# Why can we use the identity matrix when defining the characteristic polynomial?

We begin with $$\lambda v = Tv$$ where $$\lambda$$ is an eigenvalue, $$v$$ is an eigenvector, and $$T$$ is the transformation in question. We state $$\lambda v - Tv = 0$$ We then must state $$(\lambda I - T)v = 0$$What allows us to bring in $$I$$, the identity matrix? I understand that subtraction of a linear transformation (or matrix) from a scalar is ill-defined, but what permits us to multiply $$\lambda$$ by $$I$$? Would a more rigorous evaluation be \begin{align*} \lambda v - Tv &= \lambda Iv - Tv \\ &= (\lambda I - T)v \end{align*} where the identity matrix provides a multiplicative identity of some sorts? I understand that $$I \in M_{n \times n}(\mathbb{F})$$, so I am wondering how it can be the identity for $$v \in M_{n \times 1}(\mathbb{F})$$?

The reason I ask is that I am proving Cayley-Hamilton for a diagonal matrix, and I have come upon the following:

\begin{align*} \chi(t) &= det(D-tI) \\ &= \prod_{i=1}^{n}d_{ii}-t \\ &= (d_{11} - t)(d_{22} - t) \cdots (d_{nn} - t) \end{align*}

Consider \begin{align*} \chi([D]) &= (d_{11}I - D)(d_{22}I - D) \cdots (d_{nn}I - D) \end{align*}

Where does the $$I$$ come from?

The underlying question is really this:

I understand that $$I \in M_{n \times n}(\mathbb{F})$$, so I am wondering how it can be the identity for $$v \in M_{n \times 1}(\mathbb{F})$$?

• I suppose the simple answer of factoring $v$ on the right is it... May 27, 2023 at 22:52
• Guess I don't understand what you find confusing about $\,I v = v\,$.
– dxiv
May 27, 2023 at 22:59
• It is mostly the last bit, about Cayley-Hamilton @dxiv May 27, 2023 at 23:03
• "Substituting" $D$ into the characteristic polynomial is done by replacing $t^n$ with $D^n$. $D^0 = I$ so we replace $d_{ii}t^0$ with $d_{ii}I$ May 27, 2023 at 23:07
• So it is as trivial as simply writing it as $(d_{11}t^0 - t)(d_{22}t^0 - t) \cdots (d_{nn}t^0 - t)$ @OgglieOstrich ? May 27, 2023 at 23:11

You have several questions; I will just try to answer the main one:

"I understand that $$I \in M_{n \times n}(\mathbb{F})$$, so I am wondering how it can be the identity for $$v \in M_{n \times 1}(\mathbb{F})$$?"

First, in general, vector spaces (in your case $$M_{n \times 1}(\mathbb F)$$ or better known as $$\mathbb F^n$$) don't come with multiplicative identities. In fact vector spaces don't even need any multiplication within itself; it just needs to be an abelian group with a single operation reminiscent of addition. This is the case with $$\mathbb F^n$$. It has a well-defined vector addition and it satisfies a bunch of rules on how the field $$\mathbb F$$ acts on it. There is no unified notion of multiplication in $$\mathbb F^n$$.

The situation you are seeing with $$I \in M_{n \times n}(\mathbb F)$$ acting on $$v \in \mathbb F^n$$ like an "identity" is an entirely new phenomenon that $$\mathbb F^n$$ enjoys in addition to its the vector space properties. $$M_{n \times n}(\mathbb F)$$ forms a structure called a ring, which is a crippled field: both addition and multiplication are defined in it but division may not always be possible.

Just like the field $$\mathbb F$$ acts on $$\mathbb F^n$$, through scalar multiplication, to give it its vector space property, the ring $$M_{n \times n}(\mathbb F)$$ also acts on $$\mathbb F^n$$, through left matrix-vector multiplication, to give it a separate property called a left module. One of the central requirements of a left module is exactly what you noticed: that the identity of the ring must also act like an identity on the module elements.

• Can I ask if this means that an element of $\mathbb{F}$ is both a left and right module? For example, if $[A] = [a_{ij}]$ then $c[A] = [A]c$ for $c \in \mathbb{F}$? May 28, 2023 at 0:11
• @user129393192 by default scalar multiplication is defined as acting on the left but yes it can also be defined as acting on the right through that equation. May 28, 2023 at 0:21
• I see. I ask because I want to know if something like this is possible: $(d_{11}[Q]^{-1}[I][Q] - [Q]^{-1}[D][Q]) = [Q]^{-1}(d_{11}[I] - [D])[Q]$. May 28, 2023 at 0:30
• @user129393192 Don't see anything wrong with that May 28, 2023 at 0:49