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Show that $0$ is an eigenvalue of the following matrix with multiplicity at least $2$. $$M = \begin{bmatrix} 0 & 2 & 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1\\ 2 & 0 & 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1\\ 2 & 2 & 0 & 2 & 1 & 1 & 1 & 1 & 1 & 1\\ 2 & 2 & 2 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 & 0 & 1 & 1 & 2 & 1 & 1\\ 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 2 & 1\\ 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 2\\ 1 & 1 & 1 & 1 & 2 & 1 & 1 & 0 & 1 & 1\\ 1 & 1 & 1 & 1 & 1 & 2 & 1 & 1 & 0 & 1\\ 1 & 1 & 1 & 1 & 1 & 1 & 2 & 1 & 1 & 0 \end{bmatrix}$$


This question was asked in our university exam the previous year. Is there any trick involved?

I tried finding the row reduced echelon form of the matrix to find the rank which in turn can give some information about the eigenvalue $0,$ but the calculations are getting bad.

I also tried to make two rows $0$ by elementary row operations but I could not succeed.

Can someone please teach me any quick but interesting trick to find it?

NOTE: I have a request, please don't answer this like, these are eigenvectors corresponding to eigenvalue $0$ found using a software which are linearly independent and hence $0$ has multiplicity $2$

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  • $\begingroup$ See if you can spot two linearly independent vectors in the nullspace of $M$. $\endgroup$
    – JimmyK4542
    Sep 6 '20 at 6:06
  • $\begingroup$ @JimmyK4542; But finding the null space is as complicated as finding the rank , isn't it? $\endgroup$
    – Math_Freak
    Sep 6 '20 at 6:08
  • $\begingroup$ Certainly $0$ is an eigenvalue since the determinant is zero. The trace is also zero, so there should be some trick there. I've always been bad at these type of problems though. $\endgroup$
    – Elliot G
    Sep 6 '20 at 6:13
  • $\begingroup$ Also, as a general rule, if you have a big matrix with some nice property like being symmetric, you will definitely want to use that rather than doing row operations. $\endgroup$
    – Elliot G
    Sep 6 '20 at 6:19
  • $\begingroup$ @ElliotG; I dont understand how symmetric can help here, could u elaborate please? $\endgroup$
    – Math_Freak
    Sep 6 '20 at 6:21
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With a bit of inspection, you can see that $Mv_1 = Mv_2 = \vec{0}$ where $$v_1 = \begin{bmatrix}0&0&0&0&1&-1&0&1&-1&0\end{bmatrix}^T$$ and $$v_2 = \begin{bmatrix}0&0&0&0&1&0&-1&1&0&-1\end{bmatrix}^T.$$ It's easy to check that $v_1$ and $v_2$ are linearly independent, so $0$ is an eigenvalue of $M$ with multiplicity at least $2$.

The key to noticing this is to notice that columns $5$, $6$, and $7$ have the same numbers except with the $5$-th, $6$-th, and $7$-th entries permuted as well as the $8$-th, $9$-th, and $10$-th entries permuted. The same can be said about columns $8$, $9$, and $10$. So it was not too difficult to find a linear combination of these columns that added to $0$.

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  • $\begingroup$ How did you know where to put in the $0s$ and $1s$ in the the entries $5,6,7,8,9,10$ $\endgroup$
    – Math_Freak
    Sep 6 '20 at 6:42
  • $\begingroup$ The eigenvectors corresponding to $0$ can be calculated using a software but I need to find them by hand in the exam, so I want a definite procedure to answer the question $\endgroup$
    – Math_Freak
    Sep 6 '20 at 6:43
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Frankly, I prefer finding eigenvectors by inspection to other seemingly more systematic methods. I don't understand why you don't like JimmyK4542's answer. You shouldn't expect that every problem (even if it comes from an examination) has a stereotyped solution recipe.

Let $P=\pmatrix{I_4\\ &I_3&I_3\\ &&I_3}$. Then the last three columns of $MP$ are identical to each other (each of them is a column of $2$s). Hence the nullity of $MP$ is at least $2$. Since $P$ is invertible, the nullity of $M$ is also at least $2$.

Alternatively, if the matrix is real, the problem can be solved as follows. The leading principal $4\times4$ submatrix of $M$ is $A=2(E_{4\times4}-I_4)$, which is invertible with $A^{-1}=\frac12(\frac13E_{4\times4}-I_4)$. Therefore $M$ is congruent to $A\oplus S$, where $S$ is the Schur complement of $A$ in $M$, i.e. \begin{aligned} S&=\pmatrix{E_{3\times3}-I_3&E_{3\times3}+I_3\\ E_{3\times3}+I_3&E_{3\times3}-I_3}-E_{6\times4}A^{-1}E_{4\times6}\\ &=\pmatrix{E_{3\times3}-I_3&E_{3\times3}+I_3\\ E_{3\times3}+I_3&E_{3\times3}-I_3}-\frac12E_{6\times4}\left(\frac13E_{4\times4}-I_4\right)E_{4\times6}\\ &=\pmatrix{E_{3\times3}-I_3&E_{3\times3}+I_3\\ E_{3\times3}+I_3&E_{3\times3}-I_3}-\frac12\left(\frac43-1\right)(4)E_{6\times6}\\ &=\pmatrix{\frac13E_{3\times3}-I_3&\frac13E_{3\times3}+I_3\\ \frac13E_{3\times3}+I_3&\frac13E_{3\times3}-I_3}.\\ \end{aligned} Since $E_{3\times3}$ is similar to $\operatorname{diag}(3,0,0)$, $S$ is similar to \begin{aligned} \pmatrix{\frac13\operatorname{diag}(3,0,0)-I_3&\frac13\operatorname{diag}(3,0,0)+I_3\\ \frac13\operatorname{diag}(3,0,0)+I_3&\frac13\operatorname{diag}(3,0,0)-I_3} =\pmatrix{\operatorname{diag}(0,-1,-1)&\operatorname{diag}(2,1,1)\\ \operatorname{diag}(2,1,1)&\operatorname{diag}(0,-1,-1)}, \end{aligned} which, in turn, is similar to $$ \pmatrix{0&2\\ 2&0}\oplus\pmatrix{1&-1\\ -1&1}\oplus\pmatrix{1&-1\\ -1&1}. $$ As each sub-block $\pmatrix{1&-1\\ -1&1}$ is singular, $S$ has two zero eigenvalues. Hence by Sylvester's law of inertia, $M$ has two zero eigenvalues too.

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Replace the last row by one-sixth of the sum of the last six rows; so the last row is now the all $1$'s vector. Subtract it from the rows $5,6,7,8,9$. The new rows $5$ and $8$ are negatives of each other; replace row $5$ by their sum. The same is true of rows $6$ and $9$, so replace row $6$ by their sum. You need not carry the Guassian reduction any further as we now have two rows of zeros.

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