# How to calculate the multiplicity of the eigenvalue $0$?

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 $$2$$

• See if you can spot two linearly independent vectors in the nullspace of $M$. Sep 6 '20 at 6:06
• @JimmyK4542; But finding the null space is as complicated as finding the rank , isn't it? Sep 6 '20 at 6:08
• 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. Sep 6 '20 at 6:13
• 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. Sep 6 '20 at 6:19
• @ElliotG; I dont understand how symmetric can help here, could u elaborate please? Sep 6 '20 at 6:21

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 $$0$$.

• How did you know where to put in the $0s$ and $1s$ in the the entries $5,6,7,8,9,10$ Sep 6 '20 at 6:42
• 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 Sep 6 '20 at 6:43

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.

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.