 0.$ Then $b=2(1-a)=\frac{1-e}{2}$ which implies $e=1-4(1-a).$ You get finally $$a+1-4(1-a)\leq0\implies a\leq\frac{5}{3},$$ and for a such $a$ if you consider $b=2(1-a), e=1-4(1-a)$ and $c=f=i=0$ then a such matrix will satisfy what you are looking for.

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  • $\begingroup$ This appears to be a very good answer. Once I have hopefully verified it for myself, I will give it the green tick. Thanks @ Baloown $\endgroup$ – Mitch Baker Dec 6 '15 at 0:03
  • $\begingroup$ You are welcome ! $\endgroup$ – Balloon Dec 6 '15 at 0:08
  • $\begingroup$ @ Baloown Yep, now the whole question makes sense; there's a second part to it where they want one to solve a system of linear equations, taking the second eigenvector as the vector b. They specifically ask for a (underlined) solution, not "the" solution. Thanks again, Sir, you taught me something new. $\endgroup$ – Mitch Baker Dec 6 '15 at 0:33
  • $\begingroup$ You are welcome again ! ^^ $\endgroup$ – Balloon Dec 6 '15 at 0:41

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