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Need your help. I am given a matrix
$$B_4=\begin{pmatrix}2 & -1 & 0 & 0\\ -1 & 2 & -1 & 0\\ 0 & -1 & 2 & -1\\ 0 & 0 & -1 & 2\end{pmatrix}.$$

Eigenvectors are given by $$j{\rm th\ component\ of\ }v_k{\rm\ is\ }\sin(jk\pi/5)$$ for I = 1 : 4, J = 1 : 4.

I have to decide whether the vectors $v_1, v_2, v_3$, and $v_4$ are orthogonal and of norm 1.
My answer
This is what I came up with when I used the formula and came up with the matrix for the eigenvectors

octave-3.2.4.exe:84> Q7

$$Q_7=\begin{pmatrix} 0.58779 & 0.95106 & 0.95106 & 0.58779\\ 0.95106 & 0.58779 & -0.58779 & -0.95106\\ 0.95106 & -0.58779 & -0.58779 & 0.95106\\ 0.58779 & -0.95106 & 0.95106 & -0.58779\end{pmatrix}$$ octave-3.2.4.exe:85> transpose(Q7)*Q7

ans = $$\begin{pmatrix}2.50000 & 0.00000 & 0.00000 & 0.00000\\ 0.00000 & 2.50000 & 0.00000 & 0.00000\\ 0.00000 & 0.00000 & 2.50000 & 0.00000\\ 0.00000 & 0.00000 & 0.00000 & 2.50000\end{pmatrix}$$ octave-3.2.4.exe:86> inverse(Q7)

ans = $$\begin{pmatrix}0.23511 & 0.38042 & 0.38042 & 0.23511\\ 0.38042 & 0.23511 & -0.23511 & -0.38042\\ 0.38042 & -0.23511 & -0.23511 & 0.38042\\ 0.23511 & -0.38042 & 0.38042 & -0.23511\end{pmatrix}$$ octave-3.2.4.exe:87> transpose(Q7)

ans = $$\begin{pmatrix}0.58779 & 0.95106 & 0.95106 & 0.58779\\ 0.95106 & 0.58779 & -0.58779 & -0.95106\\ 0.95106 & -0.58779 & -0.58779 & 0.95106\\ 0.58779 & -0.95106 & 0.95106 & -0.58779\end{pmatrix}$$ octave-3.2.4.exe:88> norm(Q7)

ans = 1.5811

As transpose of the matrix multiplied by matrix is not $I$ and inverse is not equal to transpose I have to conclude that it is not orthogonal and norm is not 1. Please let me know if my logic and workings above are correct.

Thanks

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It is not clear to me what your question is. –  Martin Argerami Mar 7 '12 at 13:35
    
Obviously your eigenvectors do not have length 1. Of you want to find such vectors, you should multiply/divide you vectors by a suitbale constant. –  Martin Sleziak Apr 3 '12 at 15:44

1 Answer 1

Note that since $B_4$ is symmetric its eigenvectors are guaranteed to be orthogonal.

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