Proving a theorem in linear algebra

Question

Let $T:\mathbb{R}^2 \to \mathbb{R}^2$ be an invertible linear trasformation on $\mathbb{R}^2$ and Tx=x for all $x\in \mathbb{R}^2$ then $T=I$ where $I$ is the identity map on $\mathbb{R}^2$.

It seems to be trivial. But l don't know how to prove it.

Thanks for any help.

Answer 1

You could prove it by contradiction. Assume that $T \neq I$, then there exists $x\in\mathbb{R}^2$ such that $Tx \neq x$. This is impossible by hypothesis.

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PS : there is no need here for $T$ to be linear, nor invertible.

Answer 2

Take $T = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and use the condition $Tx = x$ twice: First for $x = (1,0)^t$, then for $x = (0,1)^t$. That way you get four equations that you can solve for the four unknowns $a,b,c,d$.

Answer 3

If $v\in\mathbb{R}^2$, then $T(v)=v=\operatorname{Id}(v)$. Therefore, $T=\operatorname{Id}$.

Answer 4

We have to show that the following two functions$$T:\mathbb{R}^2 \to \mathbb{R}^2$$ and $$I:\mathbb{R}^2 \to \mathbb{R}^2$$ are equal.

Notice that two functions are equal if and only if they have the same domain, the same codomain and the same value at each point of the domain.

Both T and I have the same domain and the same codomain as $\mathbb{R}^2$.

Also, $T(X)=I(X)=X$ for any $X\in \mathbb{R}^2$

Thus the functions T and I are equal.

Answer 5

As stated this is indeed trivial: Saying that $Tx=x$ for every $x$ is exactly what it means to say $T$ is the identity map; there's nothing to prove, because the conclusion is given.

That's so trivial that it's hard to believe the question was stated correctly. Was the original question asking about linear maps, or was it a question about matrices? A matrix is not literally the same thing as a linear map - in particular the following question is not the same as the question you asked:

Different question: Suppose $A$ is a $2\times 2$ matrix and $Ax=x$ for every $x\in\Bbb R^2$. Show that $A$ is the identity matrix (that is, show that $A=\begin{bmatrix}1&0\\0&1\end{bmatrix}$.)

Although this is still very easy, it's not trivial in the sense that the question you asked is trivial, here there is something to be proved. (How do you prove this? The answer by flawr actually gives a solution to this question, not to the question you askked.)