Problem:
Let $V$ be a finite-dimensional vector space, and let $\lambda$ be any scalar. For any ordered basis $\beta$ for $V$, prove that $[\lambda I_V]_{\beta}=\lambda I$. Then compute the characteristic polynomial of $\lambda I_V$ and show that $\lambda I_V$ is diagonalizable and has only one eigenvalue.
Now I proceed to compute $[\lambda I_V]_{\beta}$. I am stuck here since $V$ is not given. But I suppose $[\lambda I_V]_{\beta}=\lambda [I_V]_{\beta}=\lambda I$.
But now suppose $f_T(t)=\det ([T]_{\beta}-\lambda I_V)=1-0^n(t-\lambda)^n=(\lambda-t)^n$.
Also, note that $[\lambda I_V]_{\beta}=\lambda I$ is diagonal.
Then $\lambda I$ is diagonalizable.
Then $f_T(t)=0\implies (\lambda-t)^n=0\implies t=\lambda$.
It follows that $\lambda I$ has only one eigenvalue, which is $\lambda$.
Could anyone help me to have a look at whether my proof is legal?
You don’t have to suppose $[\lambda I_V]_{\beta}=\lambda [I_V]_{\beta}=\lambda I$. Let’s prove it for general case. Since $[(c\cdot T)(\alpha_j)]_B=[c\cdot T(\alpha_j)]_B=c\cdot [T(\alpha_j)]_B$, we have $[c\cdot T]_B=c\cdot [T]_B$. Matrix representation of an identity map on $V$ with respect to any ordered basis is $I_n$.
By definition, characteristic polynomial of $\lambda I_V$ is characteristic polynomial of $\lambda I$. Let $A=\lambda I$. Then characteristic polynomial of $A$ is $f_A:F\to F$ such that $f_A(t)=\text{det}(A-tI)$, $\forall t\in F$. By distributive property, $A-tI=\lambda I-tI=(\lambda-t)I$. So $\text{det}(A-tI)=\text{det}[(\lambda-t)I]$. Since $\text{det}$ is $n$-linear, we have $\text{det}[(\lambda-t)I]=(\lambda -t)^n\text{det}(I)=(\lambda -t)^n$. Thus characteristic polynomial of $\lambda I_V$ is $(\lambda -t)^n$.
I’m using following definition of diagonalizable, $T\in L(V,V)$ is diagonalizable$\iff$$\exists$ basis $B$ of $V$ such that $[T]_B$ is a diagonal matrix. Since $[\lambda I_V]_{\beta}=\lambda I$ and $\lambda I$ is diagonal matrix, we have $\lambda I_V$ is diagonalizable. It’s easy to check, $\lambda$ is the only eigenvalue of $\lambda I_V$.