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spruce
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Let V and W be nonzero vector spaces over the same field, and let $T:V \rightarrow W$ be linear map. Prove that $T^t$ is onto if and only if T is one-to-one.

Here is the theorem: Let V and W be finite-dimensional vector spaces over F with ordered bases $\beta$ and $\gamma$, respectively. For any linear map $T:V \rightarrow W$, the mapping $T^t:W^* \rightarrow V^*$ defined by $T^t(g)=gT$ for all $g \in W^*$ is a linear map with the property that $[T^t]^{{\beta}^*}_{{\gamma}^*}=([T]^\gamma_\beta)^t$. I

I want to prove the reverse direction.

Assume T is one-to-one, and weI want to prove that $T^t$ is onto. How am I supposed to construct a linear functional such that T is one-to-one? Any hint is appreciated.

Let V and W be nonzero vector spaces over the same field, and let $T:V \rightarrow W$ be linear map. Prove that $T^t$ is onto if and only if T is one-to-one.

Here is the theorem: Let V and W be finite-dimensional vector spaces over F with ordered bases $\beta$ and $\gamma$, respectively. For any linear map $T:V \rightarrow W$, the mapping $T^t:W^* \rightarrow V^*$ defined by $T^t(g)=gT$ for all $g \in W^*$ is a linear map with the property that $[T^t]^{{\beta}^*}_{{\gamma}^*}=([T]^\gamma_\beta)^t$. I want to prove the reverse direction.

Assume T is one-to-one, and we want to prove that $T^t$ is onto. How am I supposed to construct a linear functional such that T is one-to-one? Any hint is appreciated.

Let V and W be nonzero vector spaces over the same field, and let $T:V \rightarrow W$ be linear map. Prove that $T^t$ is onto if and only if T is one-to-one.

Here is the theorem: Let V and W be finite-dimensional vector spaces over F with ordered bases $\beta$ and $\gamma$, respectively. For any linear map $T:V \rightarrow W$, the mapping $T^t:W^* \rightarrow V^*$ defined by $T^t(g)=gT$ for all $g \in W^*$ is a linear map with the property that $[T^t]^{{\beta}^*}_{{\gamma}^*}=([T]^\gamma_\beta)^t$.

I want to prove the reverse direction.

Assume T is one-to-one, and I want to prove that $T^t$ is onto. How am I supposed to construct a linear functional such that T is one-to-one? Any hint is appreciated.

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spruce
spruce
  • 705
  • 3
  • 20

Let V and W be nonzero vector spaces over the same field, and let $T:V \rightarrow W$ be linear map. Prove that $T^t$ is onto if and only if T is one-to-one.

IHere is the theorem: Let V and W be finite-dimensional vector spaces over F with ordered bases $\beta$ and $\gamma$, respectively. For any linear map $T:V \rightarrow W$, the mapping $T^t:W^* \rightarrow V^*$ defined by $T^t(g)=gT$ for all $g \in W^*$ is a linear map with the property that $[T^t]^{{\beta}^*}_{{\gamma}^*}=([T]^\gamma_\beta)^t$. I want to prove the reverse direction.

Assume T is one-to-one, and we want to prove that $T^t$ is onto. How am I supposed to construct a linear functional such that T is one-to-one? Any hint is appreciated.

Let V and W be nonzero vector spaces over the same field, and let $T:V \rightarrow W$ be linear map. Prove that $T^t$ is onto if and only if T is one-to-one.

I want to prove the reverse direction.

Assume T is one-to-one, and we want to prove that $T^t$ is onto. How am I supposed to construct a linear functional such that T is one-to-one? Any hint is appreciated.

Let V and W be nonzero vector spaces over the same field, and let $T:V \rightarrow W$ be linear map. Prove that $T^t$ is onto if and only if T is one-to-one.

Here is the theorem: Let V and W be finite-dimensional vector spaces over F with ordered bases $\beta$ and $\gamma$, respectively. For any linear map $T:V \rightarrow W$, the mapping $T^t:W^* \rightarrow V^*$ defined by $T^t(g)=gT$ for all $g \in W^*$ is a linear map with the property that $[T^t]^{{\beta}^*}_{{\gamma}^*}=([T]^\gamma_\beta)^t$. I want to prove the reverse direction.

Assume T is one-to-one, and we want to prove that $T^t$ is onto. How am I supposed to construct a linear functional such that T is one-to-one? Any hint is appreciated.

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spruce
spruce
  • 705
  • 3
  • 20

let $T:V \rightarrow W$ be linear map. Prove that $T^t$ is onto if and only if T is one-to-one.

Let V and W be nonzero vector spaces over the same field, and let $T:V \rightarrow W$ be linear map. Prove that $T^t$ is onto if and only if T is one-to-one.

I want to prove the reverse direction.

Assume T is one-to-one, and we want to prove that $T^t$ is onto. How am I supposed to construct a linear functional such that T is one-to-one? Any hint is appreciated.