Linear Applications between Vector Spaces

Question

Let V be a vector space on R of finite dimension not null and f be an endomorphism, such that: $\frac{1}{4}f^2$ is the identity in V, in other words: $$f(f(v))=4v, \forall v\in V$$

1. To prove that $f$ is injective and surjective
2. Calculate the inverse matrix of the matrix associated with the application in any base
3. Test $f$ is diagonalizable

Could you help with this exercise? Thanks.

Answer 1

$ f \circ f$ is bijective so is $f$

$\frac{1}{4}f \circ f = id$ so if the matrix of $f$ in any basis is $A$ its inverse is $\frac{1}{4}A$

$\frac{1}{4}X^2 -1 = \frac{1}{4}(X-2)(X+2)$ so $f$ is diagonalizable

Answer 2

Let $\;g:V\to V\;$ be the map $\;g(v)=\frac14f(v)=f\left(\frac14v\right)\;$, then:

=== Prove $\;g\;$ is a linear map, and

=== $\;g\circ f(v)=g(f(v))=\frac14f(f(v))=v\;,\;\;f\circ g(v)=f(g(v))=f\left(\frac14f(v)\right)=\frac14f(f(v))=v\;$

=== Deduce $\;g=f^{-1}=\;$ the inverse map of $\;f\;$ , and get from here $\;f\;$ is bijective. The rest of the exercise is standard after this.

Answer 3

To expand a bit on Lucas' answer - the fact that $(T-I)(T+I)=0$ implies that $T$ is diagonalizable may not be familiar. It must be clear from things covered in "advanced" linear algebra courses, but it's certainly not contained in MATH 3013 here at OSU.

Theorem If $T:V\to V$ is linear and $(T-I)(T+I)=0$ then $T$ is diagonalizable.

Proof: For every $x\in V$ we have $x=x_1-x_2$, where $$x_1=\frac12(Tx+x)$$ and $$x_2=\frac12(Tx-x).$$ But $$Tx_1-x_1=\frac12(T-I)(T+1)x=0,$$so $x_1$ is an eigenvector of $T$. Similarly $x_2$ is an eigenvector, so the eigenspaces of $T$ span $V$.

One can show similarly that if $\prod_{j=1}^n(T-\alpha_jI)=0$ where the $\alpha_j$ are distinct then $T$ is diagonalizable, although it requires a bit of algebra: Define polynomials $p_k$ by $$p_k(t)=\prod_{j\ne k}(t-\alpha_j).$$A "bit of algebra" ($F[x]$ is a PID...) shows that there exist polynomials $q_j$ with $$\sum q_jp_j=1.$$So if you let $x_j=q_j(T)p_j(T)x$ then $$x=\sum x_j,$$and as above $$(T-\alpha_jI)x_j=0,$$so each $x_j$ is an eigenvector.