# Multiplicity of factors of a determinant by observation

Consider the determinant

$$\Delta = \begin{vmatrix} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{vmatrix}$$

It can be explicitly shown to be $$\Delta = (x-1)^2(x+9)$$

Suppose we want to write down this value without direct calculation.

On putting $$x=1$$, first column, $$C_1$$, becomes a scalar multiple of other columns, $$C_2$$ as well as $$C_3$$ and along the way so do $$C_2$$ and $$C_3$$, making the determinant $$\Delta = 0$$.

What can be concluded from this? Can we say that $$(x-1)^2$$ is a factor of $$\Delta$$, i.e., with multiplicity $$2$$?

But clearly $$(x-1)^3$$ not a factor of $$\Delta$$.

Can multiplicity of $$(x-x_0)$$ in $$\Delta$$ can be attempted by such an observation? If so, what is the highest power of $$(x-x_0)$$ that can be concluded correctly for $$n \times n$$ determinant when $$k$$ out of $$n$$ columns are scalar multiples of each other on substituting $$x=x_0$$?

Please explain for the above $$3 \times 3$$ case before $$n \times n$$ case. Your help is appreciated.

• What are $C_1, C_2, C_3$? Oct 18 at 16:10
• Then you should add this information to your question. And the columns are not identical, only scalar multiples. Oct 18 at 16:13
• Are you familiar with the idea of eigenvalues and their multiplicity? Oct 18 at 16:14
• @PaulFrost Yes I missed that, I edited now. Oct 18 at 16:16
• @MishaLavrov Not much, but you can explain. I'll look up/ask anything I don't understand. Oct 18 at 16:17

The rank of the matrix you get is what gives you hints about how many times a root appears. In general, the rule is:

If setting $$x=x_0$$ gives us an $$n \times n$$ matrix with rank $$n-k$$, then there is a factor of $$(x-x_0)^k$$ in the determinant.

This applies to any matrix where the entries are linear expressions in $$x$$ (such as $$x+1$$ or $$3$$).

In your example, setting $$x=1$$ gives us a $$3 \times 3$$ matrix with rank $$1$$: all rows are multiples of the same row vector. Since $$1$$ can be written as $$3-2$$, we know that there is a factor of $$(x-1)^2$$ in the determinant.

Let me try to explain the logic behind the rule in the $$3 \times 3$$ example. The argument generalizes, but the generalization is harder to read.

Suppose we use a different variable in each row: for example, $$\Delta(x,y,z) = \begin{vmatrix} x+1 & 3 & 5 \\ 2 & y+2 & 5 \\ 2 & 3 & z+4 \end{vmatrix}$$ The polynomial $$\Delta(x,y,z)$$ cannot possibly have a factor of $$x^2$$, $$y^2$$, or $$z^2$$ appear anywhere. (This is true here because there's only one of each variable. It's true in general if each variable only appears in one row.) This means that we can write $$\Delta(x,y,z)=A + B(x-1) + C(y-1) + D(z-1) + E(x-1)(y-1) + F(x-1)(z-1) + G(y-1)(z-1) + H(x-1)(y-1)(z-1)$$ where $$A,B,C,D,E,F,G,H$$ are constants. Because $$\Delta(1,1,1)=0$$, we know $$A=0$$.

Because the matrix with $$x=y=z=1$$ has rank $$1$$, we know that actually $$\Delta(1,1,z)=0$$ for any $$z$$: if we set $$x=y=1$$, the first two rows are already multiples of each other, so the determinant is $$0$$. Similarly, $$\Delta(1,y,1)=0$$ for any $$y$$ and $$\Delta(x,1,1)=0$$ for any $$x$$. This tells us that in the expansion above, $$A=B=C=D=0$$.

Therefore $$\Delta(x,y,z) = E(x-1)(y-1) + F(x-1)(z-1) + G(y-1)(z-1) + H(x-1)(y-1)(z-1)$$ which means that $$\Delta(x,x,x) = (E+F+G)(x-1)^2 + H(x-1)^3$$ which is divisible by $$(x-1)^2$$.

• "Since $1$ can be written as $3−2$". So it is useful to keep the powerful Rank-Nullity theorem in mind. Here $1$ is rank, $2$ is nullity corresponding to $x_0=1$, right? Thank you for very nice explanation! Oct 18 at 17:21
• Yep, $2$ is the nullity; I phrased this in terms of the rank because thinking of the rank comes more naturally to me, but the nullity is actually the parameter we care about. Oct 18 at 17:23
• Indeed, one looks at the rank first and then concludes about corresponding nullity. Thanks again for clever and neat explanation! Oct 18 at 18:07

Consider the matrix $$A= -\begin{bmatrix} 1 & 3 & 5\\ 2 & 2 & 5\\ 2 & 3 & 4 \end{bmatrix}\ .$$ The OP wants to compute the characteristic polynomial $$P_A$$ of this matrix, computed in $$x$$, so $$P_A(x)$$.

The factor $$(x-1)$$ in $$P_A$$ was seen because $$P_A(1)=\det(I-A)=0$$, because we have a column relation for $$I-A$$, which is explicitly $$I-A= \begin{bmatrix} 2 & 3 & 5\\ 2 & 3 & 5\\ 2 & 3 & 5 \end{bmatrix} = \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} \begin{bmatrix} 2 & 3 & 5 \end{bmatrix}\ .$$ But we can exhibit with bare eyes "not only one" linear relation between the column vectors $$C_1$$, $$C_2$$, $$C_3$$ of $$\pm A$$, but two of them, moreover two independent linear relations, e.g. $$3C_1-2C_2=0$$ and $$5C_1-2C_3=0$$. Equivalently, the two linearly independent column vectors $$v=[3\ -2\ 0]^T$$ and $$w=[5\ 0\ -2]^T$$ are eigenvectors for the eigenvalue $$1$$, i.e. $$(I-A)v=(I-A)w=0$$. So $$1$$ has an eigenspace of dimension at least $$2$$, i.e. is an eigenvalue of multiplicity at least two. So $$(x-1)^2$$ divides $$P_A(x)$$.

To see with bare eyes it is not three, there are some chances...

• Look for an other eigenvalue, we will have it soon, $$-9$$.
• The Jordan form of the quasi-projector $$(I-A)$$ is diagonal, and the kernel, the corresponding $$1$$-eigenspace does not have full dimension three, else we would have $$(I-A)=0$$.
• $$P_A(x)$$ is a polynomial in $$\Bbb Z[x]$$, so we may want to compute it modulo some small prime like two or three. It turns out that modulo three $$P_A(0)$$ is $$\pm\begin{vmatrix} 1 & 3 & 5\\ 2 & 2 & 5\\ 2 & 3 & 4 \end{vmatrix} \equiv \pm\begin{vmatrix} 1 & 0 & -1\\ * & 2 & *\\ -1 & 0 & 1 \end{vmatrix} \equiv 0\pmod3\ .$$ So $$P_A(x)=(x-1)^3$$ is excluded by passing to $$\Bbb Z/3$$. The above is easy for the bare eyes, too.

One sees more or less easily, that $$-9$$ is an eigenvector of $$A$$, because we consider $$(-9)I-A = \begin{bmatrix} -8 & 3 & 5\\ 2 & -7 & 5\\ 2 & 3 & -5 \end{bmatrix}\ ,$$ and observe that the sum on each row is zero. (So the sum of the columns is the zero vector.) This means that $$x-(-9)=x+9$$ is also a factor of $$P_A(x)$$. This characteristic polynomial is monic, so $$P_A(x)=(x-1)^2(x+9)$$.

A comment complains that the eigenvalue $$-9$$ is "not easy". Well, it "is easy" - if we manage to observe that $$A$$ has same sum in each row. But OK, assume this is too complicated. So far we know that $$P_A(x)=(x-1)^2(x-a)$$ with some eigenvalue $$a$$ to be still determined. Then just compute $$P_A(0)=\det(-A)$$ or some other value of $$P_A$$.

For the "general" case of an $$n\times n$$ matrix we have the same situation, compute (generalized) eigenspaces for a given eigenvalue. (Either with bare eyes, or by solving linear systems of equations explicitly.)

• Thank you for answering. Can you explain "To see it is not three, it is enough to take some vector like [1 0 0]$^T$ which is not in the kernel of $(A+I)$"? Also, $-9$ is not easy enough to be seen by naked eyes :) Oct 18 at 17:29
• @MyMolecules I have edited the answer... The "general case" is of course too general, the question did not make it clear, and it cannot do it. But in cases where some columns are repeated, as above then adding subtracting the same number on the diagonal, there is a corresponding multiplicity of "the number" (taken with the right sign)... Oct 20 at 2:17
• Thanks! I now understand. Oct 20 at 12:24