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Can anyone tell me a $3\times3$ matrix that has eigenvalues $1,2,3$ ( or any matrix with integer eigenvalues)?

I need to show the classroom how to calculate eigenvectors using Gauss-Jordan method. So diagonal elements just won't do it.

Please help me.

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    $\begingroup$ Conjugate your nice diagonal matrix by any invertible matrix, preferably one from $GL(n,\Bbb Z)$. Or you could successively conjugate by small transvections until you find the result looks sufficiently scambled. $\endgroup$ Mar 12, 2014 at 12:14

5 Answers 5

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For any $d_1, d_2, d_3$, the diagonal matrix $$ D = \begin{bmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{bmatrix} $$ has eigenvalues $d_1, d_2, d_3$. If you take any invertible matrix $P$, then $$ A = P D P^{-1} $$ has the same eigenvalues as $D$, and the columns of $P$ are the corresponding eigenvectors.

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$$ \begin{pmatrix} 1&0&0\\ 0&2&0\\ 0&0&3 \end{pmatrix} $$ This is a matrix with eigenvalues as required.

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  • $\begingroup$ Hint: you can use \begin{pmatrix}\end{pmatrix} environment to typeset matrices. $\endgroup$ Mar 12, 2014 at 8:30
  • $\begingroup$ @Shashank You can construct a square matrix with any eigenvalues you like by placing them along the diagonal, then "dress it up" by performing row operations to fill in non-diagonal elements. (In many applications, we often want to go the other way, reducing a square matrix to its "diagonalized" form in order to read off the eigenvalues. Such a matrix has many convenient uses.) $\endgroup$ Mar 12, 2014 at 8:32
  • $\begingroup$ Well...i know that.I think I should have been clearer on the question. I need to show the classroom how to calculate eigen vectors using Gauss-Jordan method. So diagonal elements just won't do it. $\endgroup$
    – Shashank
    Mar 12, 2014 at 8:33
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General recipe: start by choosing desired eigenvalues $\lambda_i$ and desired eigenvectors $v_i$ orthogonal to one another. Then form matrices

$$D=\begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{pmatrix} \qquad V=\begin{pmatrix} \vert & \vert & \vert \\ v_1 & v_2 & v_3 \\ \vert & \vert & \vert \end{pmatrix}$$

Using these, you can compute $M=V\cdot D\cdot V^{-1}$ which will have the desired eigenvectors and eigenvalues.

If you want $M$ itself to be integral as well, use the adjugate matrix instead of the inverse: $ M = V\cdot D\cdot \operatorname{adj}(V) $. The resulting matrix will have integral eigenvalues, but not neccessarily the ones you specified but a common multiple thereof. The common factor is $\det(V)$.

I tend to tune my eigenvectors by playing around until I have $\det(V)=\pm 1$. At that point, $V^{-1}=\pm V^T$ will be integral as well, and I can rely on the given eigenvalues. But this playing around may take some time, so if any integral eigenvalues will do, using the adjugate will likely be faster.

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Here it is$$\begin{pmatrix}5&2&-2\\2&5&-2\\-2&-2&5 \end{pmatrix}$$ and $$\begin{pmatrix}1&0&0\\0&2&-2\\1&0&3 \end{pmatrix}$$

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  • $\begingroup$ @shashank here it is with 3,3,9 eigenvalues $\endgroup$
    – Semsem
    Mar 12, 2014 at 8:51
  • $\begingroup$ Thanks. Can you tell me another with unique eigenvalues too. $\endgroup$
    – Shashank
    Mar 12, 2014 at 8:52
  • $\begingroup$ @Shashank here it is with 1 $\endgroup$
    – Semsem
    Mar 12, 2014 at 8:57
  • $\begingroup$ non-unique. and also where are you finding them? $\endgroup$
    – Shashank
    Mar 12, 2014 at 9:00
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    $\begingroup$ i google it@Shashank $\endgroup$
    – Semsem
    Mar 12, 2014 at 9:01
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(this is for $3 \times 3$ matrix example, but this method easily generalizes) Pick any set of $3$ orthonormal vectors $u,v,w$ so $u^Tu=v^Tv=w^Tw=1$ whereas $u^Tv=0$, $u^Tw=0$, $v^Tw=0$. Pick your favorite eigenvalues $a,b,c$ and write $$ A = auu^T+bvv^T+cww^T. $$ If your orthonormal basis has normalization by $\sqrt{3}$ then just pick an e-value which has $3$ as a factor and the pesky fractions vanish and you can easily verify $u,v,w$ are e-vectors with e-values $a,b,c$ respective. To make this method quick you probably want to pick a favorite orthonormal basis and calculate the rank one $uu^T$, $vv^T$ and $ww^T$ matrices carefully for repeated use. I should have done this years ago. Basically, we're just reverse-engineering the spectral theorem.

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