Suppose V is finite-dimensional. Prove that every linear map on a subspace of V can be extended to a linear map on V. In other words, show that if U is a subspace of V and S ∈ L(U,... Suppose V is finite-dimensional. Prove that every linear map on a subspace of V can be extended to a linear map on V. In other words, show that if U is a subspace of V and S ∈ L(U, W), then there exists T ∈ L(V, W) such that Tu = Su for all u ∈ U Read more

Steps to Solve

1. Choose a basis for U

Let $U$ be a subspace of the finite-dimensional vector space $V$. Let ${u_1, u_2, ..., u_k}$ be a basis for $U$.

2. Extend the basis of U to a basis of V

Since $U$ is a subspace of $V$, we can extend the basis of $U$ to a basis of $V$. Let ${u_1, u_2, ..., u_k, v_{k+1}, ..., v_n}$ be a basis for $V$, where $n$ is the dimension of $V$.

3. Define the linear map T on the basis of V

We are given a linear map $S: U \to W$. We want to define a linear map $T: V \to W$ such that $T(u) = S(u)$ for all $u \in U$. We know the values of $S$ on the basis vectors ${u_1, u_2, ..., u_k}$ of $U$. Define $T$ on the basis of $V$ as follows:

• $T(u_i) = S(u_i)$ for $i = 1, 2, ..., k$
• $T(v_i) = 0$ for $i = k+1, ..., n$ (We can choose any values for $T(v_i)$, but 0 is a simple choice.)

4. Extend T linearly to all of V

Since we have defined $T$ on a basis of $V$, we can extend it linearly to all of $V$. For any vector $v \in V$, we can write $v$ as a linear combination of the basis vectors: $$ v = a_1u_1 + a_2u_2 + ... + a_ku_k + b_{k+1}v_{k+1} + ... + b_nv_n $$ Define $T(v)$ as: $$ T(v) = a_1T(u_1) + a_2T(u_2) + ... + a_kT(u_k) + b_{k+1}T(v_{k+1}) + ... + b_nT(v_n) $$ Substituting the values of $T(u_i)$ and $T(v_i)$ we defined earlier: $$ T(v) = a_1S(u_1) + a_2S(u_2) + ... + a_kS(u_k) + b_{k+1}(0) + ... + b_n(0) $$ $$ T(v) = a_1S(u_1) + a_2S(u_2) + ... + a_kS(u_k) $$

5. Verify that T is linear

Let $v, w \in V$ and $c$ be a scalar. We need to show that $T(v + w) = T(v) + T(w)$ and $T(cv) = cT(v)$. Let $v = \sum_{i=1}^{k} a_iu_i + \sum_{i=k+1}^{n} b_iv_i$ and $w = \sum_{i=1}^{k} c_iu_i + \sum_{i=k+1}^{n} d_iv_i$. Then, $v + w = \sum_{i=1}^{k} (a_i + c_i)u_i + \sum_{i=k+1}^{n} (b_i + d_i)v_i$. So, $T(v + w) = \sum_{i=1}^{k} (a_i + c_i)S(u_i) = \sum_{i=1}^{k} a_iS(u_i) + \sum_{i=1}^{k} c_iS(u_i) = T(v) + T(w)$. And, $cv = \sum_{i=1}^{k} (ca_i)u_i + \sum_{i=k+1}^{n} (cb_i)v_i$. So, $T(cv) = \sum_{i=1}^{k} (ca_i)S(u_i) = c\sum_{i=1}^{k} a_iS(u_i) = cT(v)$. Thus, $T$ is linear.

6. Verify that T agrees with S on U

Let $u \in U$. Then $u$ can be written as a linear combination of the basis vectors of $U$: $$ u = a_1u_1 + a_2u_2 + ... + a_ku_k $$ Then, $T(u) = a_1T(u_1) + a_2T(u_2) + ... + a_kT(u_k) = a_1S(u_1) + a_2S(u_2) + ... + a_kS(u_k) = S(a_1u_1 + a_2u_2 + ... + a_ku_k) = S(u)$. Thus, $T(u) = S(u)$ for all $u \in U$.