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Exercise.

Suppose $V$ is finite-dimensional and $T \in \mathcal{L}(V)$. Prove that $T$ has the same matrix with respect to every basis of $V$ if and only if $T$ is a scalar multiple of the identity operator.

Source.

Linear Algebra Done Right, Sheldon Axler, 4th Edition, Section 3D, exercise number 19.

Where I'm stuck.

I believe I was able to prove the backward direction (see section below this one for that proof). I'm having trouble proving the forward direction. Here's what I have tried:

Assuming $T$ has the same matrix with respect to every basis of $V$, that means that the entry in row $j$, column $k$ of $\mathcal{M}(T, (v_1,\ldots,v_n))$ is equal to the entry in row $j$, column $k$ of $\mathcal{M}(T, (u_1,\ldots,u_n))$, where $v_1,\ldots,v_n$ and $u_1,\ldots,u_n$ are any bases of $V$.

This fact, along with the way the entries are defined by $T$, implies

$$ Tv_k = A_{1,k}v_1 + \cdots + A_{n,k}v_k \\ Tu_k = A_{1,k}u_1 + \cdots + A_{n,k}u_k $$

I need to somehow use this to show $T = \alpha I$ for some $\alpha \in \mathbf{F}$, or equivalently, $Tv = \alpha v$ for all $v \in V$.

What's the strategy here? I've tried every algebraic manipulation possible, but it gets messy and leads nowhere. I suspect I'm missing a key point here.

My proof for backward direction (I'm not expecting anyone to verify it's correct, but be my guest).

We prove that if $T$ is a scalar multiple of the identity operator, then $T$ has the same matrix with respect to every basis of $V$

Let $\alpha \in \mathbf{F}$ and let $T = \alpha I$ where $I$ is the identity operator. Let $v_1, \ldots v_n$ be any basis of $V$. We want to show that the entries of $\mathcal{M}(T, (v_1, \ldots,v_n))$ are fixed values, that is, they do not depend on the basis.

We have that

\begin{align} \mathcal{M}(T, (v_1,\ldots,v_n)) &= \mathcal{M}(\alpha I, (v_1,\ldots,v_n)) \\ &= \alpha\mathcal{M}(I, (v_1,\ldots,v_n)) \end{align}

The entries in column $k$ of the $n$-by-$n$ matrix $\alpha\mathcal{M}(I, (v_1,\ldots,v_n))$, and hence $\mathcal{M}(T, (v_1,\ldots,v_n))$, are defined by

$$ \alpha I(v_k) = \alpha A_{1,k}v_1 + \cdots + \alpha A_{n,k}v_n \tag{1} $$

But we know that $\alpha I(v_k) = \alpha v_k$ by definition of the identity operator. This fact, along with equation $(1)$, tells us that

$$ \alpha A_{j,k} = \begin{cases} \alpha,& j=k \\ 0, & j \neq k \end{cases} \tag{2} $$

for $1 \leq j \leq n$. The entries in $(2)$ imply that

$$ \mathcal{M}(T, (v_1,\ldots,v_n)) = \begin{pmatrix} \alpha & & 0 \\ & \ddots \\ 0 & & \alpha \end{pmatrix} $$

That is, $\mathcal{M}(T, (v_1,\ldots,v_n))$ will always have $\alpha$ in the diagonal, and $0$ everywhere else no matter which basis is chosen.

Paul Ash
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  • I suspect you have a typo in the first paragraph. With both bases ($v$’s and $u$’s it should be row $k$ column $j$. I suggested starting with a basis $v_{1},\dots,v_{n}$ and playing with it - what happens if you change the order of $v_{1},v_{2}$, say? What happens if you replace $v_{1}$ (and keep the other vectors the same) with $\lambda v_{1}$ for some $\lambda\in F$? – Espace' etale Feb 10 '24 at 07:37
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    Do you see the question is equivalent to this: https://math.stackexchange.com/q/27808/96384 – Torsten Schoeneberg Feb 10 '24 at 07:46
  • This is very confusing for me... I would take help of the equivalence relation called similarity... Matrices expressed using different bases are similar... The scalar multiples of identity are a class and they are unaffected by conjugation... – Nothing special Feb 10 '24 at 07:54
  • You are encountering difficulty with figuring out how to operationalize "$T$ is not a scalar multiple of the identity" in writing a proof. One possible strategy: see if you can show that this condition implies the existence of nonzero $v$ for which ${v, Tv}$ is linearly independent (so in particular: $Tv \neq v$ and the vectors $v$ and $Tv$ can be extended to a basis). With that in hand, examine what the matrix of $T$ would look like in a basis beginning $(v, Tv, \dots)$, versus a basis beginning $(Tv, v, \dots)$. [This hint is a more spoilery version of the hint in the first comment.] – leslie townes Feb 10 '24 at 09:00
  • @Espace'etale thank you, I fixed the typo. I will take your comment and everyone else's into account when taking another jab at the exercise. – Paul Ash Feb 11 '24 at 20:00

1 Answers1

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Hint:

  1. What happens to the matrix if you replace basis $\{v_1, v_2, \ldots, v_n\}$ with let's say $\{\alpha v_1, v_2, \ldots, v_n\}$ for $\alpha \not= 0$?

  2. What happens if you re-order the basis? In particular, let's say you exchange $v_i$ and $v_j$?

Tony Pizza
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