Prove that Diag$_n(\mathbb{R})$ is isomorphic to $\mathbb{R}^n$

Question

Prove that each of the following spaces are isomorphic to $\mathbb{R}^d$ for some $d$ and compute this number.

(a) Diag$_n(\mathbb{R})$, that is diagonal matrices of order $n$

(b) UT$_n(\mathbb{R})$, that is, upper triangular matrices of order n

(c) $\{f \in P_5(\mathbb{R}): f(x) = f(-x)\}$ that is, even polynomials of degree $\le 5$

My attempt:

For (a) I was thinking to give a linear transformation $T: Diag_n(\mathbb{R}) \rightarrow \mathbb{R}^n$ and show that $T$ is one to one and due to $Diag_n(\mathbb{R})$ and $\mathbb{R}^n$ have the same dimension this implies that $T$ is onto. In order to show that $T$ is one to one I only have to prove that $ker(T)= \{0\}$. Is this correct? Is there other simpler way to prove it?

For (b) a similar procedure with $T: UT_n(\mathbb{R}) \rightarrow \mathbb{R}^{n(n+1)/2}$

I dont know if I could prove (c) in a similar way. Any ideas?

Answer 1

You can't do that. First, finding a linear map $T:{\rm Diag}_n(\mathbb{R})\rightarrow\mathbb{R}^n$ is a good start, if you manage to show it is an isomorphism, then you're done. In order to do that, you have to prove that $T$ is one-to-one, meaning that ${\rm ker}\,T=\{0\}$ but you can't conclude using an argument of dimension, because that would mean that you use the fact that ${\rm Diag }_n(\mathbb{R})$ and $\mathbb{R}^n$ have the same dimension to conclude that indeed they have the same dimension ! You have to prove that $T$ is also onto and only then after you have proven that it is an isomorphism. As for $(b)$ you can apply the same strategy. For $(c)$, you can find a basis of the space, try to understand what is going on if $f=X^n$.

Answer 2

In general, the following fact might be useful, if you are not already familiar with it.

Fact. Let $k$ be a field. If $V$ and $W$ are $k$-vector spaces of the same finite dimension, then $V$ and $W$ are isomorphic.

Proof. Let $n$ be the common dimension of $V$ and $W.$ Let $\mathscr B = \{v_1, \dots, v_n\}$ be a basis for $V,$ and let $\mathscr B' = \{w_1, \dots, w_n\}$ be a basis for $W.$ Consider the map $T : V \to W$ defined by $T(v_i) = w_i$ (and extend linearly to determine how $T$ acts on an arbitrary vector of $V$). Observe that if $v$ is in $\ker(T),$ then $T(v) = 0.$ By writing $v = a_1 v_1 + \cdots + a_n v_n,$ this means that $$a_1 w_1 + \cdots + a_n w_n = a_1 T(v_1) + \cdots + a_n T(v_n) = T(a_1 v_1 + \cdots + a_n v_n) = T(v) = 0.$$ But the vectors $w_1, \dots, w_n$ are linearly independent, hence $v = 0.$ Consequently, $T$ is injective. By the Rank-Nullity Theorem, we conclude that $\operatorname{rank}(T) = \dim_k(W),$ i.e., $T$ is surjective. QED.

Ultimately, this reduces the problem at hand to computing a basis of each vector space. Each of the $\mathbb R$-vector spaces has finite dimension, hence each one is isomorphic to $\mathbb R^{\dim_{\mathbb R}(V)}.$

Below, I have provided some hints toward finding a basis for each of the vector spaces at hand.

Observe that any matrix in $\operatorname{Diag}_n(\mathbb R)$ can be written as $a_1 D_1 + \cdots + a_n D_n,$ where the matrix $D_i$ has a $1$ in position $(i, i)$ and $0$s elsewhere.

Use a similar argument to find a basis of $\operatorname{UT}_n(\mathbb R).$ Observe that any $n \times n$ upper-triangular matrix must have $0$s below the main diagonal, so the $1$s can only lie on the main diagonal and above. How many entries are there on the main diagonal and above? (Hint: you're already correct.)

Every even polynomial of degree $\leq 5$ can be written as $f(x) = a x^4 + b x^2 + c$ for some real numbers $a, b,$ and $c.$ (Why?)