Let X∗ denote the dual space of the vector space X. If x ∈ R
n
, define φx ∈ (R
n
)
∗
by φx(y) := hx, yi

for all y ∈ R
n
. Define T : R
n → (R
n
)
∗
by T(x) = φx. Show that T is a 1-1 linear transformation and

conclude that every φ ∈ (R
n
)
∗
is φx for a unique x ∈ R
n

To solve this problem, we need to establish the linearity and injectivity of the map $ T: \mathbb{R}^n \to (\mathbb{R}^n)^* $ defined by $ T(x) = \phi_x $, where $ \phi_x(y) = \langle x, y \rangle $ for all $ y \in \mathbb{R}^n $. Then, we'll conclude that every $ \phi \in (\mathbb{R}^n)^* $ corresponds to a unique vector $ x \in \mathbb{R}^n $.

### Step 1: Linearity of $ T $

We need to show that $ T $ is a linear transformation. That is, for all $ x_1, x_2 \in \mathbb{R}^n $ and scalars $ \alpha, \beta \in \mathbb{R} $, we must show that:
\[
T(\alpha x_1 + \beta x_2) = \alpha T(x_1) + \beta T(x_2).
\]

By the definition of $ T $ and $ \phi_x $:
\[
T(\alpha x_1 + \beta x_2) = \phi_{\alpha x_1 + \beta x_2}.
\]
Now, for any $ y \in \mathbb{R}^n $:
\[
\phi_{\alpha x_1 + \beta x_2}(y) = \langle \alpha x_1 + \beta x_2, y \rangle.
\]
Using the bilinearity of the inner product:
\[
\langle \alpha x_1 + \beta x_2, y \rangle = \alpha \langle x_1, y \rangle + \beta \langle x_2, y \rangle.
\]
Therefore:
\[
\phi_{\alpha x_1 + \beta x_2}(y) = \alpha \phi_{x_1}(y) + \beta \phi_{x_2}(y).
\]
This implies:
\[
\phi_{\alpha x_1 + \beta x_2} = \alpha \phi_{x_1} + \beta \phi_{x_2},
\]
which means:
\[
T(\alpha x_1 + \beta x_2) = \alpha T(x_1) + \beta T(x_2).
\]
Thus, $ T $ is a linear transformation.

### Step 2: Injectivity of $ T $

We now show that $ T $ is injective (one-to-one). To do this, assume $ T(x) = T(x') $ for some $ x, x' \in \mathbb{R}^n $. This implies:
\[
\phi_x = \phi_{x'}.
\]
For all $ y \in \mathbb{R}^n $, this means:
\[
\langle x, y \rangle = \langle x', y \rangle.
\]
The above equation holds for all $ y $, which implies that $ x = x' $ because the inner product $ \langle x - x', y \rangle = 0 $ for all $ y $ only if $ x - x' = 0 $, meaning $ x = x' $.

Since $ T(x) = T(x') $ implies $ x = x' $, $ T $ is injective.

### Step 3: Surjectivity of $ T $

Finally, we need to show that $ T $ is surjective, meaning that every $ \phi \in (\mathbb{R}^n)^* $ is of the form $ \phi = \phi_x $ for some $ x \in \mathbb{R}^n $.

Since $ \phi $ is a linear functional on $ \mathbb{R}^n $, by the Riesz Representation Theorem (which applies in finite-dimensional spaces like $ \mathbb{R}^n $), there exists a unique vector $ x \in \mathbb{R}^n $ such that:
\[
\phi(y) = \langle x, y \rangle
\]
for all $ y \in \mathbb{R}^n $. Thus, $ \phi = \phi_x $.

### Conclusion

We've shown that $ T $ is a bijective (both injective and surjective) linear transformation, which implies that $ T $ is an isomorphism between $ \mathbb{R}^n $ and $ (\mathbb{R}^n)^* $. Therefore, every $ \phi \in (\mathbb{R}^n)^* $ is of the form $ \phi_x $ for a unique $ x \in \mathbb{R}^n $.

This concludes the proof.

Would you like more details on any part of this explanation? Here are some related questions:

1. What is the Riesz Representation Theorem, and how does it apply here?
2. How does linearity affect the properties of vector spaces?
3. Can you provide an example of a linear map similar to $ T $ in another context?
4. What are dual spaces, and why are they important in linear algebra?
5. How would this proof change if we considered an infinite-dimensional vector space?

**Tip:** When dealing with linear transformations, always check for both injectivity and surjectivity to determine if the transformation is an isomorphism.

Explore the linear transformation $T$ from $\mathbb{R}^n$ to the dual space $(\mathbb{R}^n)^*$, showing its injectivity, surjectivity, and the application of the Riesz Representation Theorem. Understand how every linear functional in $(\mathbb{R}^n)^*$ corresponds uniquely to a vector in $\mathbb{R}^n$.

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