# 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 at 22:52
• Guess I don't understand what you find confusing about $\,I v = v\,$.
– dxiv
May 27 at 22:59
• It is mostly the last bit, about Cayley-Hamilton @dxiv May 27 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 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 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 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 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 at 0:30
• @user129393192 Don't see anything wrong with that May 28 at 0:49