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 $v \in M_{n \times 1}(\mathbb{F})$?

  • $\begingroup$ I suppose the simple answer of factoring $v$ on the right is it... $\endgroup$ May 27 at 22:52
  • $\begingroup$ Guess I don't understand what you find confusing about $\,I v = v\,$. $\endgroup$
    – dxiv
    May 27 at 22:59
  • $\begingroup$ It is mostly the last bit, about Cayley-Hamilton @dxiv $\endgroup$ May 27 at 23:03
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    $\begingroup$ "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 $d_{ii}I$ $\endgroup$ May 27 at 23:07
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    $\begingroup$ 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 ? $\endgroup$ May 27 at 23:11

1 Answer 1


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.

  • $\begingroup$ 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 $c \in \mathbb{F}$? $\endgroup$ May 28 at 0:11
  • $\begingroup$ @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. $\endgroup$
    – balddraz
    May 28 at 0:21
  • $\begingroup$ I see. I ask because I want to know if something like this is possible: $(d_{11}[Q]^{-1}[I][Q] - [Q]^{-1}[D][Q]) = [Q]^{-1}(d_{11}[I] - [D])[Q]$. $\endgroup$ May 28 at 0:30
  • $\begingroup$ @user129393192 Don't see anything wrong with that $\endgroup$
    – balddraz
    May 28 at 0:49

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